Kerala Plus Two Physics Notes

PLUS

2 NOTES
Chapter Wise

ELECTROSTATICS-CHAPTER 1
ELECTRIC CHARGES AND FIELDS
Topics Included
 Electric Charge
 Properties of Electric Charge
 Methods of Charging
 Coulomb’s Law
 Electric Field
 Electric Field Lines
 Electric Dipole
 Electric Flux
 Gauss’s Theorem
Electric charge ‘Q’
Electric charge is a fundamental property of a particle (just like its mass) due to which electric and
magnetic effects are produced in the matter.
 SI unit of charge is coulomb ( C )
Properties of electric charges
i) Additive property
Electric charge is additive like real numbers .That is total charge on a body is equal to the algebraic
sum of charges present at different parts of the body.
ii) Quantization of charges
Quantization of charge means that the total charge of a body is always an integral multiple of certain
smallest charge (charge of an electron)
ie Q= +ne
Where Q = Total charge on the body
n = 1,2,3,4……..
e = charge of an electron (1.6 x 10-19C)
+ sign stands for loss of electron and – sign stands for gain of electron.
► Number of electrons constituting 1C of charges is 6.25 x 1018
{ n = Q/e = 1C/1.6 x 10 -19 C = 6.25 x 1018 }
iii) Conservation of charges
Total charge on an isolated system remains constant. This means that there is only a transfer of
charges from one body to other, but no creation or destruction of charges
Methods of charging a body
1 . Charging by friction
2 . Charging by conduction
3 . Charging by induction

1. Charging by friction : When two insulating surfaces are rubbed against each other, due to friction,
electrons are transferred from one body to other. Hence both get charged. The electric charges so
acquired are called frictional electricity or charging by friction.
► The substance which loses electrons become positively charged and which gain electrons become
negatively charged.
► When a body is charged negatively, electrons are added hence its mass increases

► When a body is charged positively, electrons are removed hence its mass decreases.

2. Charging by conduction: When an un charged conductor is brought in to contact with a charged body,
charge flows from the charged body to the uncharged body.
 During charging by conduction, both objects acquire the same type of charge.

2.Charging by conduction

Problem 1
How many electrons are present in -1 coulomb of charge ?
Solution: q=ne (or) n= q e = 1 1.6 x 10−19=6.25 x 1018
ie , -1 C means 6.25 x 1018 electrons are excess.
Similarly +1 C means 6.25 x 1018 electrons are deficient.
Problem 2
A polythene piece rubbed with wool is found to have a negative charge of 3 × 10-7 C.
a). Estimate the number of electrons transferred (from which to which?)
b). Is there a transfer of mass from wool to polythene?
Solution: a) Q = 3 × 10-7 C
Q = ne
n = Q/e = 3 ×10-7 / 1.6× 10-19 =1.875×1012 , electrons transferred from wool to polythene
b) yes. There is a mass transfer from wool to polythene
Mass of electrons transferred =nme
12 -31
= 1.875×10 X 9.1×10 = 1.706×10-18kg
Point charges
When the size of charged bodies are much smaller than the distance separating them, their sizes can be
neglected and the charged bodies can be treated as point charges.
Coulomb’s law OR Inverse square law in electrostatics
According to this law “The magnitude of force of attraction or repulsion between any two point
charges at rest is directly proportional to product of magnitudes of their charges and inversely proportional to
the square of the distance between them”.
Consider two point charges Q1 and Q2 separated by a distance r. The force between them
F α Q 1 Q2 , F α 1/ r2
F α Q1 Q2
r2
F = Constant Q1 Q2
r2
In SI system F = 1 Q1 Q2
4 π εo ε r r2

Where εo → Permittivity of air or free space

εr → Relative permittivity of the medium
where the charges are placed
► ε o = 8.854 x 10 -12 C 2N - 1 m - 2
Therefore 1/ 4 π ε o = 9 x 109
F = 9 x 10 9 Q1 Q2
εr r2

