SCHEME OF WORK
WEEK:
1. Electromagnetic waves. Type of radiation in electromagnetic in spectrum description and uses.
2. Gravitational Field, Law, Gravitational Potential, Escape Velocity, Potential Energy in Gravitational Field .
3. Electric Field, Coulombs law, electric field intensity, electric potential, Capacitor and Capacitance.
4. Electric Cells - Current Electricity – Primary and Secondary. Defects of Simple Cells. Cells in series and Parallel.
5. Electrolysis - Electrolytes, Electrodes, Ions, Faraday’s law, electro-chemical equivalent.
6. Electric Measurement: Resistivity, Conductivity, Conversion of galvanometer to ammeter and Voltmeter, Measuring resistance ammeter – voltmeter methods,
7. Magnetic field – Magnetic and magnetic materials, temporary and permanent magnets, magnetization, demagnetization, Theory of magnetization, magnetic flux, Earth magnetic field.
8. Magnetic field around current carrying conductor – Straight conductor, circular conductor, solenoid, applications- electromagnets, uses, electromagnet-electric bell, telephone earpiece.
9. Electromagnetic field – Flemings left hand rule. Application – DC. Motor, moving coil. Galvanometers.
10. Electromagnetic induction – induced current, laws of electromagnetic induction, induction coil, A/C and D.C. Generator transformer and power transmission.
11. Revision
1ST TERM
WEEK 1
ELECTROMAGNETIC WAVES
Contents;
Definition and Concept
Types of radiation
Detectors.
Uses.
1. Definition and Concept: Electromagnetic waves are produced by electromagnetic vibrations. Electromagnetic waves have electrical origin and the ability to travel in vacuum. So, electromagnetic waves are regarded as a combination of traveling electric and magnetic forces which vary in value and are directed at right angles to each other and to the direction of travel. In other words, they are transverse waves.
https://youtu.be/7v2gs8rdQzU
https://youtu.be/hk63uUhkZH4
Types of Radiation: The electromagnetic waves consist of the following:
1. Radio waves with wavelength 10-3m to 1000m.
2. Infra-red waves with average wavelengths of 10-6m.
3. Visible spectrum, known as light waves, with wavelengths of 7 x 10-7 m for red rays.
4. Ultraviolet rays with wavelength of 10-8m
5. X- rays with wavelength of 10-10 m.
6. Gamma –rays with wavelength of 10-11m.
https://youtu.be/5oUagoF_viQ
Radio waves: Radio waves have the longest wavelengths. Radio waves are emitted from transmitters and carry radio signals to radio sets. The shortest radio waves are called microwaves. Microwaves are used in radar and in heating hence they are used in cooking.
Infra-red waves Infra red waves are found just beyond the red end of the visible spectrum. They are present in the radiation from the sun or from the filament of an electric lamp. Many manufacturing industries used infra red lamps to dry paints on painted items. They are also used for the treatment of muscular problems.
https://youtu.be/Ldnh0XlMVc0
Visible Spectrum or Light Waves: The visible spectrum is made up of red, orange, yellow, green, blue, indigo and violet rays. These are all colours of the rainbow. When these rays combine, they form a white light. In the visible spectrum, red rays have the longest wavelengths while the violet rays have the shortest wavelengths. The main source of light is the sun.
https://youtu.be/DOnXkEDQ6e8
ULTRA VIOLET RAYS
Ultra violet rays are located just beyond the violet end of the visible spectrum. Ultraviolet rays can be produced by quarts, mercury filaments, or the sun. Ultraviolet rays can cause certain materials to fluoresce (i,e glow).
X. RAYS: X. rays are produced when fast moving electrons strike a metal target, which reduces their velocity.
X- rays are used in hospitals to destroy malignment growth in the body and to produce x – ray photographs which can locate broken bones, Much of x- ray in the body is harmful and can lead to sterility and adverse change in the blood. X- rays are used in industries to locate cracks in metal castings and flows in pipes.
X-rays ionize gases and have a penetrating effect such that they pass through substances opaque to white light are diffracted by crystals and unaffected by either electric or magnetic fields.
https://youtu.be/q_CxKQC-zpg
GAMMA – RAYS
Gamma – rays are emitted by radio active substances such as cobalt. 60, uranium and polonium. Like x- rays, gamma rays ionize gases and darken photographic plates. Because of their shorter wavelengths gamma rays have a greater penetrating power.
https://youtu.be/dBFGjdgbpno
DETECTORS: The detectors of the various radiations in the electromagnetic spectrum are
i. Gamma rays - Geiger-Muller tube
ii. X- rays - Photographic films
iii. Ultraviolet rays - Photographic films, fluorescent substances
iv. Visible rays - Eye, photographic film, photo electric cell
v. Infra red rays - Skin, thermometer, photo transistor, photographic film.
vi. Radio waves - Radio set, Television set, Aerials
Uses.
a. A knowledge of infra-red rays is used in developing infer red telescopes, infra-red signaling lamps which are useful to soldiers fighting in darkness.
b. With the aid of photographic film which are sensitive to infra red, it is possible to take clear photographs through mist and haze.
c. X- rays are useful in hospitals ( e.g to inspect broken bones), industry ( to inspect metal castings), and in science to study crystal structure of matters.
d. Gamma rays are used to kill cancer cells in patients body as well as bacteria in foods and hospitals equipment.
e. A knowledge of ultraviolet rays is used in developing ultraviolet lamps, the lamps are useful in conducting experiments on photo-electric effect.
f. Radio waves are very important for effective communication especially when radio set, television set, walkie-talkie are involved.
https://youtu.be/KYDil96NR5Q
EVALUATION:
1. Give three similarities of electromagnetic waves. Mention two distinguishing
properties of infrared and ultraviolet rays.
2. Mention and describe two important uses of x-rays.
3. How can you detect the following radiations
i. X- rays
ii. Visible rays
iii. Infra red rays.
4. Electromagnetic waves are regarded as transverse waves, why?
5. What is the relationship between the direction of a radio wave and the direction of its electric and magnetic fields.?
6. A radio station transmits at a frequency of 1200KHZ. What is the wavelength of the radio wave?(c = 3.0 x 108 ms-1).
READING ASSIGNMENT.
New School Physics for Senior Secondary School - M. W. ANYAKOHA pg 364- 369.
GENERAL EVALUATION:
1. When a mass is hung on a spring, the spring stretches 6cm. determine its period of vibration, if its pulled down and released [ Π = 3.142 , g = 10m/s2]
2. A simple pendulum has a period of 17.0s. When the length is shortened by 1.5cm its period is 8.5s. Calculate the original length.
ASSIGNMENT.
1 .Arrange the following electromagnetic waves in their increasing order of wavelength: Infra-red rays; visible rays, radio waves; and x-rays.
(a) Radio waves, infra- red ray, visible rays, x-rays (b) x-rays, visible rays, infra –red rays, radio waves
( c) Visible rays, x –rays, radio waves, infra-red rays. (d) Infra-red rays, Radio waves, X- rays, Visible rays.
2. Which of the following radiation is of nuclear origin?
(a) X- rays (b) Visible –rays (c ) Radio waves (d) Gamma rays.
3. Which of the following radiations is found useful by soldiers fighting in darkness.
(a) Gamma-rays (b) x- rays ( c) infra- red rays (d ) ultra violet rays.
4. The velocity of light in vacuum is
(a) 3. 0 x 106 m/s (b) 3.0 x 107 m/s (c) 3.0 x 108 m/s (d) 3.0 x 109 m/s
5. Which of the following has the shortest wavelength?
(a ) Radio waves (b) x –rays ( c) Infra –red rays (d) microwaves.
THEORY
1.(a) What rays make up the radio waves?
(b) Arrange these radiation in their increasing order of wavelength.
Ultraviolet rays Radio waves Visible spectrum Infra-red waves X- rays Gamma –rays
2.(a) What is a radar?
(b) What type of electromagnetic radiation does it use?
(c ) How does it function?
Contents;
Definition and Concept
Types of radiation
Detectors.
Uses.
1. Definition and Concept: Electromagnetic waves are produced by electromagnetic vibrations. Electromagnetic waves have electrical origin and the ability to travel in vacuum. So, electromagnetic waves are regarded as a combination of traveling electric and magnetic forces which vary in value and are directed at right angles to each other and to the direction of travel. In other words, they are transverse waves.
https://youtu.be/7v2gs8rdQzU
https://youtu.be/hk63uUhkZH4
Types of Radiation: The electromagnetic waves consist of the following:
1. Radio waves with wavelength 10-3m to 1000m.
2. Infra-red waves with average wavelengths of 10-6m.
3. Visible spectrum, known as light waves, with wavelengths of 7 x 10-7 m for red rays.
4. Ultraviolet rays with wavelength of 10-8m
5. X- rays with wavelength of 10-10 m.
6. Gamma –rays with wavelength of 10-11m.
https://youtu.be/5oUagoF_viQ
Radio waves: Radio waves have the longest wavelengths. Radio waves are emitted from transmitters and carry radio signals to radio sets. The shortest radio waves are called microwaves. Microwaves are used in radar and in heating hence they are used in cooking.
Infra-red waves Infra red waves are found just beyond the red end of the visible spectrum. They are present in the radiation from the sun or from the filament of an electric lamp. Many manufacturing industries used infra red lamps to dry paints on painted items. They are also used for the treatment of muscular problems.
https://youtu.be/Ldnh0XlMVc0
Visible Spectrum or Light Waves: The visible spectrum is made up of red, orange, yellow, green, blue, indigo and violet rays. These are all colours of the rainbow. When these rays combine, they form a white light. In the visible spectrum, red rays have the longest wavelengths while the violet rays have the shortest wavelengths. The main source of light is the sun.
https://youtu.be/DOnXkEDQ6e8
ULTRA VIOLET RAYS
Ultra violet rays are located just beyond the violet end of the visible spectrum. Ultraviolet rays can be produced by quarts, mercury filaments, or the sun. Ultraviolet rays can cause certain materials to fluoresce (i,e glow).
X. RAYS: X. rays are produced when fast moving electrons strike a metal target, which reduces their velocity.
X- rays are used in hospitals to destroy malignment growth in the body and to produce x – ray photographs which can locate broken bones, Much of x- ray in the body is harmful and can lead to sterility and adverse change in the blood. X- rays are used in industries to locate cracks in metal castings and flows in pipes.
X-rays ionize gases and have a penetrating effect such that they pass through substances opaque to white light are diffracted by crystals and unaffected by either electric or magnetic fields.
https://youtu.be/q_CxKQC-zpg
GAMMA – RAYS
Gamma – rays are emitted by radio active substances such as cobalt. 60, uranium and polonium. Like x- rays, gamma rays ionize gases and darken photographic plates. Because of their shorter wavelengths gamma rays have a greater penetrating power.
https://youtu.be/dBFGjdgbpno
DETECTORS: The detectors of the various radiations in the electromagnetic spectrum are
i. Gamma rays - Geiger-Muller tube
ii. X- rays - Photographic films
iii. Ultraviolet rays - Photographic films, fluorescent substances
iv. Visible rays - Eye, photographic film, photo electric cell
v. Infra red rays - Skin, thermometer, photo transistor, photographic film.
vi. Radio waves - Radio set, Television set, Aerials
Uses.
a. A knowledge of infra-red rays is used in developing infer red telescopes, infra-red signaling lamps which are useful to soldiers fighting in darkness.
b. With the aid of photographic film which are sensitive to infra red, it is possible to take clear photographs through mist and haze.
c. X- rays are useful in hospitals ( e.g to inspect broken bones), industry ( to inspect metal castings), and in science to study crystal structure of matters.
d. Gamma rays are used to kill cancer cells in patients body as well as bacteria in foods and hospitals equipment.
e. A knowledge of ultraviolet rays is used in developing ultraviolet lamps, the lamps are useful in conducting experiments on photo-electric effect.
f. Radio waves are very important for effective communication especially when radio set, television set, walkie-talkie are involved.
https://youtu.be/KYDil96NR5Q
EVALUATION:
1. Give three similarities of electromagnetic waves. Mention two distinguishing
properties of infrared and ultraviolet rays.
2. Mention and describe two important uses of x-rays.
3. How can you detect the following radiations
i. X- rays
ii. Visible rays
iii. Infra red rays.
4. Electromagnetic waves are regarded as transverse waves, why?
5. What is the relationship between the direction of a radio wave and the direction of its electric and magnetic fields.?
6. A radio station transmits at a frequency of 1200KHZ. What is the wavelength of the radio wave?(c = 3.0 x 108 ms-1).
READING ASSIGNMENT.
New School Physics for Senior Secondary School - M. W. ANYAKOHA pg 364- 369.
GENERAL EVALUATION:
1. When a mass is hung on a spring, the spring stretches 6cm. determine its period of vibration, if its pulled down and released [ Π = 3.142 , g = 10m/s2]
2. A simple pendulum has a period of 17.0s. When the length is shortened by 1.5cm its period is 8.5s. Calculate the original length.
ASSIGNMENT.
1 .Arrange the following electromagnetic waves in their increasing order of wavelength: Infra-red rays; visible rays, radio waves; and x-rays.
(a) Radio waves, infra- red ray, visible rays, x-rays (b) x-rays, visible rays, infra –red rays, radio waves
( c) Visible rays, x –rays, radio waves, infra-red rays. (d) Infra-red rays, Radio waves, X- rays, Visible rays.
2. Which of the following radiation is of nuclear origin?
(a) X- rays (b) Visible –rays (c ) Radio waves (d) Gamma rays.
3. Which of the following radiations is found useful by soldiers fighting in darkness.
(a) Gamma-rays (b) x- rays ( c) infra- red rays (d ) ultra violet rays.
4. The velocity of light in vacuum is
(a) 3. 0 x 106 m/s (b) 3.0 x 107 m/s (c) 3.0 x 108 m/s (d) 3.0 x 109 m/s
5. Which of the following has the shortest wavelength?
(a ) Radio waves (b) x –rays ( c) Infra –red rays (d) microwaves.
THEORY
1.(a) What rays make up the radio waves?
(b) Arrange these radiation in their increasing order of wavelength.
Ultraviolet rays Radio waves Visible spectrum Infra-red waves X- rays Gamma –rays
2.(a) What is a radar?
(b) What type of electromagnetic radiation does it use?
(c ) How does it function?
WEEK 2
GRAVITATIONAL FIELD
CONTENTS;
Introduction
Law of Universal Gravitation
Gravitational Potential
Escape Velocity
Potential energy in Gravitational Field.
