Veerendra

Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism is part of Plus Two Physics Notes. Here we have given Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism.

Kerala Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism

Introduction; Oersted Experiment
The magnetic effect of current was discovered by Danish Physicist Hans Christians Oersted. He noticed that a current in a straight wire makes a deflection in a magnetic needle.

The deflection increases on increasing current. He also found that reversing the direction of current reverses direction of needle. Oersted concluded that current produces a magnetic field around it.

Magnetic Force
1. Sources and fields:
The static charge is the source of electric field. The source of magnetic field is current or moving charge. Both the electric and magnetic fields are vector fields and both obeys superposition principle.

2. Lorentz Force:
The force experienced by moving charge in electric and magnetic field is called Lorentz force. The Lorentz force experienced by charge ‘q’ moving with velocity ‘v’, is given by

= F_electric + F_magnetic
The features of Lorentz Force:

1. The Lorentz force on positive charge is opposite to that on negative charge because it depends on charge ‘q’.
2. The direction of Lorentz force is perpendicular to velocity and magnetic field. Its direction is given by screw rule or right hand rule.
3. Only moving charge experiences magnetic force. For static charge (v = 0), magnetic force is zero.

Note:

1. A charge particle moving parallel or antiparallel to magnetic field will not experience magnetic force and moves undeviated.
2. The work done by magnetic force is zero. Because magnetic force is always perpendicular to direction of velocity.
3. A charged particle entering perpendicular magnetic field (θ = 90°) will make a circular path.
4. The unit of B is Tesla.

3. Magnetic force on current-carrying conductor:
Consider a rod of uniform cross section ‘A’ and length ‘e’. Let ‘n’ be the number of electrons per unit volume (number density). ‘v_d’ be the drift velocity of electrons for steady current ‘I’.
Total number of electrons in the entire volume of rod = nAl
Charge of total electrons = nA l.e
‘e’ is the charge of a single electron.
The Lorentz force on electrons,

(I = neAV_d)

Motion In Magnetic Field
Case I:
The charged particle enters perpendicular to magnetic field.(\(\overrightarrow{\mathrm{V}}\) is perpendicular to \(\overrightarrow{\mathrm{B}}\))
When charged particle moves perpendicular to magnetic field, it experiences a magnetic force of magnitude, qVB and the direction of the force is perpendicular to both \(\overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{V}}\). This perpendicular magnetic field act as centripetal force and charged particle follows a circular path.

Mathematical explanation:
Let a charge ‘q’ enters into a perpendicular magnetic field B with velocity V. Let r be the radius of circular path. The centripetal force for charged particle is provided by magnetic force.

Thus radius of circle described by charged particle depends on momentum, charge and magnetic field. If ω is the angular frequency
ω = \(\frac{v}{r}\)
Thus from (1) we get

The frequency ν = \(\frac{q B}{2 \pi m}\)
Thus frequency of revolution of charge is independent of velocity (and hence energy)
The time period T = \(\frac{2 \pi \mathrm{m}}{\mathrm{qB}}\)
(ν = \(\frac{1}{T}\)).

Case II:
The charged particles enters at an angle ‘θ’ with magnetic field.
Since the charged particle enters at an angle ‘θ’ with magnetic field, its velocity will have two components; a component parallel to magnetic field, V_॥ (Vcosθ) and a component perpendicular to the magnetic field, V_⊥(Vsinθ).

The parallel component of velocity remains unaffected by magnetic field and it causes charged particle to move along the field.

The perpendicular component makes the particle to move in circular path. The effect of linear and circular movement produce helical motion.

Pitch and Helix: The distance moved along magnetic field in one rotation is called pitch ‘P’

The radius of circular path of motion is called helix.

Motion In Combined Electric And Magnetic Fields
1. Velocity selector:
A transverse electric and magnetic field act as velocity selector. By adjusting value of E and B, it is possible to select charges of particular velocity out of a beam containing charges of different speed.

Explanation:
Consider two mutually perpendicular electric and magnetic fields in a region. A charged particle moving in this region, will experience electric and magnetic force. If net force on charge is zero, then it will move undeflected. The mathematical condition for this undeviation is

The charges with this velocity pass undeflected through the region of crossed fields.

2. Cyclotron
Uses: It is a device used to accelerate particles to high energy.
Principles: Cyclotron is based on two facts

1. An electric field can accelerate a charged particle.
2. A perpendicular magnetic field gives the ion a circular path.

Constructional Details:
Cyclotron consists of two semicircular dees D₁ and D₂, enclosed in a chamber C. This chamber is placed in between two magnets. An alternating voltage is applied in between D₁ and D₂. An ion is kept in a vacuum chamber.

Working:
At certain instant, let D₁ be positive and D₂ be negative. Ion (+ve) will be accelerated towards D₂ and describes a semicircular path (inside it). When the particle reaches the gap, D₁ becomes negative and D₂ becomes positive. So ion is accelerated towards D₁ and undergoes a circular motion with larger radius. This process repeats again and again.

Thus ion comes near the edge of the dee with high K.E. This ion can be directed towards the target by a deflecting plate.

Mathematical expression:
Let V be the velocity of ion, q the charge of the ion and B the magnetic flux density. If the ion moves along a semicircular path of radius ‘r’, then we can write

[Since θ =90°, B is perpendicular to v]
or v = \(\frac{q B r}{m}\) _____(1)
Time taken by the ion to complete a semicircular path.

Eq. (2) shows that time is independent of radius and velocity.

Resonance frequency (cyclotron frequency):
The condition for resonance is half the period of the accelerating potential of the oscillator should be ‘t’. (i.e.,T/2 = t or T = 2t). Hence period of AC
T = 2t

K.E of positive ion

Thus the kinetic energy that can be gained depends on mass of particle charge of particle, magnetic field and radius of cyclotron.
Limitations:

1. As the particle gains extremely high velocit, the mass of particle will be changed from its constant value. This will affect the normal working of cyclotron as frequency depends of mass of particle.
2. Very small particles like electron can not be accelerated using cyclotron. This is because as the mass of electron is very small the cyclotron frequency required becomes extremely high which is practically difficult.
3. Neutron can’t be accelerated

Magnetic Field Due To Current Element; Biot Savart Law

The magnetic field at any point due to an element of current carrying conductor is

1. Directly proportional to the strength of the current (I)
2. Directly proportional to the length of the element (dl)
3. Directly proportional to the sine of the angle (θ) between the element and the line joining the midpoint of the element to the point.
4. Inversely proportional to the square of the distance of the point from the element

The direction of magnetic field is perpendicularto the plane containing d/and rand is given by right hand screw rule.
In the above expression \(\frac{\mu_{0}}{4 \pi}\) is the constant of proportionality and µ₀ is called the permeability of vacuum. Its value is 4π × 10^-7 TmA^-1.
Note: A magnetic field acting perpendicularly into the plane of the paper is represented by the symbol ⊗ and a magnetic field acting perpendicularly out of the paper is represented by the symbol ⵙ.