► For air or free space ε r = 1
► Since ε r > 1, Fair > Fmedium
► ε r is also called dielectric constant (K) of the medium
 For vacuum K = 1, for air k =1.008, for water K= 80
► ε r has no unit or dimensional formula.
Electric field ‘ E’
E = F
The space around a charge within which its electro static force
q
is experienced is called electric field.
► SI unit of Electric field intensity is Newton / coulomb ( N C-1 ) or volt/meter ( V m-1 )
► E is a vector quantity
► Force experienced by a charge q placed in an electric field of strength E is given by F = qE
Electric Field Lines or Electric Lines of Forces
Digramatic visualization of electric field around a charged configuration is called Electric field lines.

Electric dipole
A pair of equal and opposite charges separated by a small distance is called an electric dipole. Figure
shows an electric dipole having charges +q and –q separated by a distance 2a.

Electric dipole moment :
The strength a dipole is measured in terms of its dipole moment which is defined as the product of
magnitude of one of the charges and length of the dipole .
Ie Dipole moment P = Q 2a
► P is a vector quantity directed from negative to positive charge along the axis of the dipole
► SI unit of dipole moment is coulomb meter ( C m )
Dipole field
The electric field produced by a dipole is called dipole field. The total charge of a dipole is zero, but
the dipole field is not zero
i) Field at any point on the axial line of the dipole
E = 1 2P
4πεo r3
ii) Field at any point on the equatorial line of the dipole
E = 1 P
4πεo r3
iii) Field at any point ( general case )
E = 1 P √ 3 cos2θ + 1
4πεo r3

► E axial : E equit = 2:1

Case 1: Field at any point on the axial line of the dipole

The electric field at P due to +q

The electric field at P due to -q

Thus the total electric field at P

On simplifying we get

For r ˃˃ a ,there for

Torque on an electric dipole placed in an uniform electric field
Consider an electric dipole of length 2a placed in a uniform electric field of strength E making an angle θ with field.
Torque = Force x Perpendicular distance between forces
τ = F ( BC )
= QE (AB sinθ )
= QE (2a sinθ )
= Q 2a E sinθ
τ = P E sinθ,
τ = P x E

Electric flux
Total number of electric lines of force passing through a given surface is called flux through it. Then flux of
electric field through the area element dS
dΦ = E dS cos θ
The electric flux through the whole surface is given by
Φ = ∫s dΦ

Φ = ∫s E dS cos θ
Φ = ∫s E . dS
► If the field is uniform over the surface the total flux Φ = E S cos θ
 Maximum flux passes through the surface
when it is held perpendicular to the electric field( Here θ = 0 ) Φ max = E S
 No flux passes through the surface when it is held parallel to the electric field ( Here θ = 90 )
 Electric flux is a scalar quantity

 SI unit of flux is NC-1 m-2 or V m
Gauss’s Theorem
According to this Law “Total electric flux through any closed surface in frees pace is equal to 1/ εo
times the net charge enclosed within the surface”.
ie Φ = 1 Q ∫s E . dS = 1 Q
εo εo
Where Q is the net charge enclosed and εo is the permittivity of free space
Applications of Gauss’s Theorem
1.Electric field due to an infinitely long straight charged wire.

2.Electric Field due to uniformly charged infinite plane sheet.

3.Electric Field due to a uniformly charged thin spherical shell

Important Previous Questions
1. State Gauss’s Theorem. Find electric field due to an infinitely long straight charged wire.
2. Derive an expression for Electric field due to an electric dipole at a point on the axial line.
3. Two point charges q1 =3µC and q2 =6 µC separated by a distance 3m apart in air.Find the electrostatic

force between them.
4. Draw electric field lines around
a) An isolated negative charge
b) An electric dipole