Introduction.
Gravitational field is a region or space around a mass in which the gravitational force of the mass can be felt. Gravitation is the force of attraction exerted by a body on all other bodies in the universe . Gravitational force exists between a body and all other bodies around it. Gravitational force act between all masses and hold together planets, stars and galaxies. Each mass has a gravitational field around it.
CONCEPT OF GRAVITATIONAL FIELD
Is a region of space surrounding the earth where object of masses experience force of gravity.
F α m1m2 = F = Gm1m2
r2 r2
note that G = 6.67 x 10-11 Nm2/kg2
example: what is the magnitude of force attraction between two 5kg ball, 50mm apart?
Relationship between "G" and "g" with earth
Fg = mg; and F = Gmm
R2
Mg = GMm
R2
G = GM
R2
At an altitude i.e. g (acc. At a point) is = g 1-2h
R
Which implies that the acceleration due to gravity decreases with increased rise in altitude.
Weightlessness occurs when
h = R
2
Also when a plane descends, or a body in a lift with i.e. a = g
Where R is reaction
W - R = ma
R = W- ma = mg - ma
M (g-a)
When R = 0, mg - ma =0
Mg = ma
And a=g
https://youtu.be/ViY0-lOzbg0
Law of Universal Gravitation.
Newton’s law of universal of gravitation states that “every particle in the universe attracts every other particles with a force that is proportional to the products of their masses and inversely proportional to the square of the distance between them"
m[sub]1[/sub] and m[sub]2[/sub] are the masses of the two particles r is the distance between them and G is the universal gravitational constant.
The numerical value of G = 6.67 x 10[sup]-1[/sup]1 Nm[sup]2[/sup] kg[sup]-2[/sup].
https://youtu.be/c9shwPMpSq8
Evaluation.
1. State Newton’s law of universal gravitation.
2. Two small objects of masses 100Kg and 90Kg respectively are separated by a distance of 1.2m.
Determine the force of attraction between the two objects. (G = 6.67 x 10[sup]-1[/sup]1 Nm[sup]2[/sup] kg[sup]-2[/sup])
Gravitational Field Intensity
Gravitational field intensity at a point is the force per unit mass of an object placed at that point.
g = [sup]F[/sup]/[sub]m[/sub]
The unit is N/Kg . It is a vector quantity and it is regarded as acceleration due to gravity.
Relation Between g and G.
If the force of attraction (F) between two particles of matter separated by a distance r is given by:
This is the gravitational field intensity
https://youtu.be/_SgAv4cgAmM
Evaluation:
(1) If the mass of the earth is 5.78 x 10 24Kg and gravitational constant is 6.67 x 10-11 Nm2kg2. Calculate the gravitational field intensity due to earth. Radius of earth is 6400km
(2) The gravitational field intensity at a location X in space is two-fifth of its value on the earth surface. If the weight of an object at x is 4.8N, what is the weight on earth.
GRAVITATIONAL POTENTIAL
The gravitational potential at a point is the work done in taking a unit mass from infinity to that point. The unit is Jkg[sup]-1[/sup].
The gravitational potential, V, is given by V = [sup]- Gm[/sup]/[sub]r[/sub]
m is the mass producing the gravitational field and r is the distance of the point to the mass. The gravitational potential decreases as r increases and becomes zero when r is infinitely large. The negative sign indicates that the potential at infinity (zero) is higher than the potential close to the mass.
https://youtu.be/demkfiHM8dk
EVALUATION.
1. Calculate the gravitational potential at a point on the Earth surface.
Mass of earth is 6.0x10[sup]24[/sup]kg, radius of earth = 6400km and G = 6.67 x 10 [sup]-11[/sup]Nm[sup]2[/sup]kg[sup]-2[/sup]
2. calculate the gravitational potential at a point on the earth surface [ radius of the earth = 6.4X10[sup]4[/sup]m mass of the earth = 6.0 X 10 24Kg,
G= 6.67X 10[sup]-11[/sup] Nm[sup]2[/sup]Kg[sup]-2[/sup]]
Escape Velocity
This defined as the minimum velocity required for an astronomical body (e.g. satellite/rocket) to just escape/leave the gravitational influence or filed permanently.
This is the minimum velocity required for an object ( e.g. satellite, rocket) to just escape or leave the gravitational influence or filed of an astronomical body (e.g the earth) permanently.
This work must equal the Kinetic energy of the body of mass m at this point, having a velocity, Ve.
Let m be mass of astronomical body and me be mass of earth, of distance r
W D m m = w = f x r
F = [sup]Gmme[/sup]/[sub]R[/sub]
W = [sup]Gmme[/sup]/[sub]R[/sub]-------- (1)
Thus W.D must be equal to the K.E
K.E = ½ mr2 = w
½ mr2 = [sup]gmme[/sup]/[sub]R[/sub]
Make V subject
Ve = √[sup]2Gme[/sup]/[sub]r[/sub]
Thus r is the Radius of the earth (R)
but we know that
ge = [sup]Gme[/sup]/[sub]R2[/sub] making G subject
G = [sup]Rg[/sup]/[sub]Me[/sub]
Ve = √ [sup]R2g me x 2 ÷ R[/sup]/[sub]me[/sub]
Ve = √ [sup]2g R2 [/sup]/[sub]R[/sub]
Ve = √ 2gR
Q.E.D
https://youtu.be/6D-9i272WYI
ORBITAL VELOCITY
A satellite orbits a planet such as the earth with a velocity. It orbits a planet such as the earth with a velocity. It orbits just because the centripetal force on it is equal to the gravitational force on it.
[sup]Mv2[/sup] /[sub] r[/sub] = [sup]GMm[/sup] / [sub]r2[/sub]
[sup]V2[/sup]/[sub]r [/sub] = [sup]Gm[/sup]/[sub]r2[/sub]
V2 = [sup]Gmr[/sup]/[sub]R[/sub]
Hence V = √ [sup]Gm[/sup]/[sub]r[/sub]
where V is the orbital velocity.
https://youtu.be/hUJb35RGpY8
PARKING ORBIT
This is the orbit of a satellite having the same period as that of the earth
T = [sup]2ￗr[/sup]/[sub]V[/sub]
https://youtu.be/O3xhBjx_ikE
EVALUATION.
The following conclusions can be drawn from the equation.
i. The magnitude of the total energy is equal to that of the k.e of the satellite.
ii. The kinetic energy of a satellite in an orbit increases as the radius o the orbit decreases.
iii. The kinetic energy of a satellite in an orbit increases as the speed of the satellite increases.
iv. The potential energy of the satellite in orbit is twice its kinetic energy and of opposite sign.
EVALUATION.
Distinguish between orbital velocity and parking orbit of an astronomical body.
Derive an expression for the total energy in a gravitational field. What conclusions can you draw from the equation.
READING ASSIGNMENT.
New School Physics for Senior Secondary Schools (M.W Anyakoha) pgs 370-377.
GENERAL EVALUATION:
1. An object is projected with a velocity of 50m/s from the ground level at an angle of φ to the vertical. If the total time of flight is 5s, calculate the value of φ.
2. A ball is projected horizontally from a height of 30m above the ground with a speed of 20m/s. calculate the horizontal distance travelled by the ball when it hits the ground g = 10m/s2]
ASSIGNMENT.
I. The negative sign indicates that work done is against the gravitational field.
II. The potential at infinity is zero
III. The potential at infinity is less than that at the surface of the earth.
IV. Gravitational potential is a vector quantity
(a)I and II only (b) III and IV Only (c) III only (d) I,II and III only (e) II and IV only.
THEORY.
1. The average radius of Jupiter’s orbit round the sun of mass 2 x 10[sup]30[/sup]Kg is 7.8 x 10[sup]11[/sup]m.
If the mass of Jupiter is 1.9 x 1027, find the gravitational force the sun exerts on Jupiter , G = 6.67 x 10[sup]-11[/sup]Nm[sup]2[/sup]kg[sup]-2[/sup]
2. If the mass of a portion is 1.67 x 10[sup]-27[/sup]kg and the mass of an electron is 9.11 x 10[sup]-31[/sup]kg, calculate the force of gravitation between:
i. a proton and an electron ii. two electrons iii. two protons.
Take G= 6.67 x 10[sup]-11[/sup] Nm[sup]2[/sup]kg[sup]-2[/sup], distance between the protons = 4.0m.,
distance between the electrons = 2 x 10[sup]-2[/sup]m, distance between the proton and the electron = 5.4 x 10[sup]-11[/sup]m
3. The mass of the moon is 1/80M and its 1/5R; where M and R are mass and radius of the earth respectively.
Calculate the acceleration due to gravity on the surface of the moon. [g= 9.8m/s2]
CONTENTS;
Introduction
Law of Universal Gravitation
Gravitational Potential
Escape Velocity
Potential energy in Gravitational Field.
Introduction.
Gravitational field is a region or space around a mass in which the gravitational force of the mass can be felt. Gravitation is the force of attraction exerted by a body on all other bodies in the universe . Gravitational force exists between a body and all other bodies around it. Gravitational force act between all masses and hold together planets, stars and galaxies. Each mass has a gravitational field around it.
CONCEPT OF GRAVITATIONAL FIELD
Is a region of space surrounding the earth where object of masses experience force of gravity.
F α m1m2 = F = Gm1m2
r2 r2
note that G = 6.67 x 10-11 Nm2/kg2
example: what is the magnitude of force attraction between two 5kg ball, 50mm apart?
Relationship between "G" and "g" with earth
Fg = mg; and F = Gmm
R2
Mg = GMm
R2
G = GM
R2
At an altitude i.e. g (acc. At a point) is = g 1-2h
R
Which implies that the acceleration due to gravity decreases with increased rise in altitude.
Weightlessness occurs when
h = R
2
Also when a plane descends, or a body in a lift with i.e. a = g
Where R is reaction
W - R = ma
R = W- ma = mg - ma
M (g-a)
When R = 0, mg - ma =0
Mg = ma
And a=g
https://youtu.be/ViY0-lOzbg0
Law of Universal Gravitation.
Newton’s law of universal of gravitation states that “every particle in the universe attracts every other particles with a force that is proportional to the products of their masses and inversely proportional to the square of the distance between them"
m[sub]1[/sub] and m[sub]2[/sub] are the masses of the two particles r is the distance between them and G is the universal gravitational constant.
The numerical value of G = 6.67 x 10[sup]-1[/sup]1 Nm[sup]2[/sup] kg[sup]-2[/sup].
https://youtu.be/c9shwPMpSq8
Evaluation.
1. State Newton’s law of universal gravitation.
2. Two small objects of masses 100Kg and 90Kg respectively are separated by a distance of 1.2m.
Determine the force of attraction between the two objects. (G = 6.67 x 10[sup]-1[/sup]1 Nm[sup]2[/sup] kg[sup]-2[/sup])
Gravitational Field Intensity
Gravitational field intensity at a point is the force per unit mass of an object placed at that point.
g = [sup]F[/sup]/[sub]m[/sub]
The unit is N/Kg . It is a vector quantity and it is regarded as acceleration due to gravity.
Relation Between g and G.
If the force of attraction (F) between two particles of matter separated by a distance r is given by:
This is the gravitational field intensity
https://youtu.be/_SgAv4cgAmM
Evaluation:
(1) If the mass of the earth is 5.78 x 10 24Kg and gravitational constant is 6.67 x 10-11 Nm2kg2. Calculate the gravitational field intensity due to earth. Radius of earth is 6400km
(2) The gravitational field intensity at a location X in space is two-fifth of its value on the earth surface. If the weight of an object at x is 4.8N, what is the weight on earth.
GRAVITATIONAL POTENTIAL
The gravitational potential at a point is the work done in taking a unit mass from infinity to that point. The unit is Jkg[sup]-1[/sup].
The gravitational potential, V, is given by V = [sup]- Gm[/sup]/[sub]r[/sub]
m is the mass producing the gravitational field and r is the distance of the point to the mass. The gravitational potential decreases as r increases and becomes zero when r is infinitely large. The negative sign indicates that the potential at infinity (zero) is higher than the potential close to the mass.
https://youtu.be/demkfiHM8dk
EVALUATION.
1. Calculate the gravitational potential at a point on the Earth surface.
Mass of earth is 6.0x10[sup]24[/sup]kg, radius of earth = 6400km and G = 6.67 x 10 [sup]-11[/sup]Nm[sup]2[/sup]kg[sup]-2[/sup]
2. calculate the gravitational potential at a point on the earth surface [ radius of the earth = 6.4X10[sup]4[/sup]m mass of the earth = 6.0 X 10 24Kg,
G= 6.67X 10[sup]-11[/sup] Nm[sup]2[/sup]Kg[sup]-2[/sup]]
Escape Velocity
This defined as the minimum velocity required for an astronomical body (e.g. satellite/rocket) to just escape/leave the gravitational influence or filed permanently.
This is the minimum velocity required for an object ( e.g. satellite, rocket) to just escape or leave the gravitational influence or filed of an astronomical body (e.g the earth) permanently.
This work must equal the Kinetic energy of the body of mass m at this point, having a velocity, Ve.
Let m be mass of astronomical body and me be mass of earth, of distance r
W D m m = w = f x r
F = [sup]Gmme[/sup]/[sub]R[/sub]
W = [sup]Gmme[/sup]/[sub]R[/sub]-------- (1)
Thus W.D must be equal to the K.E
K.E = ½ mr2 = w
½ mr2 = [sup]gmme[/sup]/[sub]R[/sub]
Make V subject
Ve = √[sup]2Gme[/sup]/[sub]r[/sub]
Thus r is the Radius of the earth (R)
but we know that
ge = [sup]Gme[/sup]/[sub]R2[/sub] making G subject
G = [sup]Rg[/sup]/[sub]Me[/sub]
Ve = √ [sup]R2g me x 2 ÷ R[/sup]/[sub]me[/sub]
Ve = √ [sup]2g R2 [/sup]/[sub]R[/sub]
Ve = √ 2gR
Q.E.D
https://youtu.be/6D-9i272WYI
ORBITAL VELOCITY
A satellite orbits a planet such as the earth with a velocity. It orbits a planet such as the earth with a velocity. It orbits just because the centripetal force on it is equal to the gravitational force on it.