Comparison between Biot-Savart Law and Coulomb’s law
Similarities:

1. The two laws are based on inverse square of distance and hence they are long range.
2. Both electrostatic and magnetic fields obey superposition principle.
3. The source of magnetic field is linear; (the current element \(\overrightarrow{\mathrm{ldl}}\)). The source of electrostatic force is also linear; (the electric charge).

Differences:

Magnetic Field On The Axis Of A Circular Current Loop

Consider a circular loop of radius ‘a’ and carrying current ‘I’. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A. The magnetic field at ‘p’ due to this small element dl,

\(\mathrm{dB}=\frac{\mu_{0} \mathrm{Idl}}{4 \pi \mathrm{x}^{2}}\) _____(1)
[since sin 90° – 1]
The dB can be resolved into dB cosΦ (along Py) and dB sinΦ (along Px).
Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get.
dB sinΦ (along px) and dB cosΦ (along py¹)
dB cosΦ components cancel each other, because they are in opposite direction. So only dB sinΦ components are found at P, so total filed at P is

but from ∆AOP we get, sinΦ = a/x
∴ We get

Point at the centre of the loop: When the point is at the centre of the loop, (r = 0)
Then,

1. Magnetic field at the centre of loop:
The magnetic field at a distance x from centre of loop is given by

The direction of magnetic field due to current carrying circular loop is given by right hand thumb rule.

Thumb Rule: Curl of palm of right hand around circular coil with fingers pointing in the direction of current. Then extended thumb gives the direction of magnetic field.

Note:

1. An anticlockwise current gives a magnetic field out of the coil and a clockwise current gives a magnetic field into the coil.
2. The current carrying loop is equivalent to magnetic dipole of dipole moment m = IA

Ampere’s Circuital Law
According to ampere’s law the line integral of magnetic field along any closed path is equal to µ₀ times the current passing through the surface.

Applications Of Ampere’s Circuital Law
1. Long straight conductor:
Consider a long straight conductor carrying T ampere current. To find magnetic field at ‘P’, we construct a circle of radius r (passing through P).

According to Ampere’s circuital law we can write

[B and dl are parallel]
B∫dl = µ₀I
B2πr = µ₀I
B = \(\frac{\mu_{0} I}{2 \pi r}\)

2. Magnetic field due to long solenoid:

Consider a solenoid having radius T. Let ‘n’ be the number of turns per unit length and I be the current flowing through it.
In order to find the magnetic field (inside the solenoid) consider an Amperian loop PQRS. Let V ‘ be the length and ‘b’ the breadth
Applying Amperes law, we can write

Substituting the above values in eq (1),we get
Bl = µ₀ l_enc ____(2).
But l_enc = n l I
where ‘nl ’ is the total number of turns that carries current I (inside the loop PQRS)
∴ eq (2) can be written as
Bl = µ₀ nIl
B = µ₀nI
If core of solenoid is filled with a medium of relative permittivity µ_r. then
B = µ₀µ_rnl

3. The toroid:
Consider a toroid of average radius ‘r’. Let ‘n’ be the number of turns per unit length. Let I be the current flowing through the toroid. In order to find magnetic field inside the toroid, an camperian loop of radius ‘r’ is considered.

Applying Amperes law to the loop, we can write

Where ‘n2πr ‘ is the total number of turns of the solenoid that carries current I (inside the Amperian loop) Integrating the eq(1) we get
B 2πr = µ₀2πrI
B = µ₀n I
If the core of the solenoid is filled with a medium of relative permeability µ_r then the above equation is modified as

Note: The magnetic field due to toroid is same as that due to solenoid.

Force Between Two Parallel Currents, The Ampere Force Between Two Parallel Conductors

P and Q are two infinitely long conductors placed parallel to each other and separated by a distance r, Let the current through P and Q be l₁ and l₂ respectively.
Magnetic field at a distance ‘r’ from P is

Conductor ‘Q’ is placed in this magnetic field.
If l₂ is the length of the conductor ‘Q’, the Lorentz force on ‘Q’ is

∴ Force per unit length can be written as,

Where f = \(\frac{F}{\ell_{2}}\)
Note:

1. When currents are in the same direction, the force is attractive
2. If the currents are in the opposite direction, the force is repulsive.

Definition of ampere:
An ampere is defined as that constant current which if maintained in two straight parallel conductors of infinite lengths placed one meter apart in vacuum will produce between a force of 2 × 10^-7 Newton per meter length.

Force On A Current Loop, Magnetic Dipole
1. Torque on a rectangular current loop in uniform magnetic filed:

Considers rectangular coil PQRS of N turns which is suspended in a magnetic field, so that it can rotate (about yy 1). Let ‘l’ be the length (PQ) and ‘b’ be the breadth (QR).

When a current l flows in the coil, each side produces a force. The forces on the QR and PS will not produce torque. But the forces on PQ and RS will produce a Torque.
Which can be written as
τ = Force × ⊥ distance _______(1)
But, force = BlI ______(2)
[since θ = 90° ]
And from ∆QTR , we get
⊥ distance (QT) = b sin θ ______(3)
Substituting the vales of eq (2) and eq (3) in eq(1) we get
τ = BIl b sin θ
= BIA sin θ [since lb = A (area)]
τ = IAB sin θ
τ = mB sin θ [since m = IA]

If there are N turns in the coil, then
τ = NIAB sin θ

2. Circular Current loop as a magnetic dipole:
Current loop of any shape act as magnetic dipole.
Current loop acts as magnetic dipole:
The magnetic field due to circular loop of radius R carrying current I at a distance ‘x’ from the centre of loop (on the axis of loop) is given by,

The magnetic field at large distance (x>>R) on axis of loop is

Dividing and multiplying by π

Comparison of magnetic dipole and electric dipole:
The equation (1) is similar to electric field due to electric dipole at a distance ‘x’ from the centre of dipole on its axial line.