ELECTROSTATICS-CHAPTER 2
ELECTRIC POTENTIAL AND CAPACITANCE
Topics Included
 Electric potential
 Equipotential Surface
 Electric Capacitance
 Capacitance of PPC
 Combinations of Capacitors
 Energy stored in a capacitor
Electric Potential ‘V’
The work done in bringing a unit positive test charge (without acceleration) from infinite to a point
in an electric field gives the potential at the point.
Electric Potential
V = W/q
 SI unit of potential is J/C or volt ( V )
 Potential is a scalar quantity
 Potential due to an isolated positive charge is positive and negative charge is negative
Potential due to a point charge

V = 1 Q V α 1
4πεo r r
E = 1 Q V α 1
4 π ε o r2 r2

Equipotential surface
A surface with a constant value of potential at all points on it is called equipotential surface
Eg surface of a charged conductor
Properties of Equipotential surface
1) Potential is same at every point on the ep surface
2) Pd between any two points on the ep surface is zero
3) No work is done for moving a test charge from one point to other on an ep surface
4) Electric field lines are always perpendicular to the ep surface
5) For an isolated point charge ep surfaces are concentric spheres with charge at the centre
6) For uniform electric field ep surfaces are parallel planes right angles to the lines of force

Point Charge (+q) Uniform Field An Electric Dipole
Electric Capacitance ‘C’

It is the measure of ability of a conductor to store electric charge. When some charge is given to a
conductor its potential rises. The increase in potential is directly proportional to the charge given
Ie Q α V
Or Q = C V or C = Q / V
Where C is a constant called electrical capacitance or capacity of the conductor
 SI unit of capacitance is ‘farad’ ( F )

Capacitor
Capacitor is a device used to store large amount of electric charge in a very small space It consists of two
conducting plates separated by a dielectric medium or Air.

Capacitance of a parallel plate capacitor ( PPC )

Combination of capacitors
Capacitors can be combined in two ways
1) Series Combination
In series combination charge on each capacitor
is same but potential across each capacitor is different.

So Net applied voltage V = V1+V2+V3 ………(1)
We have V=Q/C
There for (1) Q/Cs = Q/C1 +Q/C2 +Q/C3 ….(2)
Q/Cs = Q[1/C1 +1/C2 +1/C3 ]

1/Cs = [1/C1 +1/C2 +1/C3 ]

2) Parallel Combinations
In parallel combinations Potential on each capacitor is same Charge is different.

Total Charge from Source Q =Q1+Q2+Q3

Important Previous questions:

1) Explain the Series/Parallel combination of 3 capacitors of capacitances C1,C2 and C3

2) Derive an equation for energy stored in a capacitor.
3) Give any 2 properties of equipotential surfaces.
4) Derive an expression for capacitance of a parallel plate air capacitor.

5) Draw equipotential surface around
a) An isolated positive charge
b) An electric dipole
c) An uniform electric field

CHAPTER 3
CURRENT ELECTRICITY
Ohm’s Law
At constant temperature, the current flowing through a conductor is directly proportional to the potential
difference between the ends of the conductor
V∝I
V=RI R is called the resistance of the conductor
The SI units of R is ohm (Ω.)
Conductance (K)
The reciprocal of resistance is called Conductance
unit :mho ( Ω-1)siemens
Connection Diagram to study Ohm’s Law
Voltage –Current Graph (V-I Graph)
Slope = AB/BC
Slope = 𝐕/𝐈 = R
Key
Slope of V-I graph gives Resistance.
Its reciprocal gives conductance

Factors on which the Resistance of a Conductor Depends
i)Length of the conductor R α l
ii) Area of cross section of the conductor R α A
iii) Material of the conductor
iv) Temperature
Resistance of a particular conductor at constant temperature depends only up on length and area
R α l/A R = ρ l/A ρ is called Resistivity Unit Ω m
Conductivity (σ)
Reciprocal of resistivity is called conductivity σ=1/ρ unit Ω-1m-1 or mho m-1
Electric Power

Electro motive force (emf) and Terminal potential difference of a cell

Electro motive force (emf) is the potential difference between two terminals of a cell in open
circuit. Terminal potential difference is the potential difference between two terminals of a cell
(or any two points in an electrical circuit) in a closed circuit.
Consider a resistance R connected across a cell of emf E whose internal resistance is r. If a current I is
flowing through the circuit, then according to Ohm’s law,

Current through R,

Potential difference across R,

Rearranging eq (2) we get

Kirchhoff’s Rules
(a)Kirchhoff’s First Rule - Junction Rule: ΣI=0
Kirchhoff’s junction rule is in accordance with law of conservation of
charge.