[sup]Mv2[/sup] /[sub] r[/sub] = [sup]GMm[/sup] / [sub]r2[/sub]
[sup]V2[/sup]/[sub]r [/sub] = [sup]Gm[/sup]/[sub]r2[/sub]
V2 = [sup]Gmr[/sup]/[sub]R[/sub]
Hence V = √ [sup]Gm[/sup]/[sub]r[/sub]
where V is the orbital velocity.
https://youtu.be/hUJb35RGpY8
PARKING ORBIT
This is the orbit of a satellite having the same period as that of the earth
T = [sup]2ￗr[/sup]/[sub]V[/sub]
https://youtu.be/O3xhBjx_ikE
EVALUATION.
The following conclusions can be drawn from the equation.
i. The magnitude of the total energy is equal to that of the k.e of the satellite.
ii. The kinetic energy of a satellite in an orbit increases as the radius o the orbit decreases.
iii. The kinetic energy of a satellite in an orbit increases as the speed of the satellite increases.
iv. The potential energy of the satellite in orbit is twice its kinetic energy and of opposite sign.
EVALUATION.
Distinguish between orbital velocity and parking orbit of an astronomical body.
Derive an expression for the total energy in a gravitational field. What conclusions can you draw from the equation.
READING ASSIGNMENT.
New School Physics for Senior Secondary Schools (M.W Anyakoha) pgs 370-377.
GENERAL EVALUATION:
1. An object is projected with a velocity of 50m/s from the ground level at an angle of φ to the vertical. If the total time of flight is 5s, calculate the value of φ.
2. A ball is projected horizontally from a height of 30m above the ground with a speed of 20m/s. calculate the horizontal distance travelled by the ball when it hits the ground g = 10m/s2]
ASSIGNMENT.
I. The negative sign indicates that work done is against the gravitational field.
II. The potential at infinity is zero
III. The potential at infinity is less than that at the surface of the earth.
IV. Gravitational potential is a vector quantity
(a)I and II only (b) III and IV Only (c) III only (d) I,II and III only (e) II and IV only.
THEORY.
1. The average radius of Jupiter’s orbit round the sun of mass 2 x 10[sup]30[/sup]Kg is 7.8 x 10[sup]11[/sup]m.
If the mass of Jupiter is 1.9 x 1027, find the gravitational force the sun exerts on Jupiter , G = 6.67 x 10[sup]-11[/sup]Nm[sup]2[/sup]kg[sup]-2[/sup]
2. If the mass of a portion is 1.67 x 10[sup]-27[/sup]kg and the mass of an electron is 9.11 x 10[sup]-31[/sup]kg, calculate the force of gravitation between:
i. a proton and an electron ii. two electrons iii. two protons.
Take G= 6.67 x 10[sup]-11[/sup] Nm[sup]2[/sup]kg[sup]-2[/sup], distance between the protons = 4.0m.,
distance between the electrons = 2 x 10[sup]-2[/sup]m, distance between the proton and the electron = 5.4 x 10[sup]-11[/sup]m
3. The mass of the moon is 1/80M and its 1/5R; where M and R are mass and radius of the earth respectively.
Calculate the acceleration due to gravity on the surface of the moon. [g= 9.8m/s2]
WEEK 3
ELECTRIC FIELD
Contents
Electric Field
Coulombus Law
Electric Field Intensity
Electric Potential
Capacitor and Capacitance.
Electric Field: An electric field is a region of space which surrounds a system of electric charges. Electrical forces will act on any electric charge which is placed within the region. Electric field is a vector quantity. The direction of the filed can be determined using a test charge (a small positive charge).
https://youtu.be/VFbyDCG_j18
Fundamental Law of Electrostatics
The fundamental law of electrostatic states that: “Like charge repels, unlike charges attract.
This law states that the electric force between 2 point charges q1 and q2 separated by a distance r, is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
https://youtu.be/a4fd2HhD7h4
https://youtu.be/4ubqby1Id4g
EVALUATION.
With the aid of a sketch diagram explain the following :
(i) Like charges repel
(ii) Unlike charges attract.
Electric Force between Two Charges: Coulomb’s Law
It has been pointed out that like charges repel each other while unlike charges attract each other. Charles Coulomb formulates the law that governs electrostatics forces between electric charges. This law is known as Coulomb’s law .
Coulomb’s law states that in a given medium, the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the two charges.
For charges of magnitude q1, q2, separated by distance r as shown above, Coulomb’s law can be stated mathematically as :
RELATIVE PERMITIVITY (εr)
The relative permittivity of a medium is the ratio of the permittivity of a medium to that of air.
Εr = [sup]εm[/sup]/[sub]εo[/sub]
εm = permittivity of medium εo = permittivity of air/ vaccum
https://youtu.be/q7Js6LLIxLo
EVALUATION
1. State Coulomb’s law
2. Find the force of attraction between two charges of magnitude 6UC and 20UC respectively.
If the distance between them is 0.5m (take ¼ πEo = 9.0 x 109 Bm2C[sup]-2[/sup] ).
3. Three charges +15C – 25C and -20C are distributed as shown in the diagram below. Find the resultant force acting on 15C charge.
Electric Field Intensity or Strength (E)
The elelectri field intensity, E, at any point in an electric field is the force experienced by a unit positive test charge at that point . It is a vector quantity whose S. I unit is (N/C), mathematically.
E = [sup]F[/sup]/[sub]q[/sub]
E= Electric field intensity (NC-1), F = Force, q = charge.
https://youtu.be/LWwSIebbs9I
EVALUATION.
1.calculate the electric field intensity in vacuum at a distance of 5cm from a charge of 5,0x 10[sup]4[/sup]c ,(¼πEo = 9.0 x 109 NM2C[sup]-2[/sup]).
2. Two similar but opposite point charges –q and +q each of magnitude 6UC are separated by a distance of 12cm in vacuum as shown below:
Calculate the magnitude and direction of the resultant electric field intensity at p.
Electric Potential
The electric potential (V) at a point is the work done in bring a unit positive charge from infinity tot hat point against the electrical forces of the field. It is measured in volts. It is scalar quantity.
Mathematically, V = [sup]w[/sup]/[sub]q[/sub]
V= electric potential (volts)
W= work done in joules, q = charge in coulombs
The electric potential at a point due to a charge Q at a distance r from the charge Q at a distance r from the charge is given as:
V = [sup]Q[/sup]/[sub]4πEor[/sub]
If the work done is against the field, the potential is positive. If the work done is by the field, the potential is negative. The potential an infinity is zero. Also the potential of the earth is zero. The earth is used to test the potential of the body. This is done by connecting a wire form the body to the earth (the body is said to be earthed). If electron flow from the body to the earth, the body is at a negative potential. If electron flows from the earth to the body, the body is at positive potential. Positive points are at higher potential while negative points are at lower potential.
https://youtu.be/IvAnzZgwcFQ
Potential difference:
The potential difference between any two points in an electric field is the work done is taking a unit positive charge from one point to another in the field.
If a charge Q is moved from a point at a potential V1 to another at a potential V2, the potential difference (V1 – V2) is the work done by the field.
Work done on the charge, W = Q (V1 – V2)
If Q moves from A to B, then the work done,
W = Force x distance
https://youtu.be/SNlOPxZ-Ev4
ELECTRON VOLT (eV)
The electron volt is the quantity of energy gained by an electron in accelerating through a potential difference of one volt.
Electronic charge = 1.6 x 10[sup]-19[/sup]C
i e V = 1.6 x 10[sup]-19[/sup] x 1 = 1.6 x 10[sup]-19[/sup]J
The energy acquired by a charged particle accelerated by an electronic field in a vacuum depends only on its charge and the p.d. through which it falls.
When the electron is in motion, its kinetic energy will be ½ mv[sup]2[/sup].
If the electron moves in a circle of radius r, the force towards the centre in mv[sup]2[/sup]r (centripetal force), and it is provided by the electrical force of attraction
https://youtu.be/uqAVjdg0LQA
EVALUATION:
1. Calculate the electric potential due to a positive charge of -12C at a point distance 10cm away ( [sup]1[/sup]/[sub]4π εo[/sub] = 9.0 x 10[sup]9[/sup]m/
2. A point, A, is at a potential of 120v. Determine the work done in moving an electric charge 25C from A to B.
3. Calculate the velocity of an electron as it strikes the anode of a thermionic tube if the p.d. between anode and cathode is 150v.
Mass of electron is 9.1 x 10[sup]-31[/sup]kg while its charge is -1.6 x 10[sup]-19[/sup]C.
WORKED EXAMPLE
1. Calculate the energy in eV and in Joule of an α particle (helium nucleus) accelerated through a p.d. of 4 x 10[sup]6[/sup]V.
SOLUTION
The charge on an α particle is 2e.
Ke = work done
= charge x p.d.
= 2 x 4 x10[sup]6[/sup]
= 8 x 10[sup]6[/sup] eV = 8 MeV
IeV = 1.6 x 10[sup]-19[/sup]J.
K.e. gained = 8 x 10[sup]6[/sup] x 1.6 x 10[sup]-19[/sup]
= 1.48 x 10[sup]-12[/sup]J
2. An electron gun releases an electron. The p.d. between the gun and the collector plate is 100V.
What is the velocity of the electron just before it touches the collector plate? (e = -1.6 x 10[sup]-19[/sup]C, Me = 9.1 x 10[sup]-31[/sup] kg).
SOLUTION:
Electrical energy = QV
= 100 x 1.6 x 10[sup]-19[/sup]
= 1.6 x 10[sup]-19[/sup]J.
½ (9.1 x 10[sup]-31[/sup]) v[sup]2[/sup] = 1.6 x 10[sup]-19[/sup]
V2 = 3.2 x 10[sup]-16[/sup]
9.1 x 10[sup]-31[/sup]
V2 = 0.35 x 10[sup]14[/sup]
: . V = 6 x 10[sup]6[/sup] ms[sup]-1[/sup]
CAPACITORS AND CAPACITANCES
A capacitor is a device for storing electricity (electric charges). It consists essentially of two conductors metal plates carrying opposite charges. The metal plates are separated by a small distance ‘d’ by an insulator.
Capacitance is defined as the ratio of the charge Q on either plates to the potential difference V between them.
C = Q
V
FACTORS
i. Size
ii. Shape
iii. Common area of plate
iv. Distance separating the two plates
v. Nature of material that separates them.
ENERGY STORED IN CAPACITOR
W.D in charging capacitor through p.d of 1/2V, is
w = ½ Vq = ½ qV
but since
C = [sup]q[/sup]/[sub]V[/sub] OR V= [sup]q[/sup]/[sub]c[/sub]
Therefore:
W = ½ q2 C OR ½ CV2
APPLICATIONS OF CAPACITORS
i. Radio for tuning
ii. Ignition systems of motor vehicles
iii. The elimination of sparks when a containing inductance is suddenly operated.
EVALUATION:
Write out the expression for total energy of two capacitors identical in parallel.
ASSIGNMENT: Calculate
a. The value of capacitance in the diagram shown below
b. The p.d across each capacitor
c. The charge on each capacitor
d. Energy stored in total capacitance in the diagram below.
READING ASSIGNMENT
New School Physics for Senior Secondary Schools (M.W. Anyakoha) pages 377 – 379.
GENERAL EVALUATION:
1. Three identical cells each of emf 1.5V and internal resistance 1.0Ω are connected in parallel across an external load of resistance 2.67Ω. Calculate the current in the load.
2. A radio is operated by eight cells each of emf 2.0V connected in series. If two of the cells are wrongly connected, the net emf of the radio is ?
ASSIGNMENT
1. The electric force between two point charges each of magnitude q at a distance r apart in air of permittivity εo is
3. Calculate the force acting on an electron carrying a charge of 1.6 x 10[sup]-19[/sup]C in an electric field of intensity 5.0 x 10[sup]8[/sup] N/C is
(A). 3.2 x 10[sup]-29[/sup]N (B). 8.0 x 10[sup]-11[/sup] (C). 3.1 x 10[sup]27[/sup]N D. 4.6 x 10[sup]-6[/sup]N
4. Find the electric field intensity in a vacuum at a distance of 10cm from a point charge of 15uc if ¼πε0 = 9.0 x 10[sup]9[/sup]
(A). 1.35 x 10[sup]7[/sup]NC[sup]-1[/sup] (B). 1.4 x 10[sup]10[/sup]NC[sup]-1[/sup] (C). 1.3 x 10[sup]11[/sup]NC[sup]-1[/sup] (D).1.5x10[sup]10[/sup]NC[sup]-1[/sup]
5.Which of the following statements is I are true about an isolated positively charged sphere?
I. It contains excess positive charges. II. It has an electric field associated with it.
III. It carries electric current. Iv It has excess negative charges.
A. I and II only B. I, II and III only C. II and IV only D. III and IV E. I and III only
THEORY
1.If three charges are distributed as shown in the diagram below.
Find the resultant force on the +10C charge (take [sup]1[/sup]/[sub]4πEo[/sub] = 9.0 x 109 s.I. units)
2.Two charges of +5uc and -10NC are separated by a distance of 8cm in a vacuum as shown below.
Calculate the magnitude and direction of the resultant electric field intensity due to point B.
¼πEo = 9.0 x 10[sup]9[/sup] s.I unit.
Contents
Electric Field
Coulombus Law
Electric Field Intensity
Electric Potential
Capacitor and Capacitance.
Electric Field: An electric field is a region of space which surrounds a system of electric charges. Electrical forces will act on any electric charge which is placed within the region. Electric field is a vector quantity. The direction of the filed can be determined using a test charge (a small positive charge).
https://youtu.be/VFbyDCG_j18
Fundamental Law of Electrostatics
The fundamental law of electrostatic states that: “Like charge repels, unlike charges attract.
This law states that the electric force between 2 point charges q1 and q2 separated by a distance r, is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
https://youtu.be/a4fd2HhD7h4
https://youtu.be/4ubqby1Id4g
EVALUATION.
With the aid of a sketch diagram explain the following :
(i) Like charges repel
(ii) Unlike charges attract.
Electric Force between Two Charges: Coulomb’s Law
It has been pointed out that like charges repel each other while unlike charges attract each other. Charles Coulomb formulates the law that governs electrostatics forces between electric charges. This law is known as Coulomb’s law .
Coulomb’s law states that in a given medium, the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the two charges.
For charges of magnitude q1, q2, separated by distance r as shown above, Coulomb’s law can be stated mathematically as :
RELATIVE PERMITIVITY (εr)
The relative permittivity of a medium is the ratio of the permittivity of a medium to that of air.
Εr = [sup]εm[/sup]/[sub]εo[/sub]
εm = permittivity of medium εo = permittivity of air/ vaccum
https://youtu.be/q7Js6LLIxLo
EVALUATION
1. State Coulomb’s law
2. Find the force of attraction between two charges of magnitude 6UC and 20UC respectively.
If the distance between them is 0.5m (take ¼ πEo = 9.0 x 109 Bm2C[sup]-2[/sup] ).