Comparing eq(1) and (2), we get 1

m → P
B → E
From this comparison it is clear that a circular current loop acts as a magnetic dipole.

3. The magnetic dipole moment of a revolving electron:
According to Bohr’s model of atom, electrons are revolving around nucleus in its orbit. The electron revolving in its orbit can be considered as circular current loop.

Consider an electron of charge e, revolving around nucleus of charge +ze as shown in figure. The uniform, circular motion of electron constitute current ‘I’. If T is the period of revolution e
I = \(\frac{e}{T}\) _____(1)
If r is the radius of orbit and V s the orbital speed then
T = \(\frac{2 \pi r}{v}\)
Substituting this in (1), we get
I = \(\frac{e v}{2 \pi r}\)
The magnetic moment associated orbiting electron is denoted by µ₁

A = πr², area of orbit
Dividing and multiplying by m_e (Mass of electron)

Applying Quantum Theory, Bohr has proposed that angular momentum of electron can take only discrete values given by,
l = \(\frac{\mathrm{nh}}{2 \pi}\) (Bohr’s quantization condition where n = 1, 2, 3, ……..etc) where h is Plank’s constant. Thus

The orbital magnetic moment of electron is given by

Bohr Magneton: We get the minimum value of magnetic moment, when n = 1 ie

(when n = 1)
Its value is 9.27 × 10^-24 Am². This is called Bohr magneton.
Gyromagnetic Ratio:
The orbital magnetic moment of electron is related to orbital angular momentum ‘l’ as

The ratio of orbital magnetic moment to orbital angular momentum is constant. This constant is called gyromagnetic ratio. Its value is 8.8 × 10¹⁰ c/kg for an electron.

The Moving Coil Galvanometer
It is an instrument used to measure small current.
Principle: A conductor carrying current when placed in a magnetic field experiences a force, (given by Fleming’s left hand rule).
Construction:

A moving coil galvanometer consists of rectangular coil of wire having area ‘A’ and number of turns ‘n’ which is wound on metallic frame and is placed between two magnets. The magnets are concave in shape, which produces radial field.

Working: Let ‘l’ be the current flowing the coil, Then the torque acting on the coil.
τ = NIAB Where A is the area of coil and B is the magnetic field.
This torque produces a rotation on coil, thus fiber is twisted and angle (Φ). Due to this twisting a restoring torque (τ = KΦ) is produced in spring.
Under equilibrium, we can write
Torque on the coil = restoring torque on the spring
or NIAB = kΦ
or Φ = (\(\frac{\mathrm{BAN}}{\mathrm{K}}\))I
The quantity inside the bracket is constant for a galvanometer.
Φ α I
The above equation shows that the deflection depends on current passing through galvanometer.

1. Ammeter and voltmeter:
For measuring large current, the galvanometer can be converted in to ammeter and voltmeter.
Ammeter:
Ammeter is an instrument used to measure current in the circuit.

A galvanometer can be converted into an ammeter by a low resistance (shunt) connected parallel to it.

Theory:
Let G be the resistance of the galvanometer, giving full deflection fora current Ig.

To convert it into an ammeter, a suitable shunt resistance ‘S’ is connected in parallel. In this arrangement Ig current flows through Galvanometer and remaining (I – Ig) current flows through shunt resistance.
Since G and S are parallel
P.d Across G = p.d across
Ig × G = (I – Ig)S

Connecting this shunt resistance across galvanometer we can convert a galvanometer into ammeter.

2. Conversion of galvanometer into voltmeter:
To convert a galvanometer into a voltmeter, a high resistance is connected in series with it.

Theory:
Let Ig be the current flowing through the galvanometer of resistance G. Let R be the high resistance co connected in series with G.

From figure we can write
V = IgR + IgG
V – IgG = IgR

Using this resistance we can covert galvanometer in to voltmeter.

Current sensitivity:
The current sensitivity of galvanometer is the deflection produced by unit current.

The current sensitivity can be increased by increasing number of turns.

The voltage sensitivity:
The voltage sensitivity of galvanometeris the deflection produced by unit voltage.

The increase in number of turns will not change voltage sensitivity.
When number of turns double (N → 2N), the resistance of the wire will be double (ie. R → 2R). Hence the voltage sensitivity does not change.

We hope the Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism help you. If you have any query regarding Plus Two Physics Notes Chapter 4 Moving Charges and Magnetism, drop a comment below and we will get back to you at the earliest.

Plus Two Physics Notes Chapter 3 Current Electricity is part of Plus Two Physics Notes. Here we have given Plus Two Physics Notes Chapter 3 Current Electricity.

Kerala Plus Two Physics Notes Chapter 3 Current Electricity

Introduction
In the present chapter, we shall study some of the basic laws concerning steady electric currents.

Electric Current
Rate of flow of electric charge is called electric current. or

Electric Currents In Conductors

• Conductors: Free electrons are found in conductors. The electric current in conductors is due to the flow of electrons.
• Electrolytes: The current in electrolytes is due to the flow of ions.
• Semiconductor: The current in semiconductors is due to the flow of both holes and electrons.

Ohm’s Law
At constant temperature, the current through a conductor is directly proportional to the potential difference between its ends.
V α I (or)

where R is constant, called resistance of materials.

1. Resistance of a material:
Factors Affecting Resistance of Resistor:
For a given material resistance is directly proportional to the length and inversely proportional to the area of cross-section.
R ∝ \(\frac{\mathrm{L}}{\mathrm{A}}\)

where ρ is the constant of proportionality and is called resistivity of material.

Resistivity (coefficient of specific resistance) of a Material:
The resistance per unit length for unit area of the cross-section will be a constant and this constant is known as the resistivity of the material. The resistivity or coefficient of specific resistance is defined as the resistance offered by a resistor of unit length and unit area of cross-section.

Resistivity is a scalar quantity and its unit is Ω-m.

Conductance and conductivity:
The reciprocal of resistance is called conductance and the reciprocal of resistivity is called conductivity. The SI unit of conductance is seimen and that for conductivity is seimen per meter. The unit of conductance can also be expressed as Ω^-1.

Current density: Current per unit area is called current density
current density j = \(\frac{I}{A}\)
Vector form

The mathematical expression of Ohm’s law in terms of j and E:
Considers conductor of length i. Let V’ be the potential difference between the two ends of a conductor.
According to ohms law
We know V= IR

This potential difference produces an electric field E in the conductor. The p.d. across the conductor also can be written as
V = El _____(2)
Comparing (1) and (2), we get
El = jρl
E = jρ
The above relation can be written in vector form as

where σ is called conductivity of the material.