(b)Kirchhoff’s Second Rule –Loop Rule: ΣIR= ΣE
Wheatstone Bridge

For a balanced Wheatstone’s bridge , the resistors are such that the
current through the galvanometer Ig = 0.
Apply Kirchhoff’s junction rule to junction B & junction D
𝐈𝟐 = 𝐈𝟒-------------(1)
𝐈𝟏 = 𝐈𝟑 -------------(2)
Apply Kirchhoff’s loop rule to closed loop ABDA
𝐈𝟏 𝐑𝟏= 𝐈𝟐𝐑𝟐----------------(3)
Apply Kirchhoff’s loop rule to closed loop CBDC
𝐈𝟑 𝐑𝟑= 𝐈𝟒 R𝟒---------------(4)

This is the balance condition for the
galvanometer to give zero or null deflection

CHAPTER 4
MOVING CHARGES AND MAGNETISM
Topics Included
 Magnetic Field
 Magnetic Lorentz force
 Lorentz force
 Biot–Savart law
 Magnetic field on the axis of a circular current loop
 Ampere’s circuital law
 Moving Coil Galvanometer (MCG)
 Conversion of galvanometer in to ammeter
 Conversion of galvanometer in to voltmeter
Magnetic Field ‘B’
It is the space around a magnet or a current carrying conductor or a moving charge with in which its
magnetic effect can be experienced.
Magnetic Lorentz force
Force experienced by a charged particle moving through a magnetic field is called Magnetic Lorentz
force ,which is given by
F = q(vxB)
The magnitude of the force is given by
F = q v B sin θ
where q → magnitude of the charge
v → velocity of the particle
B → strength of magnetic field
θ → angle between direction of motion (v) and direction of magnetic field (B)
 The direction of the force is perpendicular to both velocity and magnetic field
Case I : If θ = 0 or 180, then F = 0 path of the particle is straight line
Case II : If θ = 90 ,then F = q v B sin90 ,F = q v B ( maximum force ). the path of the
charged particle is circular.
Case III : When the charged particle is moving inclined to the magnetic field (θ ≠ 0 , 90 ,180 )
the path of the particle is helical.
Case IV : If the charged particle is at rest ( v = 0 ) then F = 0
Direction of Magnetic Lorentz force

Fleming’s right hand palm rule
If the thumb of the right hand point in the direction of v, the fingers in the direction
of B, and the force F is directed perpendicular to the right hand palm.

Lorentz Force
Force experienced by a charged particle moving through an electromagnetic field is called Lorentz
force.
F = qE + q(vxB)

Magnetic field due to a current element-Biot –Savart Law
Consider a conductor XY of finite length carrying current I. To find the strength of magnetic field at a
point p at a distance r, consider an infinitesimal element dl of the conductor.
According to Biot –savart‘s law “The strength of Magnetic field at p due to this current element is
dB α I
dB α dl
dB α sin θ
dB α 1/r2
in general, dB α I dl sinθ
r2
dB= μo I dl sinθ
4π r2 μo = 4π x 10-7 is a constant called permeability of air or free space.
In vector form
dB = μo I (dl x r)
4π r3

Amperes circuital law
“ The line integral of magnetic field around any closed loop (path) is equal to μo times net current
enclosed by the path”.
∫ B.dl = μo I
Where I refer to current enclosed by the path
Moving coil galvanometer
It is a device used to measure or detect small current in an electric circuit
Principle : It is based on the fact that a current carrying coil behaves like a magnetic dipole and experience
torque when placed in an external magnetic field .
NIAB = kΦ
I = kΦ
NAB