3. Three charges +15C – 25C and -20C are distributed as shown in the diagram below. Find the resultant force acting on 15C charge.
Electric Field Intensity or Strength (E)
The elelectri field intensity, E, at any point in an electric field is the force experienced by a unit positive test charge at that point . It is a vector quantity whose S. I unit is (N/C), mathematically.
E = [sup]F[/sup]/[sub]q[/sub]
E= Electric field intensity (NC-1), F = Force, q = charge.
https://youtu.be/LWwSIebbs9I
EVALUATION.
1.calculate the electric field intensity in vacuum at a distance of 5cm from a charge of 5,0x 10[sup]4[/sup]c ,(¼πEo = 9.0 x 109 NM2C[sup]-2[/sup]).
2. Two similar but opposite point charges –q and +q each of magnitude 6UC are separated by a distance of 12cm in vacuum as shown below:
Calculate the magnitude and direction of the resultant electric field intensity at p.
Electric Potential
The electric potential (V) at a point is the work done in bring a unit positive charge from infinity tot hat point against the electrical forces of the field. It is measured in volts. It is scalar quantity.
Mathematically, V = [sup]w[/sup]/[sub]q[/sub]
V= electric potential (volts)
W= work done in joules, q = charge in coulombs
The electric potential at a point due to a charge Q at a distance r from the charge Q at a distance r from the charge is given as:
V = [sup]Q[/sup]/[sub]4πEor[/sub]
If the work done is against the field, the potential is positive. If the work done is by the field, the potential is negative. The potential an infinity is zero. Also the potential of the earth is zero. The earth is used to test the potential of the body. This is done by connecting a wire form the body to the earth (the body is said to be earthed). If electron flow from the body to the earth, the body is at a negative potential. If electron flows from the earth to the body, the body is at positive potential. Positive points are at higher potential while negative points are at lower potential.
https://youtu.be/IvAnzZgwcFQ
Potential difference:
The potential difference between any two points in an electric field is the work done is taking a unit positive charge from one point to another in the field.
If a charge Q is moved from a point at a potential V1 to another at a potential V2, the potential difference (V1 – V2) is the work done by the field.
Work done on the charge, W = Q (V1 – V2)
If Q moves from A to B, then the work done,
W = Force x distance
https://youtu.be/SNlOPxZ-Ev4
ELECTRON VOLT (eV)
The electron volt is the quantity of energy gained by an electron in accelerating through a potential difference of one volt.
Electronic charge = 1.6 x 10[sup]-19[/sup]C
i e V = 1.6 x 10[sup]-19[/sup] x 1 = 1.6 x 10[sup]-19[/sup]J
The energy acquired by a charged particle accelerated by an electronic field in a vacuum depends only on its charge and the p.d. through which it falls.
When the electron is in motion, its kinetic energy will be ½ mv[sup]2[/sup].
If the electron moves in a circle of radius r, the force towards the centre in mv[sup]2[/sup]r (centripetal force), and it is provided by the electrical force of attraction
https://youtu.be/uqAVjdg0LQA
EVALUATION:
1. Calculate the electric potential due to a positive charge of -12C at a point distance 10cm away ( [sup]1[/sup]/[sub]4π εo[/sub] = 9.0 x 10[sup]9[/sup]m/
2. A point, A, is at a potential of 120v. Determine the work done in moving an electric charge 25C from A to B.
3. Calculate the velocity of an electron as it strikes the anode of a thermionic tube if the p.d. between anode and cathode is 150v.
Mass of electron is 9.1 x 10[sup]-31[/sup]kg while its charge is -1.6 x 10[sup]-19[/sup]C.
WORKED EXAMPLE
1. Calculate the energy in eV and in Joule of an α particle (helium nucleus) accelerated through a p.d. of 4 x 10[sup]6[/sup]V.
SOLUTION
The charge on an α particle is 2e.
Ke = work done
= charge x p.d.
= 2 x 4 x10[sup]6[/sup]
= 8 x 10[sup]6[/sup] eV = 8 MeV
IeV = 1.6 x 10[sup]-19[/sup]J.
K.e. gained = 8 x 10[sup]6[/sup] x 1.6 x 10[sup]-19[/sup]
= 1.48 x 10[sup]-12[/sup]J
2. An electron gun releases an electron. The p.d. between the gun and the collector plate is 100V.
What is the velocity of the electron just before it touches the collector plate? (e = -1.6 x 10[sup]-19[/sup]C, Me = 9.1 x 10[sup]-31[/sup] kg).
SOLUTION:
Electrical energy = QV
= 100 x 1.6 x 10[sup]-19[/sup]
= 1.6 x 10[sup]-19[/sup]J.
½ (9.1 x 10[sup]-31[/sup]) v[sup]2[/sup] = 1.6 x 10[sup]-19[/sup]
V2 = 3.2 x 10[sup]-16[/sup]
9.1 x 10[sup]-31[/sup]
V2 = 0.35 x 10[sup]14[/sup]
: . V = 6 x 10[sup]6[/sup] ms[sup]-1[/sup]
CAPACITORS AND CAPACITANCES
A capacitor is a device for storing electricity (electric charges). It consists essentially of two conductors metal plates carrying opposite charges. The metal plates are separated by a small distance ‘d’ by an insulator.
Capacitance is defined as the ratio of the charge Q on either plates to the potential difference V between them.
C = Q
V
FACTORS
i. Size
ii. Shape
iii. Common area of plate
iv. Distance separating the two plates
v. Nature of material that separates them.
ENERGY STORED IN CAPACITOR
W.D in charging capacitor through p.d of 1/2V, is
w = ½ Vq = ½ qV
but since
C = [sup]q[/sup]/[sub]V[/sub] OR V= [sup]q[/sup]/[sub]c[/sub]
Therefore:
W = ½ q2 C OR ½ CV2
APPLICATIONS OF CAPACITORS
i. Radio for tuning
ii. Ignition systems of motor vehicles
iii. The elimination of sparks when a containing inductance is suddenly operated.
EVALUATION:
Write out the expression for total energy of two capacitors identical in parallel.
ASSIGNMENT: Calculate
a. The value of capacitance in the diagram shown below
b. The p.d across each capacitor
c. The charge on each capacitor
d. Energy stored in total capacitance in the diagram below.
READING ASSIGNMENT
New School Physics for Senior Secondary Schools (M.W. Anyakoha) pages 377 – 379.
GENERAL EVALUATION:
1. Three identical cells each of emf 1.5V and internal resistance 1.0Ω are connected in parallel across an external load of resistance 2.67Ω. Calculate the current in the load.
2. A radio is operated by eight cells each of emf 2.0V connected in series. If two of the cells are wrongly connected, the net emf of the radio is ?
ASSIGNMENT
1. The electric force between two point charges each of magnitude q at a distance r apart in air of permittivity εo is
3. Calculate the force acting on an electron carrying a charge of 1.6 x 10[sup]-19[/sup]C in an electric field of intensity 5.0 x 10[sup]8[/sup] N/C is
(A). 3.2 x 10[sup]-29[/sup]N (B). 8.0 x 10[sup]-11[/sup] (C). 3.1 x 10[sup]27[/sup]N D. 4.6 x 10[sup]-6[/sup]N
4. Find the electric field intensity in a vacuum at a distance of 10cm from a point charge of 15uc if ¼πε0 = 9.0 x 10[sup]9[/sup]
(A). 1.35 x 10[sup]7[/sup]NC[sup]-1[/sup] (B). 1.4 x 10[sup]10[/sup]NC[sup]-1[/sup] (C). 1.3 x 10[sup]11[/sup]NC[sup]-1[/sup] (D).1.5x10[sup]10[/sup]NC[sup]-1[/sup]
5.Which of the following statements is I are true about an isolated positively charged sphere?
I. It contains excess positive charges. II. It has an electric field associated with it.
III. It carries electric current. Iv It has excess negative charges.
A. I and II only B. I, II and III only C. II and IV only D. III and IV E. I and III only
THEORY
1.If three charges are distributed as shown in the diagram below.
Find the resultant force on the +10C charge (take [sup]1[/sup]/[sub]4πEo[/sub] = 9.0 x 109 s.I. units)
2.Two charges of +5uc and -10NC are separated by a distance of 8cm in a vacuum as shown below.
Calculate the magnitude and direction of the resultant electric field intensity due to point B.
¼πEo = 9.0 x 10[sup]9[/sup] s.I unit.
WEEK 4
FIELD POTENTIAL, CAPACITORS AND CAPACITANCE
Arrangement of capacitors
Types of capacitors
Energy stored in capacitors
Every conductor may possess one or more of the following properties:
1. It could be mainly a resistor, which means that if a current is passed through it, heat energy is mainly produced
2. It could be a capacitor which means that when a current passes through it electrical energy or charges are stored.
3. Finally, it could be an inductor which stores mainly magnetic energy when a current is passed through it.
A Capacitor is essentially a device for storing electrical energy or charges. In general, capacitors can be in the form of two conductors which are insulated electrically from the surroundings. However, most common types of capacitors are in the form of two parallel plate conductors which are separated by a very small distance, d. the two plates of the capacitor can be made to carry equal and opposite charges by connecting the capacitor across the terminals of a battery such that the p.d across the plate is V.
This is called “charging”. For a charged capacitor the electric charge on one plate is +q while on the other plate it is - q
https://youtu.be/X4EUwTwZ110
https://youtu.be/L6cgSxpGmDo
EVALUATION
1. Explain the term capacitor
2. Explain the process of charging and discharging a capacitor
CAPACITANCE.
Experiment shows that the magnitude of the charge q on any of the plate is directly proportional to the potential difference, V across the capacitor, that is q α v
q = cv………………………….. 1
Where c is a constant of proportionality known as the capacitance the farad (F) is capacitance unit.
(F). Practical units are micro ( F ) and pico ( pF ) farad
CAPACITANCE IN SERIES AND IN PARALLEL.
If two or more capacitors, c1, c2 --- are connected in series , it can be shown that the equivalent or net capacitance, c of the combination is given by
If they are connected in parallel the net capacitance C in this is given by
C = c1 + c2 + c3 ------- 3
Note that the opposite is the case if these were resistance.
https://youtu.be/BIPi0vXdssE
https://youtu.be/f_MZNsEqyQw
SIMPLE PROBLEMS
E.g. A capacitor contain a charge of 4 .0 x 10– 4 coulomb when a potential difference of 400 v is applied across it. Calculate the capacitance of the capacitor
The capacitance C = q/v
= 4.0 x 10-4 = 10 – 6 F = 1. 0 F
400
EVALUATION.
1. What do you understand by the term capacitance?
2. How is it related to potential difference and charge?
ENERGY STORED IN CAPACITOR
A charged is a store of electrical energy. When a charge , q , is moved through a p.d , the work done is given by
W = average p.d x charge
= ½ v q
But v = q /c
W = ½ q/c x q = ½ q2/c
W = ½q2/C
0r if instead we use q=cv
W =1/2 cv2
Therefore the work done is either
W = ½ CV2 = 1/2 q2/c
This work is stored in the capacitor as electrical potential energy.
https://youtu.be/SlvXDIaFDjU
EVALUATION
1.The net charge on capacitor which is charged to a p.d of 200 is 1.0 x 10-4 coulomb. what is the capacitance of capacitor and the energy stored in the capacitor ?
2. Derive an expression for the energy lost on sharing charges.
PARALLEL PLATE CAPACITOR
For a parallel plate capacitor with plates each of area (A), separated by distance (d) the capacitance (C), of the capacitor is given by
C = εoA/d
Where εo is the permittivity of the dielectric material between the plates of the capacitor.
It therefore means the capacitance of a parallel plate capacitor is dependent ov several factors namely:
1. The separation or distance (d) between the plates.
2. The permittivity of the medium εo
3. The area of the plates (A).
https://youtu.be/LO56WRcg6jM
READING ASSIGNMENT
New School Physics pg 380 -385 New School Physics By M.W Anyakoha
GENERAL EVALUATION
1 A string of length 4cm is extended by 0.02cm when a load of 0.4Kg is suspended at its end. What will be the length of the string when a load of 15N is applied?
2 A spring of force constant 500N/m is compressed such that its length is shortened by 5cm. the energy stored in the spring is.
ASSIGNMENT
1. Calculate the potential at a point 1m from a charge of 10-9c in vacuum (assume = 1/4πεo = 9x 109 m/f) (a). 1/9 v (b) 9 x 10-18v (c) 9v (d) 9 x 1018v
2. Two plate conductors are placed parallel and 50cm apart. One of the plate is earthed and the other is at potential of + 10kv. What is the electric intensity between them? (a) 2x 104v (b) 2 x 10-4v (c) 2v (d) 10v
3. Which of the following statements about a capacitor is incorrect. (a) It is used for storing electric charges (b) It is mostly made of two parallel plates (c) It is charged by connecting a battery across the terminals (d) the charge on it is inversely proportional to the p.d across the terminals
4. Two capacitor of capacitance 3uF and 6uF are connected in series. Calculate the equivalent capacitance (a) 9uF (b) 6uF (c) 2uF (d) ½ uF.
5. A capacitor stores 10-4c of charge when the p.d between the plates is 1kv. What is the capacitance? (a) 10-4 uF (b) 0.1 uF (c) 4uF (d) 100uF.
THEORY
1. Define the term: capacitance of a capacitor:”
2. A capacitor stores 8 x 10-4c of charge when the potential difference between the plate is 100v. what is capacitance?
Arrangement of capacitors
Types of capacitors
Energy stored in capacitors
Every conductor may possess one or more of the following properties:
1. It could be mainly a resistor, which means that if a current is passed through it, heat energy is mainly produced
2. It could be a capacitor which means that when a current passes through it electrical energy or charges are stored.
3. Finally, it could be an inductor which stores mainly magnetic energy when a current is passed through it.
A Capacitor is essentially a device for storing electrical energy or charges. In general, capacitors can be in the form of two conductors which are insulated electrically from the surroundings. However, most common types of capacitors are in the form of two parallel plate conductors which are separated by a very small distance, d. the two plates of the capacitor can be made to carry equal and opposite charges by connecting the capacitor across the terminals of a battery such that the p.d across the plate is V.
This is called “charging”. For a charged capacitor the electric charge on one plate is +q while on the other plate it is - q
https://youtu.be/X4EUwTwZ110
https://youtu.be/L6cgSxpGmDo
EVALUATION
1. Explain the term capacitor
2. Explain the process of charging and discharging a capacitor
CAPACITANCE.