Drift Of Electrons And The Origin Of Resistivity
Random thermal motion of electrons in a metal:
Every metal has a large number of free electrons. Which are in a state of random motion within the conductor. The average thermal speed of the free electrons in random motion is of the order of 10⁵m/s.

Does random thermal motion produce any current? The directions of thermal motion are so randomly distributed. Hence the average thermal velocity of the electrons is zero. Hence current due to thermal motion is zero.

(a) Drift Velocity (vd):
The average velocity acquired by an electron under the applied electric field is called drift velocity.

Explanation: When a voltage is applied across a conductor, an electric filed is developed. Due to this electric field electrons are accelerated. But while moving they collide with atoms, lose their energy and are slowed down. This acceleration and collision are repeated through the motion. Hence electrons move with a constant average velocity. This constant average velocity is called drift velocity.

(b) Relaxation time (τ):
Relaxation time is the average of the time between two successive collisions of the free electrons with atoms.

(c) Expression for drift velocity:
Let V be the potential difference across the ends of a conductor. This potential difference makes an electric field E. Under the influence of electric field E, each free electron experiences a Coulomb force.
F = -eE
or ma = -eE
a = \(\frac{-e E}{m}\) _____(1)
Due to this acceleration, the free electron acquires an additional velocity. A metal contains a large number of electrons.
For first electron, additional velocity acquired in a time τ,
v₁ = u₁ + aτ₁
where u₁ is the thermal velocity and τ is the relaxation time.
Similarly the net velocity of second, third,……electron

v₂ = u₂ + aτ₂
v₃ = u₃ + aτ₃
v_n = u_n + aτ_n
∴ Average velocity of all the ‘n’ electrons will be

V_av = 0 + aτ (∴ average thermal velocity of electron is zero)
where τ = \(\frac{\tau_{1}+\tau_{2}+\ldots \ldots \ldots+\tau_{n}}{n}\)
where V_av is the average velocity of electron under an external field. This average velocity is called drift velocity.
ie. drift velocity V_d = aτ _____(2)

(d) Relation between electric current and drift speed:
Consider a conductor of cross-sectional area A. Let n be the number of electrons per unit volume. When a voltage is applied across a conductor, an electric filed is developed. Let v_d be the drift velocity of electron due to this field.
Total volume passed in unit time = Av_d
Total number of electrons in this volume = Av_dn
Total charge flowing in unit time=Av_dne
But charge flowing per unit time is called current I ie. current, I = Av_dne
I = neAv_d
Now current density J can be written as
J = nev_d (J= I/A)
Deduction of Ohm’s law: (Vector form)
We know current density
J = nv_de

1. Mobility: Mobility is defined as the magnitude of the drift velocity per unit electric field.

Limitations Of Ohm’s Law
Certain materials do not obey Ohm’s law. The deviations of Ohm’s law are of the following types.
1. V stops to be proportional to I.

Metal shows this type behavior. When current through metal becomes large, more heat is produced. Hence resistance of metal increases. Due to increase in resistance the V-I graph becomes nonlinear. This nonlinear variation is shown by solid line in the above graph.

2. Diode shows this type behavior. We get different values of current for same negative and positive voltages.

3. This type behavior is shown by materials like GaAs. there is more than one value of V for the same current I.

Note: The materials which do not obey ohms law are mainly used in electronics.

Resistivity Of Various Materials
The materials are classified as conductors, semiconductors, and insulators according to their resistivities. Commercially produced resistors are of two types.

1. Wire bound resistors
2. Carbon resistors

1. Wire bound resistor:
Wire wound resistors are made by winding the wires of an alloy.
Eg: Manganin, Constantan, Nichrome.

2. Carbon resistors:
Resistors in the higher range are made mostly from carbon. Carbon resistors are compact. Carbon resistors are small in size. Hence their values are given using a colourcode.

(i) Colourcode of resistors:
The resistance value of commercially available resistors are usually indicated by certain standard colour coding.

The resistors have a set of coloured rings on it. Their significance is indicated in the table The first two bands from the end indicated the first two significant digits and the third band indicates the decimal multiplier. The last metallic band indicates the tolerance.
Value of colours:

Illustration:

The colour code indicated in the given sample is Red, Red, Red with a silver ring at the right end. Then the value of given resistance is 22 × 10² ±10%.

Temperature Dependence Of Resistivity:
The resistivity of a material is found to be dependent on the temperature. The resistivity of a metallic conductor is approximately given by,
ρ_T = ρ_o[1 + α(T – T_o)]
where ρ_T is the resistivity at a temperature T and ρ_o is the resistivity at temperature T_o. α is called the temperature coefficient of resistivity.
Variation of resistivity in metals:

The temperature coefficient (α) of metal is positive. Which means the resistivity of metal increases with temperature. The variation of resistivity of copper is as shown in the above figure.
Eg: Silver, copper, nichrome, etc.
Variation of resistivity in a semiconductor:

The temperature coefficient (α) of semiconductors is negative. This means that resistivity decreases with an increase in temperature. The variation of resistivity with temperature for a semiconductor is shown in the above figure.
Eg: Carbon, Germanium, silicon
Variation of resistivity in standard resistors:

standard resistors, the variation of resistivity will be very little with temperatures. The variation of resistivity with temperature for standard resistors is shown above.
Eg: Manganin and constantan.
Explanation for the variation of resistivity:
The resistivity of a material is given by

The above equation shows that resistivity depends inversely on number density and relaxation time τ.
Metals:
Number density in metal does not change with temperature. But the average speed of electrons increases. Hence the frequency of collision increases. The increase in the frequency of collision decreases the relaxation time τ. Hence the resistivity of metal increases with temperature.

Insulators and semiconductors:
For insulators and semiconductors, the number density n increases with temperature. Hence resistivity decreases with temperature.

Electrical Energy, Power

Consider two points A and B in a conductor. Let V_A and V_B be the potentials at A and B respectively. The potential At A is greater than that B and difference in potential is V.

If ∆Q charge flows from A to B in time ∆t. The potential energy of charge will be decreased. The decrease in potential energy due to charge flow from A to B,
= V_A∆Q – V_B∆Q
= (V_A – V_B)∆Q
= V∆Q (V_A – V_B = V)
This decrease in PE appeared as KE of flowing charges. But we know, the kinetic energy of charge carriers do not increase due to the collisions with atoms. During collisions, the kinetic energy gained by the charge carriers is shared with the atoms.