Now k / NAB is a constant called Galvanometer constant G
therefore I =G Φ IαΦ

Conversion of galvanometer in to ammeter

Ig = IS / G+S

Conversion of galvanometer in to voltmeter

V = Ig ( R + G )

Important Previous Questions.
1. State Biot-Savart law. Derive an expression for magnetic field due to a circular current carrying coil at a
point on the axis of coil.
2. State Ampere’s circuital law.
3. How can we convert a moving coil galvanometer in to ammeter and voltmeter.
4. What is the path of charged particle when it moves perpendicular with magnetic field?

CHAPTER-5
MAGNETISM AND MATTER

Topics

• Magnetic dipole
• Magnetic dipole moment.
• Gauss’s Law in magnetism.
• Magnetisation.
• Magnetic intensity.

Important points.
1. Magnetic dipole : Two unlike poles of equal strength separated by a small distance.
2. S.I. unit of pole strength (p) is Am .It is a scalar quantity.
3. Magnetic dipole moment (m) = pole strength(p) x distance between poles (2L)
m = p 2L
4. S.I. unit of magnetic dipole moment is Am2. It is a vector quantity.
5. Gauss’s theorem : The total magnetic flux through any closed surface is zero.
Σ B. ΔS = 0
6. Behaviour of magnetic field lines near a
(a) diamagnetic,
(b) Ferromagnetic or paramagnetic substance.

6. Magnetisation M of a sample to be equal to the net magnetic moment per unit volume . M = m.
V
7. Magnetic intensity H = B.
Μ
8. Magnetic susceptibility ( χm ) : It is the ratio of intensity of magnetisation and the magnetic intensity
of the magnetic field
Important Questions:

1. State gauss’s law for magnetism.
2. Which physical quantity has the unit Wb m-2 ? Is it a scalar or a vector quantity ?

3. S.I. unit of magnetic induction is ..................
**************

CHAPTER 6
ELECTROMAGNETIC INDUCTION

Magnetic flux ϕ

ϕ = NBA Cos θ

N-no of turns of coil,
B – magnetic field,
A – Area of coil
θ -Angle between B and A

Faraday’s Law of electromagnetic Induction
I Law: The definition of electromagnetic induction- Whenever the magnetic flux linked with a coil changes
an emf is induced at the two ends of the coil.This phenomenon is called electromagnetic induction, emf so
generated is called induced emf.
II law: the magnitude of the induced emf is equal to the rate of change of magnetic flux linked with the coil.

Lenz’s Law
The direction of induced current is such that which is always opposing the cause producing it.
Lenz’s Law is in accordance with law
of Conservation of Energy

Motional emf
When a conductor of length 𝓁 moving with velocity v perpendicular to magnetic field B , an emf is
induced at two ends of the conductor. It is called motional emf
ε =B 𝓁 v
Self-Induction
The phenomenon of production of induced emf in a coil itself when current through the coil changes
is called self induction
The magnetic flux linked with the coil is proportional to the current through the coil.
?α I
𝝓=LI L is called self-inductance of the coil. Its unit is Henry (H)

When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. This
induced emf is also called back emf
−d ϕ dI
ε= 𝝓 = L Iε =− L
dt dt

Self Inductance of a Solenoid
Consider a solenoid of cross sectional area A and length 𝓁, having n turns per unit length and N no of
turns, N=n 𝓁 or n=N/𝓁

if I be the current flows through the solenoid
The magnetic flux linked with solenoid
𝝓 = L I -----------------------(1)

Also we have 𝝓 = NBA Cos θ here θ=0 and Cos θ=1 and inside solenoid B = 𝜇0 n I

𝝓 = N (𝜇0 n I) A ------------(2)

From equations (1) & (2) LI = N (𝜇0 n I) A

L = 𝜇0 N n A ------------(3)

Put N=n 𝓁 in equation (3) L = 𝜇0 n2 A 𝓁 -----------------(4)