Experiment shows that the magnitude of the charge q on any of the plate is directly proportional to the potential difference, V across the capacitor, that is q α v
q = cv………………………….. 1
Where c is a constant of proportionality known as the capacitance the farad (F) is capacitance unit.
(F). Practical units are micro ( F ) and pico ( pF ) farad
CAPACITANCE IN SERIES AND IN PARALLEL.
If two or more capacitors, c1, c2 --- are connected in series , it can be shown that the equivalent or net capacitance, c of the combination is given by
If they are connected in parallel the net capacitance C in this is given by
C = c1 + c2 + c3 ------- 3
Note that the opposite is the case if these were resistance.
https://youtu.be/BIPi0vXdssE
https://youtu.be/f_MZNsEqyQw
SIMPLE PROBLEMS
E.g. A capacitor contain a charge of 4 .0 x 10– 4 coulomb when a potential difference of 400 v is applied across it. Calculate the capacitance of the capacitor
The capacitance C = q/v
= 4.0 x 10-4 = 10 – 6 F = 1. 0 F
400
EVALUATION.
1. What do you understand by the term capacitance?
2. How is it related to potential difference and charge?
ENERGY STORED IN CAPACITOR
A charged is a store of electrical energy. When a charge , q , is moved through a p.d , the work done is given by
W = average p.d x charge
= ½ v q
But v = q /c
W = ½ q/c x q = ½ q2/c
W = ½q2/C
0r if instead we use q=cv
W =1/2 cv2
Therefore the work done is either
W = ½ CV2 = 1/2 q2/c
This work is stored in the capacitor as electrical potential energy.
https://youtu.be/SlvXDIaFDjU
EVALUATION
1.The net charge on capacitor which is charged to a p.d of 200 is 1.0 x 10-4 coulomb. what is the capacitance of capacitor and the energy stored in the capacitor ?
2. Derive an expression for the energy lost on sharing charges.
PARALLEL PLATE CAPACITOR
For a parallel plate capacitor with plates each of area (A), separated by distance (d) the capacitance (C), of the capacitor is given by
C = εoA/d
Where εo is the permittivity of the dielectric material between the plates of the capacitor.
It therefore means the capacitance of a parallel plate capacitor is dependent ov several factors namely:
1. The separation or distance (d) between the plates.
2. The permittivity of the medium εo
3. The area of the plates (A).
https://youtu.be/LO56WRcg6jM
READING ASSIGNMENT
New School Physics pg 380 -385 New School Physics By M.W Anyakoha
GENERAL EVALUATION
1 A string of length 4cm is extended by 0.02cm when a load of 0.4Kg is suspended at its end. What will be the length of the string when a load of 15N is applied?
2 A spring of force constant 500N/m is compressed such that its length is shortened by 5cm. the energy stored in the spring is.
ASSIGNMENT
1. Calculate the potential at a point 1m from a charge of 10-9c in vacuum (assume = 1/4πεo = 9x 109 m/f) (a). 1/9 v (b) 9 x 10-18v (c) 9v (d) 9 x 1018v
2. Two plate conductors are placed parallel and 50cm apart. One of the plate is earthed and the other is at potential of + 10kv. What is the electric intensity between them? (a) 2x 104v (b) 2 x 10-4v (c) 2v (d) 10v
3. Which of the following statements about a capacitor is incorrect. (a) It is used for storing electric charges (b) It is mostly made of two parallel plates (c) It is charged by connecting a battery across the terminals (d) the charge on it is inversely proportional to the p.d across the terminals
4. Two capacitor of capacitance 3uF and 6uF are connected in series. Calculate the equivalent capacitance (a) 9uF (b) 6uF (c) 2uF (d) ½ uF.
5. A capacitor stores 10-4c of charge when the p.d between the plates is 1kv. What is the capacitance? (a) 10-4 uF (b) 0.1 uF (c) 4uF (d) 100uF.
THEORY
1. Define the term: capacitance of a capacitor:”
2. A capacitor stores 8 x 10-4c of charge when the potential difference between the plate is 100v. what is capacitance?
WEEK 5
TOPIC: ELECTRIC CELLS
Content
Electric circuit
Types of electric cells
ELECTRIC CIRCUIT
Electric current is simply electric charge in motion. In conductors such as cables or wire, the current consist of swam of moving electron. Electric cells are chemical devices, which are capable of causing an electric current to flow. This produces electric force, which pushes the current along. If there is a complete circuit of conductors by which current can leave from one end to terminal of the cell and travel round to the other terminal, a current will flow. This current will be the at any point round the circuit and of the line is broken, the current is stopped or switched off. The electrons flow from the negative terminal or cathode of the cell to the procedure terminal or anode
TYPES OF ELECTRIC CELLS
Electric cells are divided into two namely: the primary cells and the secondary cells
PRIMARY CELLS: These are those cells in which current is produced as a result of an irreversible chemical charge.
SECONDARY CELLS: These cells are those which can be recharged when they run down by passing current backwards through them .
There are three component in a cell .Two of them are electrodes in the primary cell, the two electrodes are of different metals (graphite is often used) the third item is the container bearing the electrolyte. Examples of electrolyte are strips of aluminum, Carbons (graphite) copper, iron lead and zinc.
THE SIMPLE PRIMARY CELL (VOLTAIC CELL) A simple cell can be made by placing two different electrodes (metals) in an electrolyte. Two wire are then used to connect these metals to a voltmeter, an instrument which measure the potential different between any two point in an electric circuit. If a deflection is noticed it means that the cell creates a voltage. If the deflection is done to the right it mean that the electrode, or anode, which is connected to the positive terminal of the voltmeter is the positive electrode, or anode, while the one is connected to the negative terminal is the negative electrode or cathode. If the deflection is however done to the left, a reconnection should be done .
The two major deflects of a simple all are polarization and local action
Polarization: The cell is characterized by the release of “hydrogen bubbles.” The bubbles collect at the positive electrode and insulate it. This show down and eventually stops the working of the cell. This defect is called polarization.
This defect can be corrected either by occasionally brushing the plates, which is highly in convenient, or by using a depolarize e.g. manganese oxide. This oxides hydrogen to form and so removes the hydrogen bubble.
Local action: This occurs when pure zinc is not in use. The impurities in the zinc results in the gradual wearing away of the zinc plates. This can be prevented by cleaning the zinc with H2SO4 and then rubbed with mercury. The mercury amalgamates the zinc by covering the impurities thereby preventing it from coming into contact with electrolyte.
EVALUATION
1. What is a cell?
2. Explain the defects of a named cell.
LECLANCHE CELL
Leclanche cells are of two types : the wet and the dried types. The wet leclanche cell consists of a zinc rod at the cathode in solution of ammonium chloride contained in a glass vessel. The anode is a carbon rod contained in a porous pot and is surrounded by manganese chloride as a depolarize
An e.m.f. Is set up by the zinc, the carbon and the electrolyte, which drives a current from zinc to carbon through the cell. This carbon is at a higher potential than the zinc. When an external circuit is connected to the cell, current flows from carbon to zinc out side. The e.m.f is set up because zinc reacts with the ammonium chloride to form zinc chloride, ammonia and hydrogen, and electrons are released. These electrons flow from the zinc plate to the carbon plate out side the cell.
Hydrogen reacts with the manganese dioxide and oxidizes it to form water. The e.m.f of a leclanche cell is 1.5 v.Its defect include
When the cell has worked for sometime, the rate of hydrogen production becomes greater than rate at which it is oxidized by the manganese dioxide, hence the formation of polarization. Therefore the cell must be allowed to rest from time. This primary cells are restricted to intermittent current supply because they do not give continuous service.
They are two heavy to carry about without spilling the liquid
For the dry leclanche cell, the defect of heaviness is overcome
The ammonium chloride electrolyte is a jelly-like material and not aligned solution. The positive electrode is a carbon rod surrounded by a packed mixture of manganese dioxide and powered carbon, inside a zinc container, which is the negative electrode.
The dry can be carry about easily E.g. torch batteries, and transistor radio batteries. Due to local action, they deteriorate after sometime.
THE DANIEL CELL
This is also a primary cell invented to counter the problem of polarization. The zinc rod is the negative electrode while the positive electrode is the container. The electrolyte is dilute tetrasulphate (vi) acid contained in a porous pot around the zinc rod, and the depolarize is copper tetraoxosulphate (vi) in the surrounding copper container. The diaphoreses is mush more efficient than the leclanche cell. The e.m.f. is of a constant value of l..08V.
Secondary cell are of two main type: lead acid accumulator, and the alkaline or Nife accumulation.
The lead-acid accumulator. This is the most common one. It consist of lead oxide as the positive electrode, lead as the negative electrode and tetra oxosulphate (vi) acid as the electrolyte. During the discharge, when the cell is given out current both plates gradually charge to lead tetraoxosulphate (vi) while the acid gradually becomes more dilute and the density decreases. When fully charge the relative density and e.m.f. of the cell are 1.25 and 2.2v respectively. But when discharge they are reduced to 1.5 and less than 2.0v respectively. The rod density of the cell should not be allowed to drop 1.15 before it is recharged. Maintenance of lead acid accumulators
1 The liquid level must be maintained by using distilled H2O
2. The cell should be charge if relative. Density of acid falls below 1.15. it is fully charged when relative density of acid is 1.25. it is tested with a special hydrometer.
3 .If the cell is not in use for a long time, it should be discharge from time to time or the acid remove and the cell dried.
4. The battery should be kept clean so that current dose not leaks away across the casing between the terminals.
THE ALKALINE OR NIFE ACCUMULATORS.
The name is gotten from the chemical symbol nickel (Ni) and iron (Fe). The positive electrode is made of nickel hydrogen while the negative plate is either of iron or calcium. The electrolyte is potassium hydroxide dissolve in water. This cell last longer and lead acid cells, keep their charge longer and they require less maintenance. They are used for emergencies in factories and hospitals. They are expensive and bulky with a small e.m.f value, about 1.25v.
EVALUATION
1. What is the advantage of dry leclanche. Cell over wet leclanche cell.?
2. How can polarization and local action be prevented.
Summary
Cell Positive Terminal Negative Terminal Electrolyte Depolariser
Simple Copperplate Zinc rod Dilute sulfuric acid _
Daniell Copper container Zinc rod Dilute sulfuric acid Copper sulphate solution
Leclanché (wet) Carbon rod Zinc rod Ammonium chloride solution Manganese dioxide
Leclanché (dry) Carbon rod Zinc container Ammonium chloride paste Manganese dioxide
Lead-acid accumulator Lead oxide Lead Dilute sulfuric acid _
READING ASSIGNMENT
New School Physics for Senior Secondary Schools (M.W. ANYAKOHA pgs 397 – 402).
GENERAL EVALUATION:
1. A block of mass 2.0Kg resting on a smooth horizontal plane is acted upon simultaneously by two forces , 10N due north and 10N due east. The magnitude of the acceleration produced by the forces on the block is.
2. Two forces 3N and 4N act on a body due north and due east respectively. Calculate the equilibrant.
WEEKEND ASSIGNMENT
(1) The energy transformation taking place when a cell supplies current to a bulb is from (a)light energy to heat energy (b) mechanical energy to light energy (c)solar energy to electrical energy (d) chemical energy to light energy
(2) which of the following devices convert sheet energy to electric current?
(a)photo cell (b) battery (c) voltmeter (c) thermocouple
(3). during an activity, is coulombs of charge passed though an ammeter in 2second what is the reading of the ammeter ? (a) 2A (b) 5A (c) 8A (d)10A
(4). which of the following devices coverts mechanical energy to electric current (a) battery (b) photocell (c) thermopile (d)dynamo
(5). the rheostat could serve the following except. (a)as a variable resistor (b)as a potential divider (c)as a means of varying the current in a circuit .(d) as a converter of solar energy to electrical energy
THEORY
1. Briefly differentiate between primary cell and secondary cells.
2. list two defects of a simple cell
CONTENT: ELECTRIC CELLS
An electric cell is a device which supplies an electric current. It consists of electrolyte, and the electrodes as the basic components.
There are three types of cells: Simple, primary and secondary cell. The simple cell has a p.d of approximately 1V.
DEFECTS OF SIMPLE CELLS
i. Polarization: formation of H2 bubble at positive electrode.
ii. Local action
It cannot be left assembled as the dilute acid will react with the Zinc electrode.
Primary cells are those in which current is produced by an irreversible chemical change e.g Leclanche', Daniel and Nichel-Iron (Nife)
Secondary cells: these are those which can be recharged when they run down recharging is done by passing electric current through the cell.
i.e. discharging (+ve -ve, -ve +ve)
and recharging (+ve +ve, -ve -ve)
a good electric cell should have
i. A long life
ii. A minimum of polarization
iii. Cheap and safe electrodes
iv. Cheap and safe electrolytes
CELLS IN SERIES AND PARALLEL
Series: E1 = E1 + E2 + E3
Parallel: E = E1 = E2 = E3 in short the cell with the highest e.m.f
E.M.F AND TERMINAL P.D
The e.m.f of a cell is the p.d across the terminals of that cell when it is not delivering any current to an external circuit.
Potential difference: p.d is the e.m.f of a cell (or circuit) when it is delivering current to an external circuit.
V = E-v; V= terminal p.d supplied into R and v is the lost volt.
Therefore:
E = V +r V= E-r or E-Irs
Ex.
A battery made up of 2 cells joined in series supplies current to an external resistance of 5Ω. If the e.m.f and internal resistance of each cell is 0.6v and 3Ω respectively, calculate the current flowing in the external resistor, terminal p.d and lost voltage.
EVALUATION:
Two cells each having e.m.f of 1.5v and an internal resistance of 2Ω are connected in parallel. Find the current flowing when the cells are connected to a 11Ω resistor.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
Describe the secondary cell and explain how it is charged.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: SECONDARY CELLS
These are cells that are rechargeable e.g. car battery. During discharge, the lead in the - ve plate and PbO2 are in +ve platesboth change into insoluble lead (II) sulphate (pbSO4)
EVALUATION:
Briefly describe how a car battery can be recharged.
ASSIGNMENT:
Calculate the
i. Total R
ii. P.d V across the 2 and 4Ω
iii. Current in each resistor
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
Explain cells in series and parallel and solve simple questions in cells.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CELLS IN SERIES AND PARALLEL
Series: and parallel connections
Example:
If the terminal p.d across a circuit is 15V with internal resistance of 5Ω, resistance is 23Ω with p.d of 40V, calculate the terminal p.d.