Hence the atoms vibrate more vigorously ie. The conductor heats up. According to the conservation of energy, heat developed in between A and B in time ∆t
∆H = decrease in potential energy
∆H = V∆Q
∆H = VI∆t (∵ ∆Q=I∆t)
\(\frac{\Delta \mathrm{H}}{\Delta \mathrm{t}}\) = VI
Rate of work done is power, ie

using Ohm’s law V = IR

It is this power which heats up the conductor.
Power transmission:
The electric power from the electric power station is transmitted with high voltage. When voltage increases, the current decreases. Hence heat loss decreases very much.

Combination Of Resistors – Series And Parallel Combination
Resistors in series:
Consider three resistors R₁, R₂ and R₃ connected in series and a pd of V is applied across it.

In the circuit shown above the rate of flow of charge through each resistor will be same i.e. in series combination current through each resistor will be the same. However, the pd across each resistor are different and can be obtained using ohms law.
pd across the first resistor V₁ = I R₁
pd across the second resistor V₂, = I R₂
pd across the third resistor V₃ = I R₃
If V is the effective potential drop and R is the effective resistance then effective pd across the combination is V = IR
Total pd across the combination = the sum pd across each resistor, V = V₁ + V₂ + V₃
Substituting the values of pds we get IR = IR₁ + IR₂ + IR₃
Eliminating I from all the terms on both sides we get

Thus the effective resistance of series combination of a number of resistors is equal to the sum of resistances of individual resistors.

1. Resistors in parallel:
Consider three resistors R₁, R₂ and R₃ connected in parallel across a pd of V volt. Since all the resistors are connected across same terminals, pd across all the resistors are equal.

As the value of resistors are different current will be different in each resistor and is given by Ohm’s law
Current through the first resistor
I₁ = \(\frac{V}{R_{1}}\)
Current through the second resistor
I₂ = \(\frac{V}{R_{2}}\)
Current through the third resistor
I₃ = \(\frac{V}{R_{3}}\)
Total current through the combination is
I = \(\frac{V}{R}\), where R is the effective resistance of parallel combination.
Total current through the combination = the sum of current through each resistor
I = I₁ + I₂ + I₃
Substituting the values of current we get

Eliminating V from all terms on both sides of the equations, we get

Thus in parallel combination reciprocal of the effective resistance is equal to the sum of reciprocal of individual resistances. The effective resistance in a parallel combination will be smaller than the value of smallest resistance.

Cells, Emf, Internal Resistance
Electrolytic cell:
An electrolytic cell is a simple device to maintain a steady current in an electric circuit. A cell has two electrodes. They are immersed in an electrolytic solution.

E.M.F:
E.M.F. is the potential difference between the positive and negative electrodes in an open circuit. ie. when no current is flowing through the cell.

Voltage:
Voltage is the potential difference between the positive and negative electrodes when current is flowing through it.

Internal resistance of cell:
Electrolyte offers a finite resistance to the current flow. This resistance is called internal resistance (r).

Relation between ε, V and internal drop:
Consider a circuit in which cell of emfs is connected to resistance R. Let r be the internal resistance and I be the current following the circuit.

According to ohms law,
current flowing through the circuit

V is the voltage across the resistor called terminal voltage. Ir is potential difference across internal resistance called internal drop.

Cells In Series And Parallel

Consider two cells in series. Let ε₁, r₁ be the emf and internal resistance of first cell. Similarly ε₂, r₂ be the emf and internal resistance of the second cell. Let I will be the current in this circuit.
From the figure, the P.d between A and B
V_A – V_B = ε₁ – Ir₁ _____(1)
Similarly P.d between B and C
V_B – V_C = ε₂ – Ir₂ ______(2)
Hence, P.d between the terminals A and C
V_AC = V_A – V_C = V_A – V_B + V_B – V_C
V_AC = [V_A – V_B] + [V_B – V_C]
when we substitute eqn. (1) and (2) in the above equation.
V_AC = ε₁ – Ir₁ + ε₂ – Ir₂ V_AC = (ε₁ – ε₂) – I(r₁ + r₂)
V_AC = ε_eq – Ir_eq
where ε_eq = ε₁ + ε₂, and r_eq = r<sub1 + r₂
The rule of series combination:

1. The equivalent emf of a series combination of n cells is the sum of their individual emf.
2. The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances.

Cells in parallel:

Consider two cells connected in parallel as shown in figure. ε₁, r₁ be the emf and internal resistance of first cell and ε₂, r₂ be the emf and internal resistance of second cell. Let I₁ and I₂ be the current leaving the positive electrodes of the cells.
Total current flowing from the cells is the sum of I₁ and I₂.
ie. I = I₁ + I₂ ______(1)
Let V_B1 and V_B2 be the potential at B₁ and B₂ respectively. Considering the first cell, P.d between B₁ and B₂.

Considering the second cell, P.d between B₁ and B₂

Substituting the values I₁ and I₂ in eq.(1), we get


If we replace the combination by a single cell between B₁ and B₂, of emf and ε_eq and internal resistance r_eq, we have
V = ε_eq – Ir_eq ______(3)
The eq(2) and eq(3) should be the same.

The above equation can be put in a simpler way.

If there are n cells of emf ε₁, ε₂,………..ε_n and internal
resistance r₁, r₂………..r_n respectively, connected in parallel.

Kirchoff’s Rules
1. First law (Junction rule): The total current entering the junction is equal to the total current leaving the junction.
Explanation:

Consider a junction ‘O’. Let I₁ and I₂ be the incoming currents and I₁, I₄ and I₅ be the outgoing currents.
According to Kirchoff’s first law,

2. Second law (loop rule): In any closed circuit the algebraic sum of the product of the current and resistance in each branch of the circuit is equal to the net emf in that branch.

OR

Total emf in a closed circuit is equal to the sum of voltage drops

Explanation: Consider a circuit consisting of two cells of emf E₁ and E₂ with resistances R₁, R₂ and R₃ as shown in figure. Current is flowing as shown in figure
Applying the second law to the closed-circuit ABCDE₁A.
-I₃R₃ + E₁ + -I₁R₁ = 0
Similarly for the closed-loop ABCDE₂A.
-I₂R₂ + -I₃R₃ + E₂ = 0
For the closed loop AE₂DE₁A
-I₁R₁ + I₂R₂ + -E₂ + E₁ = 0
Note:

• Voltage drop in the direction of the current is taken as negative (and vice versa).
• emf is taken as positive if we go -ve to +ve terminal (and vice versa)

Wheatstone’s Bridge
Four resistances P, Q, R, and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂, I₃ and I₄ be the four currents passing through P, R, Q, and S respectively.