Put n=N/ 𝓁 in equation (3) L = 𝜇0 N2 A /𝓁 --------------------(5)

Equations (3),(4) & (5) are different forms of self inductance of solenoid

Mutual induction
The phenomenon of production of induced emf in a coil by varying the current through a
neighbouring coil is called mutual-induction
The magnetic flux linked with the coil is proportional to the current through the neighbouring coil.
ϕαI
ϕ=MI M is called mutual-inductance of the coil

When the current in the neighbouring coil is varied, the flux linked with the first coil changes and an emf is
induced in the coil.
dI
ε =− M
dt

Mutual inductance of two co-axial solenoids

Energy stored in an inductor
Small work done dw=P dt ---------------------(1)

dI
(electric Power P=ε I&Induced emf ε =− L )
dt

dw = L( dIdt I ) dt

dw = L I dI ----------------------------(2)

Total work done W = ∫ dw

W = L∫ I dI -------------------------(3)

LI2
W=
-----------------------------------(4)
2
This work done is stored in the magnetic field of inductor as potential energy

LI2
Energy stored in an inductor U =
2

Ac Generator
An AC generator is a device that converts mechanical energy into electrical energy in form of
alternative emf or alternating current. AC generator works on the principle of Electromagnetic Induction
When coil rotates with constant angular ferquency ω
magnetic flux linked with coil changes and induced emf is
produced.
ϕ = NBA Cos θ
−d ϕ
θ = ωt ε=
dt

d ( cos ωt )
ε =− N BA
dt

ε =N BA ω sin ωt

ε =v ∧ NBA ω= V0—Peak value of alternating voltage

V=V0 Sin ωt
Graphical representation of Alternating Voltage

CHAPTER-7
ALTERNATING CURRENT
Topics
• AC Voltage applied to a Resistor.
• Representation of AC Current and Voltage by rotating vectors – PHASORS.
• AC Voltage applied to an Inductor.
• AC Voltage applied to a Capacitor.
• AC Voltage applied to a series LCR Circuit.
• Power in an AC circuit and the power factor.
• Transformers

Important points.
1. The rms current ( Root Mean Square Current) or Effective Current:
I rms or I = im
√2
The rms voltage , V rms or V = Vm
√2
2. AC Voltage Applied to a Resistor :

Vm sin ωt - i R = 0
Vm sin ωt = iR
i = Vm sin ωt
R
i = i m sin ωt
where i m = Vm is called maximum value of current or peak value of current.
R
Graphical representation of v and i versus ωt

Phasor diagram

3. AC Circuit containing an Inductor:

v m sin ωt – L di = 0
dt
Vm sin ωt = L di
dt

di = Vm sin ωt
dt L

di = Vm sin ωt dt
L
i = Vm ∫ sin ωt dt
L

i = − Vm cos ωt
ωL
i = im sin (ωt − π/2 )
where im = Vm is the peak value of current.
ωL
In a pure inductor, the current lags the voltage by π/2 .

Inductive Reactance (XL ):

The current amplitude, im = Vm
ωL
The quantity ω L is analogous to the resistance and is called inductive reactance,
denoted by XL
XL = ωL =2πfL
Graphical representation of v and i versus ωt

Phasor diagram

4. AC Circuit containing Capacitor :

Vm sin ωt - q = 0
C
Vm sin ωt = q
C
q = CVm sin ωt
i = d (CVm sin ωt)
dt
i = C Vm ω cos ωt
i = C ω Vm cos ωt
i = im cos ωt
i = im sin (ωt + π/2 )
im = Vm is the peak value of current.
(1/ωC )
In a purely capacitive circuit, the current leads the voltage by π/2.
The quantity (1/ωC) = X C is analogous to the resistance and is called capacitive
reactance.
X C = (1/ωC) = 1
2πfC

Graphical representation of v and i versus ωt :

Phasor diagram :