Solution:
E = 15v(p.d across the whole circuit) (R +r)
r = 5Ω, R= 23Ω
from I = E = 15 = 5.357 x 10-1A
R + r 23 + 5
p.d across the load = IR = 12.3 x 10 volts
MEASUREMENT OF RESISTANCE
Voltmeter method:
Plot graph of v against I and obtain the slope of graph as resistance.
The disadvantage of this method is that some current is always directed through the voltmeter to make it work.
OTHER METHODS
i. Resistivity
ii. Wheatstone Bridge
iii. Metre Bridge
iv. Using measuring instruments
EVALUATION:
State Ohm's Law.
ASSIGNMENT:
i. Use diagram to describe and state Ohm's Law.
ii. Use diagram to describe cathode ray oscilloscope.
TOPIC: ELECTRIC CELLS AND DEFECTS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Mention at least 2 defects of simple cells.
ii. Describe the defects of simple cells
iii. Mention corrective technique for defects in simple cells.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: DEFECTS OF SIMPLE CELLS
1. Polarization
This is the release of hydrogen bubbles at the positive electrode. Thereby insulating the electrodes. This causes back e.m.f. polarization can be minimized by brushing the electrodes and also using depolarizers e.g. manganese dioxide.
2. Local Action:
Local action occurs due to improve zinc useage as electrodes. It can be minimized or corrected by cleaning zinc with H2SO4 and then rubbing or amalgamating with mercury.
DISADVANTAGE OF SIMPLE CELL OVER SECONDARY CELLS.
It does not allow for a steady supply of current.
It is not rechargeable after discharging.
EVALUATION:
List the defects of simple cells
ASSIGNMENT:
Use diagram to describe Leclanche' and lead-acid accumulator.
SUB-TOPIC: E.M.F AND TERMINAL P.D
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Solve simple problems involving cells and e.m.f of cell
ii. Distinguish between two points on a circuit.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: E.M.F OF A CELL
The e.m.f of a cell is the p.d across the terminals of a cell when it is in an open circuit; i.e. not supplying current to an external circuit.
E supplies current I to external R. therefore the p.d across R is called terminal p.d V while v is the p.d across internal resistance "r".
Therefore E= V + r
E = IR + Ir
E= I (R +r)
Therefore I= E
R + r
Where Ir is called lost volts.
Ex.
If the terminal p.d of a battery is 15v and the external resistance of 23Ω is connected and when the lost volt is 1.65v. Find the e.m.f and the internal of the cell.
Solution:
When p.d = 15v, R = 23Ω
Also p.d = 1.65v, R = 48Ω
From I = E, but E = V + v
R + r
(i) E.m.f is given by E = 15 + 1.65
E = 16.65V
(ii) R =?
Since terminal p.d across external resistor = 15V
Therefore: using Ohm's Law,
V = IR
Therefore: I = V = I = 15 = 0.652A
R 23
Using E = I (R + r)
Or E = I (R +r) gives
16.65 = 0.652 (23 + r)
Therefore: r = 16.65 - (0.652 x 23)
0.652
R = 1.654; r = 2.54Ω
0.652
EVALUATION:
Distinguish between e.m.f of a cell and terminal p.d of that cell.
ASSIGNMENT:
Calculate the e.m.f of a cell of an internal resistance of 15Ω connected to a 100Ω resistance. If the terminal p.d is 10V.
Content
Electric circuit
Types of electric cells
ELECTRIC CIRCUIT
Electric current is simply electric charge in motion. In conductors such as cables or wire, the current consist of swam of moving electron. Electric cells are chemical devices, which are capable of causing an electric current to flow. This produces electric force, which pushes the current along. If there is a complete circuit of conductors by which current can leave from one end to terminal of the cell and travel round to the other terminal, a current will flow. This current will be the at any point round the circuit and of the line is broken, the current is stopped or switched off. The electrons flow from the negative terminal or cathode of the cell to the procedure terminal or anode
TYPES OF ELECTRIC CELLS
Electric cells are divided into two namely: the primary cells and the secondary cells
PRIMARY CELLS: These are those cells in which current is produced as a result of an irreversible chemical charge.
SECONDARY CELLS: These cells are those which can be recharged when they run down by passing current backwards through them .
There are three component in a cell .Two of them are electrodes in the primary cell, the two electrodes are of different metals (graphite is often used) the third item is the container bearing the electrolyte. Examples of electrolyte are strips of aluminum, Carbons (graphite) copper, iron lead and zinc.
THE SIMPLE PRIMARY CELL (VOLTAIC CELL) A simple cell can be made by placing two different electrodes (metals) in an electrolyte. Two wire are then used to connect these metals to a voltmeter, an instrument which measure the potential different between any two point in an electric circuit. If a deflection is noticed it means that the cell creates a voltage. If the deflection is done to the right it mean that the electrode, or anode, which is connected to the positive terminal of the voltmeter is the positive electrode, or anode, while the one is connected to the negative terminal is the negative electrode or cathode. If the deflection is however done to the left, a reconnection should be done .
The two major deflects of a simple all are polarization and local action
Polarization: The cell is characterized by the release of “hydrogen bubbles.” The bubbles collect at the positive electrode and insulate it. This show down and eventually stops the working of the cell. This defect is called polarization.
This defect can be corrected either by occasionally brushing the plates, which is highly in convenient, or by using a depolarize e.g. manganese oxide. This oxides hydrogen to form and so removes the hydrogen bubble.
Local action: This occurs when pure zinc is not in use. The impurities in the zinc results in the gradual wearing away of the zinc plates. This can be prevented by cleaning the zinc with H2SO4 and then rubbed with mercury. The mercury amalgamates the zinc by covering the impurities thereby preventing it from coming into contact with electrolyte.
EVALUATION
1. What is a cell?
2. Explain the defects of a named cell.
LECLANCHE CELL
Leclanche cells are of two types : the wet and the dried types. The wet leclanche cell consists of a zinc rod at the cathode in solution of ammonium chloride contained in a glass vessel. The anode is a carbon rod contained in a porous pot and is surrounded by manganese chloride as a depolarize
An e.m.f. Is set up by the zinc, the carbon and the electrolyte, which drives a current from zinc to carbon through the cell. This carbon is at a higher potential than the zinc. When an external circuit is connected to the cell, current flows from carbon to zinc out side. The e.m.f is set up because zinc reacts with the ammonium chloride to form zinc chloride, ammonia and hydrogen, and electrons are released. These electrons flow from the zinc plate to the carbon plate out side the cell.
Hydrogen reacts with the manganese dioxide and oxidizes it to form water. The e.m.f of a leclanche cell is 1.5 v.Its defect include
When the cell has worked for sometime, the rate of hydrogen production becomes greater than rate at which it is oxidized by the manganese dioxide, hence the formation of polarization. Therefore the cell must be allowed to rest from time. This primary cells are restricted to intermittent current supply because they do not give continuous service.
They are two heavy to carry about without spilling the liquid
For the dry leclanche cell, the defect of heaviness is overcome
The ammonium chloride electrolyte is a jelly-like material and not aligned solution. The positive electrode is a carbon rod surrounded by a packed mixture of manganese dioxide and powered carbon, inside a zinc container, which is the negative electrode.
The dry can be carry about easily E.g. torch batteries, and transistor radio batteries. Due to local action, they deteriorate after sometime.
THE DANIEL CELL
This is also a primary cell invented to counter the problem of polarization. The zinc rod is the negative electrode while the positive electrode is the container. The electrolyte is dilute tetrasulphate (vi) acid contained in a porous pot around the zinc rod, and the depolarize is copper tetraoxosulphate (vi) in the surrounding copper container. The diaphoreses is mush more efficient than the leclanche cell. The e.m.f. is of a constant value of l..08V.
Secondary cell are of two main type: lead acid accumulator, and the alkaline or Nife accumulation.
The lead-acid accumulator. This is the most common one. It consist of lead oxide as the positive electrode, lead as the negative electrode and tetra oxosulphate (vi) acid as the electrolyte. During the discharge, when the cell is given out current both plates gradually charge to lead tetraoxosulphate (vi) while the acid gradually becomes more dilute and the density decreases. When fully charge the relative density and e.m.f. of the cell are 1.25 and 2.2v respectively. But when discharge they are reduced to 1.5 and less than 2.0v respectively. The rod density of the cell should not be allowed to drop 1.15 before it is recharged. Maintenance of lead acid accumulators
1 The liquid level must be maintained by using distilled H2O
2. The cell should be charge if relative. Density of acid falls below 1.15. it is fully charged when relative density of acid is 1.25. it is tested with a special hydrometer.
3 .If the cell is not in use for a long time, it should be discharge from time to time or the acid remove and the cell dried.
4. The battery should be kept clean so that current dose not leaks away across the casing between the terminals.
THE ALKALINE OR NIFE ACCUMULATORS.
The name is gotten from the chemical symbol nickel (Ni) and iron (Fe). The positive electrode is made of nickel hydrogen while the negative plate is either of iron or calcium. The electrolyte is potassium hydroxide dissolve in water. This cell last longer and lead acid cells, keep their charge longer and they require less maintenance. They are used for emergencies in factories and hospitals. They are expensive and bulky with a small e.m.f value, about 1.25v.
EVALUATION
1. What is the advantage of dry leclanche. Cell over wet leclanche cell.?
2. How can polarization and local action be prevented.
Summary
Cell Positive Terminal Negative Terminal Electrolyte Depolariser
Simple Copperplate Zinc rod Dilute sulfuric acid _
Daniell Copper container Zinc rod Dilute sulfuric acid Copper sulphate solution
Leclanché (wet) Carbon rod Zinc rod Ammonium chloride solution Manganese dioxide
Leclanché (dry) Carbon rod Zinc container Ammonium chloride paste Manganese dioxide
Lead-acid accumulator Lead oxide Lead Dilute sulfuric acid _
READING ASSIGNMENT
New School Physics for Senior Secondary Schools (M.W. ANYAKOHA pgs 397 – 402).
GENERAL EVALUATION:
1. A block of mass 2.0Kg resting on a smooth horizontal plane is acted upon simultaneously by two forces , 10N due north and 10N due east. The magnitude of the acceleration produced by the forces on the block is.
2. Two forces 3N and 4N act on a body due north and due east respectively. Calculate the equilibrant.
WEEKEND ASSIGNMENT
(1) The energy transformation taking place when a cell supplies current to a bulb is from (a)light energy to heat energy (b) mechanical energy to light energy (c)solar energy to electrical energy (d) chemical energy to light energy
(2) which of the following devices convert sheet energy to electric current?
(a)photo cell (b) battery (c) voltmeter (c) thermocouple
(3). during an activity, is coulombs of charge passed though an ammeter in 2second what is the reading of the ammeter ? (a) 2A (b) 5A (c) 8A (d)10A
(4). which of the following devices coverts mechanical energy to electric current (a) battery (b) photocell (c) thermopile (d)dynamo
(5). the rheostat could serve the following except. (a)as a variable resistor (b)as a potential divider (c)as a means of varying the current in a circuit .(d) as a converter of solar energy to electrical energy
THEORY
1. Briefly differentiate between primary cell and secondary cells.
2. list two defects of a simple cell
CONTENT: ELECTRIC CELLS
An electric cell is a device which supplies an electric current. It consists of electrolyte, and the electrodes as the basic components.
There are three types of cells: Simple, primary and secondary cell. The simple cell has a p.d of approximately 1V.
DEFECTS OF SIMPLE CELLS
i. Polarization: formation of H2 bubble at positive electrode.
ii. Local action
It cannot be left assembled as the dilute acid will react with the Zinc electrode.
Primary cells are those in which current is produced by an irreversible chemical change e.g Leclanche', Daniel and Nichel-Iron (Nife)
Secondary cells: these are those which can be recharged when they run down recharging is done by passing electric current through the cell.
i.e. discharging (+ve -ve, -ve +ve)
and recharging (+ve +ve, -ve -ve)
a good electric cell should have
i. A long life
ii. A minimum of polarization
iii. Cheap and safe electrodes
iv. Cheap and safe electrolytes
CELLS IN SERIES AND PARALLEL
Series: E1 = E1 + E2 + E3
Parallel: E = E1 = E2 = E3 in short the cell with the highest e.m.f
E.M.F AND TERMINAL P.D
The e.m.f of a cell is the p.d across the terminals of that cell when it is not delivering any current to an external circuit.
Potential difference: p.d is the e.m.f of a cell (or circuit) when it is delivering current to an external circuit.
V = E-v; V= terminal p.d supplied into R and v is the lost volt.
Therefore:
E = V +r V= E-r or E-Irs
Ex.
A battery made up of 2 cells joined in series supplies current to an external resistance of 5Ω. If the e.m.f and internal resistance of each cell is 0.6v and 3Ω respectively, calculate the current flowing in the external resistor, terminal p.d and lost voltage.
EVALUATION:
Two cells each having e.m.f of 1.5v and an internal resistance of 2Ω are connected in parallel. Find the current flowing when the cells are connected to a 11Ω resistor.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
Describe the secondary cell and explain how it is charged.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: SECONDARY CELLS
These are cells that are rechargeable e.g. car battery. During discharge, the lead in the - ve plate and PbO2 are in +ve platesboth change into insoluble lead (II) sulphate (pbSO4)
EVALUATION:
Briefly describe how a car battery can be recharged.
ASSIGNMENT:
Calculate the
i. Total R
ii. P.d V across the 2 and 4Ω
iii. Current in each resistor
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
Explain cells in series and parallel and solve simple questions in cells.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CELLS IN SERIES AND PARALLEL
Series: and parallel connections
Example:
If the terminal p.d across a circuit is 15V with internal resistance of 5Ω, resistance is 23Ω with p.d of 40V, calculate the terminal p.d.
Solution:
E = 15v(p.d across the whole circuit) (R +r)
r = 5Ω, R= 23Ω
from I = E = 15 = 5.357 x 10-1A
R + r 23 + 5
p.d across the load = IR = 12.3 x 10 volts
MEASUREMENT OF RESISTANCE
Voltmeter method:
Plot graph of v against I and obtain the slope of graph as resistance.
The disadvantage of this method is that some current is always directed through the voltmeter to make it work.
OTHER METHODS
i. Resistivity
ii. Wheatstone Bridge
iii. Metre Bridge
iv. Using measuring instruments
EVALUATION:
State Ohm's Law.