Working:
The voltage across R
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q __(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₄S
The value of R is adjusted to get zero deflection in galvanometer. Underthis condition,
I₁ = I₃ and I₂ = I₄ _____(2)
Using Kirchoffs second law in loopABDA and BCDB, weget
V_AB = V_AD ______(3)
and V_BC = V_DC _______(4)
Substituting the values from eq(1) into (3) and (4), we get
I₁P = I₂R ______(5)
and I₃Q = I₄S _____(6)
Dividing Eq(5) by Eq(6)

[since I₁ = I₃ and I₂ = I₄]
This is called a Wheatstone condition.

Meter Bridge
Uses: Meter Bridge is used to measure unknown resistance.
Principle: It works on the principle of Wheatstone bridge condition (P/Q=R/S).

Circuit details:
Unknown resistance X’ is connected in between A and B. Known resistance (box) is connected in between B and C. Voltage is applied between A and C. A100cm wire is connected between A and C. Let r be the resistance per unit length. Jockey is connected to ‘B’ through galvanometer.
Working: A suitable resistance R is taken in the box. The position of jockey is adjusted to get zero deflection.
If ‘l’ is the balancing length from A, using Wheatstone’s condition,

knowing R and l, we can find X (resistance of wire)
Resistivity: Resistivity of unknown resistance (wire) can be found from the formula
\(\rho=\frac{\pi r^{2} X}{l}\)
Where r (the radius of wire) is measured using screw gauge. l (the length of wire) is measured using meter scale
Note: Meter bridge is most sensitive when all the four resistors are of the same order

Potentiometer
(a) Comparison of e.m.f of two cells using potentiometer:

Principle: Potential difference between two points of a current-carrying conductor (having uniform thickness) is directly proportional to the length of the wire between two points.

Circuit details: A battery (B₁), Rheostat and key are connected in between A and B. This circuit is called primary circuit. Positive end of E₁ and E₂ are connected to A and other ends are connected to a two-way key. Jockey is connected to a two key through galvanometer. This circuit is called secondary circuit.

Working and theory: Key in primary circuit is closed and then E₁ is put into the circuit and balancing length l₁ is found out.
Then, E₁ α l₁ ______(1)
Similarly, E₂ is put into the circuit and balancing length (l₂ ) is found out.
Then, E₂ α l₂ _______(2)
Dividing Eq(1) by Eq(2),
\(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}\) _____(3)

(b) Measurement of internal resistance using potentiometer:
Principle: Potential difference between two points of a current carrying conductor (having uniform thickness) is directly proportional to the length of the wire between two points.

Circuit details: Battery B₁, Rheostat and key K1 are connected in between A and B. This circuit is called primary.

In the secondary circuit a battery E having internal resistance ‘r’ is connected. A resistance box (R) is connected across the battery through a key (K^2).  Jockey is connected to battery through a galvanometer.

Working and theory: The key (K₁) in the primary circuit is closed and the key is the secondary (K₂) is open. Jockey is moved to get zero deflection in the galvanometer. The balancing length l₁ (from A) is found out.
Then we can write.
E₁ α l₁ _____(1)
Key K₂ is put in the circuit, corresponding balancing length (l₂) is found out. Let V be the applied voltage, then
V₁ α l₁ _____(2)
‘V’ is the voltage across resistance box.
Current through resistance box
ie, voltage across resistance,
V = \(\frac{E R}{(R+r)}\) ____(3)
Substituting eq (3) in eq (2),
\(\frac{E R}{(R+r)} \alpha l_{2}\) ____(4)
Dividing eq (1) by eq (4),

Question 1.
Why is potentiometer superior to voltmeter in measuring the e.m.f of a cell?
Answer:
Voltmeter takes some current while measuring emf. So actual emf is reduced. But potentiometer does not take current at null point and hence measures actual e.m.f. Hence potentiometer is more accurate than voltmeter.

We hope the Plus Two Physics Notes Chapter 3 Current Electricity help you. If you have any query regarding Plus Two Physics Notes Chapter 3 Current Electricity, drop a comment below and we will get back to you at the earliest.

Plus Two Physics Notes Chapter 2 Electric Potential and Capacitance is part of Plus Two Physics Notes. Here we have given Plus Two Physics Notes Chapter 2 Electric Potential and Capacitance.

Kerala Plus Two Physics Notes Chapter 2 Electric Potential and Capacitance

Introduction
The electric field strength is a vector quantity, while the electric potential is a scalar quantity. Both these quantities are interrelated.

Electrostatic Potential

1. Electric potential: The electric potential at a point is the work done by an external agent in moving a unit positive charge from infinity to that point against the electric field (without acceleration)
Explanation: If W is the work done in moving a charge ‘q’ from infinity to a point, then the potential at
that point is, V = \(\frac{w}{q}\)

Potential difference: Electric potential difference between two points is the work done in moving a unit positive charge from one point to another.
The potential difference between points A and B is
V_AB = V_A – V_B
V_A and V_B are the potentials at points A and B respectively.

Potential energy difference: Potential energy difference is the work done to bring a q charge from one point to another point without acceleration.
Relation between potential difference and potential energy difference:

where U_A and U_B are the potential energies at points A and B respectively.

Electric field is conservative: Electric field is conservative. A conservative field is defined as the field in which work done is zero in a complete round trip.

(or)

A conservative field is one in which work done is independent of path.

Potential Due To A Point Charge

Let P be a point at a distance Y from a charge +q. Let A be a point at a distance ‘x’ from q , and E is directed along with PA. Consider a positive charge at A. Then the electric field intensity at A’ is given by

If this unit charge is moved (opposite to E) through a distance dx, the work done dw = – Edx
[-ve sign indicates that dx is opposite to E ]
So the potential at ‘P’ is given by

Potential due to an electric dipole

Consider dipole of length ‘2a’. Let P be a point at distance r₁ from +q and r₂ from -q. Let ‘r’ be the distance of P from the centre ‘O’ of the dipole. Let θ be angle between dipole and line OP.