5. AC Voltage Applied to a Series LCR Circuit:

VR be the potential difference across R , and in phase with I.
VC be the potential difference across C , which lags the current I by π/2 or 90ᵒ
VL be the potential difference across L , which leads the current I by π/2 or 90ᵒ

Vm 2 = VR 2+ ( VC − VL ) 2
Vm 2 = (im R) 2 + (im X C − im X L ) 2
Vm 2 = im 2 [(R) 2 + ( X C − X L ) 2 ]
im 2 = Vm2
(R) 2 +( X C − X L ) 2

im = Vm2

√ (R) 2 +( X C − X L ) 2

im = Vm2
Z
The quantity √ (R) +( X C − X L ) is analogous to resistance and is called impedance Z in an
2 2

ac circuit.

Impedance, Z = √(R) 2 +( X C − X L ) 2
Impedance Triangle

The phase difference φ between voltage and current is ,

φ = tan −1 X C − X L
R

6. Resonance
When X C = X L, Z = R and φ = 0. This is condition for resonance.
Resonance frequency f = 1
2π√(LC)
7. Power in an AC circuit
Power In AC Circuit , P = V I
P = Vm sin ωt x im sin (ωt + φ)
P = Vrms Irms cos φ
The quantity cosɸ is called the power factor.

8. Transformers
Principle − It works on the principle of electromagnetic Induction When current in one circuit
changes, an induced current is set up in the neighbouring circuit.

Step-up Transformer Step-down transformer
Primary secondary Primary secondary

Working : Alternating emf is supplied to the primary coil PP’. The resulting current produces
an induced current in secondary. Magnetic flux linked with primary is also linked
with the secondary. The induced emf in each turn of the secondary is equal to that
induced in each turn of the primary.

Alternating emf applied to primary ,

where np be the number of turns in the primary

Induced emf in the secondary,

where nS be the number of turns in the primary

Dividing eq (2) in (1) we get

Transformation ratio.

For step-up transformer, K > 1 ∴Es>Ep
For step-down transformer, K < 1 ∴Es<Ep

According to law of conservation of energy, , Input power = output power

Important questions:

1. A series LCR circuit connected to an AC source .
i) Write an expression for impedance offered by the circuit .
ii) Draw an impedance diagram and write the expression for the power factor from the diagram.
2. i) What are the condition for resonance in a series LCR circuit. Write the expression for
resonance frequency.
ii) At resonance in an LCR circuit the emf and current are .............
3. The S.I. unit of inductive reactance is...........
4. A transformer steps down 220V to 11V. What is the transformer ratio ?
************

CHAPTER-8
ELECTROMAGNETIC WAVES

Topics
• Displacement current.
• Electromagnetic waves.
Important points.

1. Displacement current:
Displacement current is that current which comes into play in vacuum or dielectric when electric
field is changing with time.
Id = ɛ0 dφ
dt
Maxwell introduced the idea of displacement current for the consistency of Amperes equation.

Characteristics of displacement current :
i) Id = 0 in a conductor
ii) Id not equal to zero in vacuum and dielectric
iii) In series circuit Id = Ic where Ic is the conduction current.
iv) Id is produced by the rate of change of electric field.

2. Electromagnetic waves:
Oscillating charges can produce electromagnetic waves.

For a plane electromagnetic wave propagating along x- direction is represented by
Ex = E0 sin (kz – wt )
Magnetic field along y-direction is given by
By = B0 sin (kz – wt )
Properties