ASSIGNMENT:
i. Use diagram to describe and state Ohm's Law.
ii. Use diagram to describe cathode ray oscilloscope.
TOPIC: ELECTRIC CELLS AND DEFECTS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Mention at least 2 defects of simple cells.
ii. Describe the defects of simple cells
iii. Mention corrective technique for defects in simple cells.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: DEFECTS OF SIMPLE CELLS
1. Polarization
This is the release of hydrogen bubbles at the positive electrode. Thereby insulating the electrodes. This causes back e.m.f. polarization can be minimized by brushing the electrodes and also using depolarizers e.g. manganese dioxide.
2. Local Action:
Local action occurs due to improve zinc useage as electrodes. It can be minimized or corrected by cleaning zinc with H2SO4 and then rubbing or amalgamating with mercury.
DISADVANTAGE OF SIMPLE CELL OVER SECONDARY CELLS.
It does not allow for a steady supply of current.
It is not rechargeable after discharging.
EVALUATION:
List the defects of simple cells
ASSIGNMENT:
Use diagram to describe Leclanche' and lead-acid accumulator.
SUB-TOPIC: E.M.F AND TERMINAL P.D
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Solve simple problems involving cells and e.m.f of cell
ii. Distinguish between two points on a circuit.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: E.M.F OF A CELL
The e.m.f of a cell is the p.d across the terminals of a cell when it is in an open circuit; i.e. not supplying current to an external circuit.
E supplies current I to external R. therefore the p.d across R is called terminal p.d V while v is the p.d across internal resistance "r".
Therefore E= V + r
E = IR + Ir
E= I (R +r)
Therefore I= E
R + r
Where Ir is called lost volts.
Ex.
If the terminal p.d of a battery is 15v and the external resistance of 23Ω is connected and when the lost volt is 1.65v. Find the e.m.f and the internal of the cell.
Solution:
When p.d = 15v, R = 23Ω
Also p.d = 1.65v, R = 48Ω
From I = E, but E = V + v
R + r
(i) E.m.f is given by E = 15 + 1.65
E = 16.65V
(ii) R =?
Since terminal p.d across external resistor = 15V
Therefore: using Ohm's Law,
V = IR
Therefore: I = V = I = 15 = 0.652A
R 23
Using E = I (R + r)
Or E = I (R +r) gives
16.65 = 0.652 (23 + r)
Therefore: r = 16.65 - (0.652 x 23)
0.652
R = 1.654; r = 2.54Ω
0.652
EVALUATION:
Distinguish between e.m.f of a cell and terminal p.d of that cell.
ASSIGNMENT:
Calculate the e.m.f of a cell of an internal resistance of 15Ω connected to a 100Ω resistance. If the terminal p.d is 10V.
WEEK 6
TOPIC: CONDUCTION OF ELECTRICITY THROUGH MATERIALS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Mention at least two defects of simple cell.
ii. State Ohm's Law and apply it in calculations involving cells, p.d and e.m.f
iii. Verify Ohm's Law and apply Ohm's Law
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT:
Ohm's law states that the current passing a metallic conductor at constant temperature is directly proportional to the p.d between its ends.
V α I
Therefore: V = IR
V = p.d, I = current flowing between ends of conductor and R is the constant (resistance)
R = V
I
VERIFICATION OF OHM'S LAW
In an experiment to verify Ohm's Law, the slope of graph of V against I will give the resistance.
DEFECTS OF CELLS
i. Polarization
ii. Local Action
Ex.
A circuit consists of 1Ω wire in series with a parallel arrangement of 6Ω and 3Ω and a p.d of 12V is connected across the whole circuit.
Calculate:
i. Total circuit resistance
ii. Current in each resistor
iii. P.d across each resistor
EVALUATION:
State Ohm's law
ASSIGNMENT:
As shown in the diagram, calculate:
i. Effective resistance
ii. Current in each resistor
iii. P.d across each resistor.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Distinguish between electrical energy and electrical power.
ii. Solve simple problems on electrical energy and power.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTRICAL ENERGY AND POWER
Electrical energy is when quantity charge moves between two points, which have potential difference, while electrical power is the rate of energy transferred.
i.e.
Power = Energy transferred
Time taken
Energy (W) = QV put Q - It
W = IVt, put I = V
R
W = V2, put V = IR,
R2
W = I2Rt
Unit [Joules]
From power = Energy
Time
P = IV or r2t - t = v2
R R R
Or
P = I2Rt = I2R
T
ILLUSTRATION:
How much heat is produced in 2minutes by a heater using 500mA and a potential difference of 60V.
Solution:
+ = 2min = 2 x60 = 120s, V= 60V, I = 500mA = 5 x 10 -1
Using heat energy =IVt,
= it = 0.5 x 60 x 120 = 3600J
ENERGY CONSUMPTION
Commercial power is consumed in kilowatt hour (kwh) meaning energy supplied by a rate of working of 1000watts for 1 hour.
Illustration:
A house consumed 400w for lighting and 200w for refrigerator for 24hours. Calculate the total of the tariff is 50k/unit.
EVALUATION:
Distinguish between electrical energy and electrical power.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Mention at least two defects of simple cell.
ii. State Ohm's Law and apply it in calculations involving cells, p.d and e.m.f
iii. Verify Ohm's Law and apply Ohm's Law
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT:
Ohm's law states that the current passing a metallic conductor at constant temperature is directly proportional to the p.d between its ends.
V α I
Therefore: V = IR
V = p.d, I = current flowing between ends of conductor and R is the constant (resistance)
R = V
I
VERIFICATION OF OHM'S LAW
In an experiment to verify Ohm's Law, the slope of graph of V against I will give the resistance.
DEFECTS OF CELLS
i. Polarization
ii. Local Action
Ex.
A circuit consists of 1Ω wire in series with a parallel arrangement of 6Ω and 3Ω and a p.d of 12V is connected across the whole circuit.
Calculate:
i. Total circuit resistance
ii. Current in each resistor
iii. P.d across each resistor
EVALUATION:
State Ohm's law
ASSIGNMENT:
As shown in the diagram, calculate:
i. Effective resistance
ii. Current in each resistor
iii. P.d across each resistor.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Distinguish between electrical energy and electrical power.
ii. Solve simple problems on electrical energy and power.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTRICAL ENERGY AND POWER
Electrical energy is when quantity charge moves between two points, which have potential difference, while electrical power is the rate of energy transferred.
i.e.
Power = Energy transferred
Time taken
Energy (W) = QV put Q - It
W = IVt, put I = V
R
W = V2, put V = IR,
R2
W = I2Rt
Unit [Joules]
From power = Energy
Time
P = IV or r2t - t = v2
R R R
Or
P = I2Rt = I2R
T
ILLUSTRATION:
How much heat is produced in 2minutes by a heater using 500mA and a potential difference of 60V.
Solution:
+ = 2min = 2 x60 = 120s, V= 60V, I = 500mA = 5 x 10 -1
Using heat energy =IVt,
= it = 0.5 x 60 x 120 = 3600J
ENERGY CONSUMPTION
Commercial power is consumed in kilowatt hour (kwh) meaning energy supplied by a rate of working of 1000watts for 1 hour.
Illustration:
A house consumed 400w for lighting and 200w for refrigerator for 24hours. Calculate the total of the tariff is 50k/unit.
EVALUATION:
Distinguish between electrical energy and electrical power.
WEEK 7
TOPIC: ELECTROLYSIS AND ITS APPLICATIONS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Explain the conditions under which liquids and gases conduct electricity.
ii. Explain the behaviour of charge carriers in liquids and gases.
iii. State Faraday's laws and calculate the charge transferred in the process.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CONDUCTION OF ELECTRICITY THROUGH LIQUIDS
Electrolysis is the chemical change in a liquid due to the flow of electric current through the liquid.
Voltametre is used for studying the flow of current through a liquid.
IONIC THEORY
An electrolyte is a liquid/molten substance that conducts electric current.
In an electrolyte, there are positively and negatively charged particles called ions. The molecules that constitute the electrolyte are split in solution into these ions.
EXAMPLES OF ELECTROLYSIS
1. Electrolysis of acidulated water
2. Electrolysis of CUSO4
APPLICATIONS OF ELECTROLYSIS
1. Electroplating: this is the process of coating a metallic object with another metal in order to improve its appearance or to safe guard it against corrosion.
2. Calibration of an Ammetre: the copper is weighed before the current is switched on. Let mass of copper be M and the e.c.e of copper be Z(i.e. electrochemical)
The electrochemical equivalent (e.c.e) of a substance is the mass of that substance deposited by one coulomb of electrolysis (unit = g/coulomb)
3. Extraction of metals such as aluminum and sodium.
4. Manufacture of electrolytic capacitors used in electronic appliances.
5. Manufacture of some pure metals e.g. pure cooper which is used in the manufacture of cables.
EVALUATION:
Define electrolysis and electrolytes
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. State Faraday's Law of electrolysis
ii. Solve simple problems on electrolysis and
iii. Describe conduction of electricity through gases.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: FARADAY'S LAWS OF ELECTROLYSIS
1st law: states that the mass m of a substance librated in electrolysis is directly proportional to the quantity of electricity Q, which has passed through the electrolyte.
2nd Law: states that the masses of different substances deposited by the same quantity of electricity are directly proportional to the chemical equivalent of the substance.
M α Q - 1
M α It
And m = Zit
Where Z is electrochemical equivalent.
CONDUCTIONS OF ELECTRICITY THROUGH GAS
The conduction of electricity through gases is studied using discharge tubes. Gases conduct electricity are:
1. Low pressure (0.01mmHg)
2. High voltage (1000V); they break into ions
NATURE AND PROPERTIES OF CATHODE RAYS
1. They consist of streams of fast moving particles of negative electricity (ELECTRON)
2. They cause glass and other materials to fluoresce or glow in the green colour.
3. They travel in straight lines.
4. They are deflated by magnetic and electric fields.
5. They can ionize a gas.
6. They will turn a light paddle wheel in tube to move.
7. They produce intense heat.
8. They affect photographic plates.
9. They produce x-rays
10. They have a high penetrating power.
THERMONIC EMISSION:
This is the emission of electrons from the surface of a hot metal.
The cathode ray oscilloscope is used for studying all types of wave forms especially the a.c waveform and to measure frequency, amplification of voltages of electronic devices.
EVALUATION: The teacher evaluates the lesson with the following questions:
State Faraday's Laws of electrolysis.
ASSIGNMENT:
A piece of metal has an area of 20cm2 and it takes one hour to deposit a layer of silver. What current is needed to do this? (Assume that 1cm2 of silver has a mass of 10.5g and that 1 coulomb of electricity deposits 0.0012g of silver.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Explain the conditions under which liquids and gases conduct electricity.
ii. Explain the behaviour of charge carriers in liquids and gases.
iii. State Faraday's laws and calculate the charge transferred in the process.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CONDUCTION OF ELECTRICITY THROUGH LIQUIDS
Electrolysis is the chemical change in a liquid due to the flow of electric current through the liquid.
Voltametre is used for studying the flow of current through a liquid.
IONIC THEORY
An electrolyte is a liquid/molten substance that conducts electric current.
In an electrolyte, there are positively and negatively charged particles called ions. The molecules that constitute the electrolyte are split in solution into these ions.
EXAMPLES OF ELECTROLYSIS
1. Electrolysis of acidulated water
2. Electrolysis of CUSO4
APPLICATIONS OF ELECTROLYSIS
1. Electroplating: this is the process of coating a metallic object with another metal in order to improve its appearance or to safe guard it against corrosion.
2. Calibration of an Ammetre: the copper is weighed before the current is switched on. Let mass of copper be M and the e.c.e of copper be Z(i.e. electrochemical)
The electrochemical equivalent (e.c.e) of a substance is the mass of that substance deposited by one coulomb of electrolysis (unit = g/coulomb)
3. Extraction of metals such as aluminum and sodium.
4. Manufacture of electrolytic capacitors used in electronic appliances.
5. Manufacture of some pure metals e.g. pure cooper which is used in the manufacture of cables.
EVALUATION:
Define electrolysis and electrolytes
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. State Faraday's Law of electrolysis
ii. Solve simple problems on electrolysis and
iii. Describe conduction of electricity through gases.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: FARADAY'S LAWS OF ELECTROLYSIS
1st law: states that the mass m of a substance librated in electrolysis is directly proportional to the quantity of electricity Q, which has passed through the electrolyte.
2nd Law: states that the masses of different substances deposited by the same quantity of electricity are directly proportional to the chemical equivalent of the substance.
M α Q - 1
M α It
And m = Zit
Where Z is electrochemical equivalent.
CONDUCTIONS OF ELECTRICITY THROUGH GAS
The conduction of electricity through gases is studied using discharge tubes. Gases conduct electricity are:
1. Low pressure (0.01mmHg)
2. High voltage (1000V); they break into ions
NATURE AND PROPERTIES OF CATHODE RAYS
1. They consist of streams of fast moving particles of negative electricity (ELECTRON)
2. They cause glass and other materials to fluoresce or glow in the green colour.
3. They travel in straight lines.
4. They are deflated by magnetic and electric fields.
5. They can ionize a gas.
6. They will turn a light paddle wheel in tube to move.
7. They produce intense heat.
8. They affect photographic plates.
9. They produce x-rays
10. They have a high penetrating power.
THERMONIC EMISSION:
This is the emission of electrons from the surface of a hot metal.
The cathode ray oscilloscope is used for studying all types of wave forms especially the a.c waveform and to measure frequency, amplification of voltages of electronic devices.
EVALUATION: The teacher evaluates the lesson with the following questions:
State Faraday's Laws of electrolysis.
ASSIGNMENT:
A piece of metal has an area of 20cm2 and it takes one hour to deposit a layer of silver. What current is needed to do this? (Assume that 1cm2 of silver has a mass of 10.5g and that 1 coulomb of electricity deposits 0.0012g of silver.
WEEK 8
TOPIC: MAGNETIC FIELD AND MAGNETISM
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Plot magnetic field around magnets and magnetic substances.
ii. Describe the working principle of electric bell and the telephone earpiece.
iii. State the relation between magnetic force and the motion of a charge in a magnetic field.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: MAGNETIC FIELD
The area around a magnet in which it can attract or repel objects or in which a magnetic force can be detected is called magnetic field of the magnet.
Magnet lines of force:
These are many lines along which a free- North pole would tend to move if placed in the field. A line of force may also be considered as a line such that the tangent to it at any point gives the direction of the field at that point.
PATTERNS OF MAGNETIC FIELD
Magnetic field around a straight conductor carrying current.