Therefore total potential,

From ∆ABC , we get (r₂ – r₁) = 2a cosθ
we can also take r₂ = r₁ = r (since ‘2a’ is very small) Substituting these values in equation (1), we get

Case 1: If the point lies along the axial line of the dipole, then θ = 0°

Case 2: If the point lies along the equatorial line of the dipole, then θ = 90°

V = 0

Potential Due To A System Of Charges

Consider a system of charges q₁, q₂,……,q_n with position vectors r_1P, r_2P……..,r_nP relative to some origin. The potential V₁ at P due to the charge q₁ is

where r_1P is the distance between q₁ and P. Similarly, the potential V₂ at P due to q₂,

where r_2P is the distances of P from charges q₂. By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges
ie. V = V₁ + V₂ +……+ V_n

Equipotential Surface
The surface over which the electric potential is same is called an equipotential surface.
Properties:

1. Direction of electric field is perpendicular to the equipotential surface.
2. No work is done to move a charge from one point to another along the equipotential surface.

Example:

1. Surface of a charged conductor.
2. All points equidistant from a point charge.

Equipotential surfaces for a uniform electric filed:

1. Relation Between Electric Field And Potential:

Consider two points A and B, separated by very small distance dx. Let the potential at A and B be V+ dV and V respectively. The electric field is directed from A to B.
If a unit +ve charge is moved through a distance ‘dx’ against this field, work done,
dw = -Edx _____ (1)
For unit charge dw = dv
∴ dv = – Edx
or

Electric field intensity at a point is the negative rate of change of potential with distance.

Potential Energy Of System Of Charges

1. Potential Energy of System of Two Charges:
The potential energy of a system of two charges is defined as the work done in assembling this system of charges at the given position from infinite separation.

Consider two charges q₁ and q₂ separated by distance r. Imagine q₁ to be at A and q₂ at infinity. Electric potential at B due to charge q₁ is given by

which is the work done in bringing unit positive charge from infinity to B. Therefore the work done in bringing charge q₂ from infinity to B is
W = potential difference × charge
W = (V₁ – V_∞)q₂
potential at inf infinity. V_∞ = 0
W = V₁ × q₂
\(W=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{12}}\)
This work done is stored as potential energy. Hence potential energy between the charges q₁ and q₂

2. Potential Energy of System of Three Charges:
Consider three charges q₁, q₂, and q₃ separated by distances r₁₂, r₂₃ and r₁₃.

The electric potential energy of this system is the sum of potential of each pair. Hence we can write
U = U₁₂ + U₂₃ + U₁₃

Potential Energy In An External Field

1. Potential energy of a single charge:
Consider a point O in an electric field. Let V be the electric potential at O. Hence work done in bringing a charge q from infinity to the point O is,
W = Vq.
This work done is stored in the form of electrostatic potential energy (U) of the charge q.
∴ The potential energy of the charge q in an electric field is U = Vq
Where V is the potential at that point.

2. Potential energy of a system of two charges in an electric field:
Consider an electric field. Let 1 and 2 be two points in the field and V₁ and V₂ be the potential at these points. Two charges q₁ and q₂ are located at 1 and 2.
Potential energy of the charge q₁ in the external filed is, U₁ = V₁ q₁
Potential energy of the charge q₂ in the external field is, U₂ = V₂q₂
Potential energy between the system of two charges q₁ and q₂

where r₁₂ is the distance between the charges According to the principle of superposition, the potential energy of the system of two charges in an electric field is
U = U₁ + U₂ + U₁₂

3. Potential energy of a dipole in an external field:
Consider a dipole of dipole moment ‘P’ suspended in a uniform electric field of intensity ‘E’. Let θ be the angle between P and E.
Then we know torque τ = PE sinθ
Let the dipole be turned through an angle dθ
then work done dw = τdθ
= PE sinθ dθ
Total work done in rotating the dipole from θ₁ to θ₂

W = PE (cosθ₁ – cosθ₂)
This work done is stored as potential energy.

Electrostatics Of Conductors
The electrostatic properties of conductors are given below:
1. Inside a conductor, the electrostatic field is zero:
In the static situation, there is no current found inside the conductor. Hence we conclude that the electric field is zero inside the conductor The vanishing of the electric field inside the metal cavity is called electrostatic shielding.

2. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.

3. The interior of a conductor can have no excess charge in the static situation.

4. Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface.

5. Electric field at the surface of a charged conductor
\(\bar{E}=\frac{\sigma}{\varepsilon_{0}} \hat{n}\)
where σ is the surface charge density and is a unit vector normal to the surface in the outward direction.

6. Electric field inside a metal cavity is zero. Vanishing of electric inside a metal cavity is called electrostatic shielding. Sensitive electrical instruments can be protected from external electricfield by placing it in a metal cavity.

Dielectrics And Polarization
Dielectrics (insulator): Dielectrics are non-conducting substances. They have no charge carriers. The molecules of dielectrics may be classified into two classes.

1. Polar molecule
2. Nonpolar molecule

Electric field due to a dipole at a point on the perpendicular bisector of the dipole (at a point on the equatorial line).

Consider a dipole of dipole moment P = 2aq. Let ‘S’ be a point on its equatorial line at a distance ‘r’ from its centre. The magnitudes of electric field at ‘S’ due to +q and -q are equal and acts as shown in figure. To find the resultant electric field resolve.

Their normal components cancel each otherwhere as their horizontal components add up to give the resultant field at ‘S’.
E = E_Acos θ + E_Bcos θ = 2 E_B cos θ

The direction of the field due to the dipole at a point on the equatorial line is opposite to the direction of dipole moment.

1. Dielectrics in external electric field
(a) Nonpolar dielectrics in external field:

Considers nonpolar dielectrics in an external electricfield. In electricfield, the positive and negative changes of a nonpolar molecule are displaced in opposite directions. Thus dipole moment is induced in a nonpolar molecule. The induced dipole moments of different molecules add up giving a net dipole moment.

(b) Polar dielectrics in external electric field:

The permanent dipoles in a polar dielectric are arranged randomly. So total dipole moment is zero. But when we apply external electric field, the individual dipole tends to align in the direction of electricfield. The induced dipole moments of different molecules add up giving a net dipole moment.

Electric susceptibility: Non-polar dielectrics and polar dielectrics can produce net dipole moment in the external electric field. The dipole moment per unit volume is called polarization and is denoted as P. For linear isotropic dielectrics.
\(\bar{P}=\chi_{0} \bar{E}\)
where χ_e is a constant and is known as the electric susceptibility of the dielectric medium.
How does the polarized dielectric modify the external field inside it?

Consider a dielectric slab placed inside a uniform external electric field E₀. This field produces a uniform polarization as shown in the figure. Any region inside the dielectric, the net charge is zero.

This is due to the cancellation of positive charge of one dipole with negative charge of adjacent dipole. But the positive ends of the dipole do not cancel at right surface and the negative ends at the left surface.

This surface charges (-σ_p and +σ_p) produce a field \(\left(\vec{E}_{i n}\right)\) opposite to the external field. Hence total electric field is reduced inside the dielectric field which is shown in the figure below.
ie; E₀ + E_in ≠ 0

How does a metal modify the external electric field applied on it?

When a conductor placed in a electric field, the free charges are moved in opposite direction as shown in figure. This rearrangement of charges in a metal produce an internal field (E_in) inside the metal. This internal field cancels the external field. Thus the net electric field inside the metal becomes zero.
ie; E₀ + E_in = 0

Capacitors And Capacitance
Capacitor: Capacitor is a system of two conductors separated by an insulator for storing electric charges.
Capacitance of a capacitor:

Consider two-conductor having charges +Q and -Q and potentials V₁ and V₂. The amount of charge Q on a plate is directly proportional to the potential difference (v₁ – v₂) between the plates,
ie. Q α V₁ – V₂
(or) Q α V (where V = V₁ – V₂)

The constant C is called the capacitance of the capacitor. If V = 1, we get Q = C. Hence capacitance of a capacitor may be defined as the amount of charge required to raise the potential difference between two plates by one volt.
Dielectric strength:
What happens to the charge stored in capacitor when the p.d. between two plates increases?
When the p.d. between two plates increases, electric field in between two plates increases. This high electric field can ionize the surrounding air (or medium) and accelerate the charges to the oppositely charged plates and neutralize the charge on the plate. This is called electric break down.

The maximum electric field that a dielectric medium can withstand without break down (of its insulating property) is called its dielectric strength. The dielectric strength of air is 3 × 10⁶ v/m.

The Parallel Plate Capacitor
Electric field due to a capacitor:

Consider a parallel plate capacitor consists of two large conducting plates 1 and 2 separated by a small distance d. Let +σ and -σ be the surface charge densities of first and second plates respectively. (Here, we take, electric field towards right is taken as positive and left as negative.)
Region I: This region lies above plate 1.
E = E₊ + E_– ie.

Region II: This region lies below plate 2.
E = E_– + E₊ ie.

E = 0
Electric field in between two plates: In the inner region between plates 1 and 2, the electric field due to two charged plates add up.

(a) Expression for capacitance of a capacitor: Potential difference between two plates
V= Ed

Effect Of Dielectric On Capacitance

Consider a capacitor of area A and charge densities +σ and -σ. Let d be the distance between the plates. If a dielectric slab is placed inside this capacitor, it undergoes polarization. Let +σ_p and -σ_p be polarized charge densities due to polarization.
Due to polarization electric field in between the plate becomes
\(E=\frac{\sigma}{K \varepsilon_{0}}\) _____(1)
The potential difference between the plates,
V = Ed _____(2)
Sub (1) in (2)
\(\mathrm{V}=\frac{\sigma}{\mathrm{K} \mathrm{s}_{0}} \mathrm{d}\)
Then the capacitance of capacitor

The product ε₀ K is the permittivity of the medium.

Combination Of Capacitors
1. Capacitors in series: Let three capacitors C₁, C₂ and C₃ be connected in series to p.d of V. Let V₁, V₂ and V₃ be the voltage across C₁, C_2, and C₃.

The applied voltage can be written as
V = V₁ + V₂ + V₃ ______(1)
Charge ‘q’ is same as in all the capacitor. So,

Substituting these values in (1),

If these capacitors are replaced by a equivalent capacitance ‘C’, then
V = \(\frac{q}{C}\)
Hence eq(2) can be written as

Effective capacitance is decreased by series combination.

2. Capacitors in parallel:

Let three capacitors C₁, C₂ and C₃ be connected in parallel to p.d of V. Let q₁, q₂, and q₃ be the charges on C₁, C₂ and C₃.
If ‘q’ is the total charge, then’q’can be written as
q = q₁ + q₂ + q₃
But q₁ = C₁V, q₂ = C₂V and q₃ = C₃V
Hence eq (2) can be written as
CV = C₁V + C₂V + C₃V

Effective capacitance increases in parallel connection.

Energy Stored In A Capacitor
Energy of a capacitor is the work done in charging it. Consider a capacitor of capacitance ‘C’. Let ‘q’ be the charge at any instant and ‘V’ be the potential. If we supply a charge ‘dq’ to the capacitor, then work done can be written as,
dw = Vdq
dw = \(\frac{q}{C}\)dq (since V = \(\frac{q}{C}\))
∴ Total work done to charge the capacitor (from ‘0’ to ‘Q’) is

But Q = CV
W = \(\frac{1}{2}\) CV²
This work done is stored in the capacitor as electric potential energy.
∴ Energy stored in the capacitor is,

Van De Graff Generator
Van de Graff generator is used to produce very high voltage.
Principle: If two charged concentric hollow spheres are brought in to contact, charge will always flow from inner sphere to the outer sphere.
Construction and working:

The vande Graff generator consists of a large spherical metal shell, placed on an insulating stand. Let p₁ and p₂ be two pulleys. Pulley p₁ is at the center of the spherical shell S. A belt is wound around two pulleys p₁ and p₂. This belt is rotated by a motor. Positive charges are sprayed by belt. Brush B₂ transfer these charges to the spherical shell. This process is continued. Hence a very high voltage is produced on the sphere.
Why does the charge flow from inner sphere to outer sphere?

Let ‘r’ and ‘R’ be the radius of inner sphere and outer sphere carrying charges q and Q respectively.
The potential on the outer sphere,
V(R) = Potential due to outer charge + potential due to inner charge

Potential on the inner sphere. V(r) = Potential due to outer charge + Potential due to inner charge

∴ Potential difference between the two spheres

The above equation shows that the inner sphere will be always at a higher potential. Hence, the charge always flows from the inner sphere to the outer sphere.

We hope the Plus Two Physics Notes Chapter 2 Electric Potential and Capacitance help you. If you have any query regarding Plus Two Physics Notes Chapter 2 Electric Potential and Capacitance, drop a comment below and we will get back to you at the earliest.