The direction of the field depends on the direction of flow of current. Such a direction can always be obtained by applying the R-H Grip or CLENCHED FIST RULE. this states that if the straight wire is grasped with the right hand so that the thumb points in the direction of the current, then the direction in which the fingers are curled indicates the direction of the magnetic field.
MAGNETIC CORKSCREW RULE B
This rule states that if a right handed corkscrew is turned so that its tip travels along the direction of current, the direction of rotation of the corkscrew gives the direction of the magnetic field or magnetic line of force.
MAGNETIC FIELD AROUND A SOLEMOID
A solemoid is a long cyclindrical coil of wire whose turns are usually wound close together.
METHODS OF MAKING MAGNETS
1. Electrical method: The magnetic effect of current on magnets . polarity of magnet depends on direction of flow of current in solemoid. If current flows anti clockwise in coil when one side is viewed, then the specimen at that end has a N pole.
2. By constant method:
i. Single touch
ii. Double touch
At the end of each repeated stroking, each end of the specimen will be found to have opposite polarity to that of the stroking pole.
3. Demagnetization:
i. Slant method (a.c current)
ii. Mechanical method : pointing in E-w direction.
iii. Heating method
MAGNETIC PROPERTIES OF IRON AND STEEL
1. Iron is easily magnetized than steel but also loses its magnetism more quickly than steel.
2. Iron produces a stronger magnet than steel, thus steel
i. Steel is used for making p.magnets
ii. Iron is used for making + magnets
iii. Iron nails are often used for lab experiments on magnetization and demagnetization.
ELECTROMAGNETS AND ITS APPLICATIONS OF ELECTRIC FIELD
They are soft iron core in a current carrying solemoid.
Electromagnets are used for:
i. Producing intense magnetic fields(e.g. electric motors)
ii. For lifting and transporting heavy pieces of iron and steel.
iii. Used in separating iron from mix of non magnetic substances.
iv. In construction of electromagnetic devices.
THE ELECTRIC BELL
The instrument works on a "make and break" device. The making of the contact between spring and screw at c causes the clapper to move and strike gong and consequently breaking of the contact causes the clapper to leave the gong.
TELEPHONE EAR-PIECE
The variation of electric current due to changes in the speech energy causes a variation in the magnetism of the electromagnet. Vibration of the diaphragm creates sound waves of the same frequency in a speech current.
THE EARTH'S MAGNETIC FIELD
The earth acts like a magnet. The N-geographical pole has south magnetic polarity since it attracts North pole of a magnet, the south geographical pole has North magnetic polarity.
Angle of Declination/variation:
This is the angle between North and geographical north direction at the place concerned.
Angle of Dip/inclination:
Is the angle between the direction of the earth's resultant magnetic field at the horizontal.
The angle of dip at points - component of Earth's magnetic field
Is given by tanθ = V
H
Patterns of Bar magnet in Earth's field
At the neutral point, the magnet force due to the magnetic field is equal and opposite to the force due to the magnetic force due to the field of the bar magnet.
Magnetic force on a charge moving in a magnetic field
Magnetic field exerts a force on a charge moving in the field. If electrons want to move with average velocity v, and the wires lie in the magnetic field b, with force F, on each electron, then
F = Bevsinθ
But moving q
F = qvBsinθ, becomes
F = vvB i.e. v cross B
If angle between B and v is 90∘, then sinθ = 1
EVALUATION:
Distinguish between angle of dip and declination.
ASSIGNMENT:
A beam of ions passes undeflated through two parallel horizontal plates 2.5m apart having a p.d of 2500v when the perpendicular magnetic field is 2 x 10-2 T, and in the magnetic field along radius 3m. calculate the charge per mass ratio of the ions.
SUB-TOPIC: CALCULATIONS ON ELECTRIC FIELD AND E.M.F
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Solve simple problems on force on a material in a magnetic field.
ii. Show a relationship between electric and magnetic field, charge and motion.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CLASS WORK
1. A charge of 1.6 x 10-1C enters a magnetic field of 10T with and of 2.5 x 102m/s. calculate the angle between the magnetic field and the direction of the motion at a force of 10 x 10-12N.
2. Obtain the magnetic force on an electron of charge q = 1.6 x 10-19C in a field of flux 20 tesla, with a speed of 2 x 107m/s in the direction (a) 90∘ (b) 60∘ (c) parallel to the magnetic field.
ELECTROMAGNETIC FIELD
In a field representing the joint interaction of electric and magnetic forces. It exerts a force on a charged particle; the force in a charge " moving with velocity " is given by
F= q (E + V + B)
Where f = force that occurs in both fields
q= charge, E = Electric field
B= magnetic field in tesla, or weber/m2
It can occur in an electric motor moving coil galvanometer, electromagnetic induction, generator, transformers etc.
EVALUATION:
Define electromagnetic field and give the relationship between force, charge, electric and magnetic field and motion.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Plot magnetic field around magnets and magnetic substances.
ii. Describe the working principle of electric bell and the telephone earpiece.
iii. State the relation between magnetic force and the motion of a charge in a magnetic field.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: MAGNETIC FIELD
The area around a magnet in which it can attract or repel objects or in which a magnetic force can be detected is called magnetic field of the magnet.
Magnet lines of force:
These are many lines along which a free- North pole would tend to move if placed in the field. A line of force may also be considered as a line such that the tangent to it at any point gives the direction of the field at that point.
PATTERNS OF MAGNETIC FIELD
Magnetic field around a straight conductor carrying current.
The direction of the field depends on the direction of flow of current. Such a direction can always be obtained by applying the R-H Grip or CLENCHED FIST RULE. this states that if the straight wire is grasped with the right hand so that the thumb points in the direction of the current, then the direction in which the fingers are curled indicates the direction of the magnetic field.
MAGNETIC CORKSCREW RULE B
This rule states that if a right handed corkscrew is turned so that its tip travels along the direction of current, the direction of rotation of the corkscrew gives the direction of the magnetic field or magnetic line of force.
MAGNETIC FIELD AROUND A SOLEMOID
A solemoid is a long cyclindrical coil of wire whose turns are usually wound close together.
METHODS OF MAKING MAGNETS
1. Electrical method: The magnetic effect of current on magnets . polarity of magnet depends on direction of flow of current in solemoid. If current flows anti clockwise in coil when one side is viewed, then the specimen at that end has a N pole.
2. By constant method:
i. Single touch
ii. Double touch
At the end of each repeated stroking, each end of the specimen will be found to have opposite polarity to that of the stroking pole.
3. Demagnetization:
i. Slant method (a.c current)
ii. Mechanical method : pointing in E-w direction.
iii. Heating method
MAGNETIC PROPERTIES OF IRON AND STEEL
1. Iron is easily magnetized than steel but also loses its magnetism more quickly than steel.
2. Iron produces a stronger magnet than steel, thus steel
i. Steel is used for making p.magnets
ii. Iron is used for making + magnets
iii. Iron nails are often used for lab experiments on magnetization and demagnetization.
ELECTROMAGNETS AND ITS APPLICATIONS OF ELECTRIC FIELD
They are soft iron core in a current carrying solemoid.
Electromagnets are used for:
i. Producing intense magnetic fields(e.g. electric motors)
ii. For lifting and transporting heavy pieces of iron and steel.
iii. Used in separating iron from mix of non magnetic substances.
iv. In construction of electromagnetic devices.
THE ELECTRIC BELL
The instrument works on a "make and break" device. The making of the contact between spring and screw at c causes the clapper to move and strike gong and consequently breaking of the contact causes the clapper to leave the gong.
TELEPHONE EAR-PIECE
The variation of electric current due to changes in the speech energy causes a variation in the magnetism of the electromagnet. Vibration of the diaphragm creates sound waves of the same frequency in a speech current.
THE EARTH'S MAGNETIC FIELD
The earth acts like a magnet. The N-geographical pole has south magnetic polarity since it attracts North pole of a magnet, the south geographical pole has North magnetic polarity.
Angle of Declination/variation:
This is the angle between North and geographical north direction at the place concerned.
Angle of Dip/inclination:
Is the angle between the direction of the earth's resultant magnetic field at the horizontal.
The angle of dip at points - component of Earth's magnetic field
Is given by tanθ = V
H
Patterns of Bar magnet in Earth's field
At the neutral point, the magnet force due to the magnetic field is equal and opposite to the force due to the magnetic force due to the field of the bar magnet.
Magnetic force on a charge moving in a magnetic field
Magnetic field exerts a force on a charge moving in the field. If electrons want to move with average velocity v, and the wires lie in the magnetic field b, with force F, on each electron, then
F = Bevsinθ
But moving q
F = qvBsinθ, becomes
F = vvB i.e. v cross B
If angle between B and v is 90∘, then sinθ = 1
EVALUATION:
Distinguish between angle of dip and declination.
ASSIGNMENT:
A beam of ions passes undeflated through two parallel horizontal plates 2.5m apart having a p.d of 2500v when the perpendicular magnetic field is 2 x 10-2 T, and in the magnetic field along radius 3m. calculate the charge per mass ratio of the ions.
SUB-TOPIC: CALCULATIONS ON ELECTRIC FIELD AND E.M.F
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Solve simple problems on force on a material in a magnetic field.
ii. Show a relationship between electric and magnetic field, charge and motion.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: CLASS WORK
1. A charge of 1.6 x 10-1C enters a magnetic field of 10T with and of 2.5 x 102m/s. calculate the angle between the magnetic field and the direction of the motion at a force of 10 x 10-12N.
2. Obtain the magnetic force on an electron of charge q = 1.6 x 10-19C in a field of flux 20 tesla, with a speed of 2 x 107m/s in the direction (a) 90∘ (b) 60∘ (c) parallel to the magnetic field.
ELECTROMAGNETIC FIELD
In a field representing the joint interaction of electric and magnetic forces. It exerts a force on a charged particle; the force in a charge " moving with velocity " is given by
F= q (E + V + B)
Where f = force that occurs in both fields
q= charge, E = Electric field
B= magnetic field in tesla, or weber/m2
It can occur in an electric motor moving coil galvanometer, electromagnetic induction, generator, transformers etc.
EVALUATION:
Define electromagnetic field and give the relationship between force, charge, electric and magnetic field and motion.
WEEK 9
TOPIC: TYPES OF RADIATIONS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Distinguish between electromagnetic waves and mechanical waves.
ii. Compare and contrast in terms of frequency, wavelength, six radiations in electromagnetic spectrum.
iii. State some uses of electromagnetic waves.
iv. Apply the formula v= f⋋ to solve simple problems relating to electromagnetic waves.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTROMAGNETIC WAVES
Mechanical waves and electromagnetic waves.
Electromagnetic waves are those waves that do not require a material medium for their propagation. They arise from the vibration of electric and magnetic field. The combination of electric and magnetic field waves is called ELECTROMAGNETIC WAVES.
Mechanical waves are waves that require a material medium for their propagation E.m travel at a sped less than that of light.
Mechanical waves may be transverse or longitudinal, but e.m is always transverse.
Equation:
V = f⋋ or
C= f⋋
e.g. estimate the wavelength of a radio station (Ray power) whose frequency is transmission is 100.5MHZ. [Take c = 3 x 108m/s]
EVALUATION:
Distinguish between mechanical and electromagnetic waves.
ASSIGNMENT:
List out 5 uses of electromagnetic waves.
SUB-TOPIC: USES OF E.M WAVE
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. List out some types of radiation
ii. Outline the various uses of electromagnetic waves.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: USES OF ELECTROMAGNETIC WAVES
Uses of e.m waves are numerous since there are different kinds. Hence the use of any one type is paramount important for the uses of electromagnetic waves.
1. Gamma waves: they are waves of ⋋10-8cm. with very great penetration. About x 10 of x-rays. They are used in medical field to sterilize supplies. They are also used to treat cancer and tumours.
2. Radio and TV waves: these are e.m waves with ⋋ - 1mm a few km. they are used in radio and television transmissions
- Radar detection around 1cm(⋋)
3. X-rays: (⋋ between 10-10 - 10-7cm). since they are blocked/stopped by some solids such as isomers and metals, they are equally used in hospitals to make pictures of bones and radiographs of internal body organs.
- Used in detecting flaws in metal industry.
- Used to study couptal structures.
- In air ports used as scanners to detect
EVALUATION:
List 5 uses of e-m waves.
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. Distinguish between electromagnetic waves and mechanical waves.
ii. Compare and contrast in terms of frequency, wavelength, six radiations in electromagnetic spectrum.
iii. State some uses of electromagnetic waves.
iv. Apply the formula v= f⋋ to solve simple problems relating to electromagnetic waves.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: ELECTROMAGNETIC WAVES
Mechanical waves and electromagnetic waves.
Electromagnetic waves are those waves that do not require a material medium for their propagation. They arise from the vibration of electric and magnetic field. The combination of electric and magnetic field waves is called ELECTROMAGNETIC WAVES.
Mechanical waves are waves that require a material medium for their propagation E.m travel at a sped less than that of light.
Mechanical waves may be transverse or longitudinal, but e.m is always transverse.
Equation:
V = f⋋ or
C= f⋋
e.g. estimate the wavelength of a radio station (Ray power) whose frequency is transmission is 100.5MHZ. [Take c = 3 x 108m/s]
EVALUATION:
Distinguish between mechanical and electromagnetic waves.
ASSIGNMENT:
List out 5 uses of electromagnetic waves.
SUB-TOPIC: USES OF E.M WAVE
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
i. List out some types of radiation
ii. Outline the various uses of electromagnetic waves.
REFERENCE: NEW SCHOOL PHYSICS FOR SS by ANYAKOHA, AFP 2006.
CONTENT: USES OF ELECTROMAGNETIC WAVES
Uses of e.m waves are numerous since there are different kinds. Hence the use of any one type is paramount important for the uses of electromagnetic waves.
1. Gamma waves: they are waves of ⋋10-8cm. with very great penetration. About x 10 of x-rays. They are used in medical field to sterilize supplies. They are also used to treat cancer and tumours.
2. Radio and TV waves: these are e.m waves with ⋋ - 1mm a few km. they are used in radio and television transmissions
- Radar detection around 1cm(⋋)
3. X-rays: (⋋ between 10-10 - 10-7cm). since they are blocked/stopped by some solids such as isomers and metals, they are equally used in hospitals to make pictures of bones and radiographs of internal body organs.
- Used in detecting flaws in metal industry.
- Used to study couptal structures.
- In air ports used as scanners to detect
EVALUATION:
List 5 uses of e-m waves.