Velocity of electromagnetic waves in vacuum is given by

ELECTROMAGNETIC SPECTRUM

a) Radio waves
Radio waves are produced by the accelerated motion of charges in conducting wires. They are used in radio and
television communication systems. Used in mobile communication.
b) Micro waves
Due to their short wavelengths, they are suitable for the radar systems used in aircraft
navigation ,Microwave ovens , communication satellite, cell phone network, etc.
c) Infrared waves
Infrared waves are produced by hot bodies and molecules. Infrared lamps are used in physical
therapy. Infrared rays are widely used in the remote switches of household electronic systems such as
TV, video recorders etc. Used to take photographs in darkness. Used in solar water heaters
d) Visible light
It is the part of the spectrum that is detected by the human eye. Visible light emitted or reflected
from objects around us provides us information about the world.
e) UV rays
Sun is an important source of ultraviolet light. UV lamps are used to kill germs. Used for eye
surgery , preserve food etc.
f) X- rays
X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of
cancer. Used for study of atomic structure. Used to detect fractures.
g) Gamma rays
They lie in the upper frequency range of the electromagnetic spectrum .They are used in medicine
to destroy cancer cells, radiation therapy. Used for inspection of material.

Important Questions

1. Which of the following is not an electromagnetic wave.
i) X -rays ii) γ -rays iii) β- rays iv) Microwaves.
2. An electromagnetic wave propagates through a medium of permittivity ɛ and permeability μ.
What is the speed of this wave through the medium ?
3. Arrange the following radiations in an ascending order in respect of their frequencies
X-rays , microwaves, UV rays and microwaves.
4. Name the following constituent radiations of electromagnetic spectrum which

i) produce intense heating effect ii) is absorbed by the ozone layer in the atmosphere
iii) isn used for studying crystal structure. iv) is used in satellite communication
5. What is displacement current ?

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CHAPTER-9
RAY OPTICS AND OPTICAL INSTRUMENTS

Topics
• Laws of Reflection.
• Refraction of light by spherical mirrors.
• Relation between focal length and Radius of curvature.
• The mirror equation.
• The linear magnification.
• Laws of refraction.
• Refraction at a spherical surface.
• Lens makers formula.
• Power of a lens.
• Refraction through a prism.
• Simple microscope.
• Compound microscope.
Important points.

1. Laws of reflection:
i) The incident ray , reflected ray and the normal at the point of incidence all lie in a same plane.
ii) The angle of incidence is equal to angle of reflection. i = r.

2. Reflection of light by spherical mirrors :

3. Relation between focal length and Radius of curvature.

From ΔMDC , tanƟ = MD -------------- (1)
DC
ΔMDF , tan2Ɵ = MD -------------- (2)
DF
(1) 1 = DF = f . ;
(2) 2 DC R

R = 2f or f = R .
2
4. The mirror equation : 1. + 1. = 1.
u v f

5. Linear magnification : m = – v
u
6. Laws of Refraction:
i) The incident ray , refracted ray and the normal to the surface at the point of incidence lie in the
same plane.
ii) Snell’s law : The ratio of sin of angle of incidence to the sin of angle of refraction is a constant for
a given pair of media and for a given colour of light.
Sin i = n2
sin r n1
If c is the velocity of light in vacuum and v is that in a medium, the absolute refractive index of
the medium is n = c.
v
7. Refraction at a spherical surface :

Fig shows formation of image I of an object O on the principle axis of a spherical surface with
centre of curvature C and radius of curvature R.
By snell’s law , n1 sin i = n2 sin r

For small angles , n 1 i = n2 r
For Δ ONC , i=∞+γ
For Δ NCI , γ=r+β or r=γ–β
n1 ( ∞ + γ ) = n2 (γ – β )

n2 β + n1 ∞ = ( n2 – n1 ) γ
From Δ OMN , tan∞ = ∞ = NM = NM
OM u
From Δ MNC , tanγ = γ = NM = NM
MC R
From Δ MNI , tanβ = β = NM = NM
MI v
n2 + n1 = ( n2 - n1 )

v u R
Applying sign convention u = -ve , v = +ve and R = +ve we get ,

n2 – n1 = ( n2 – n1 )
v u R

8. Lens makers formula :

For refraction at first surface APB, we get n2 – n1 = ( n2 – n1 ) --------- (1)
v’ u R1

v’ = PI’ and R1 = Radius of curvature of 1st surface.

For refraction at second surface AP’B, we get n1 – n2 = ( n1 – n2 ) --------- (2)
v v’ R2
v = PI nd
and R2 = Radius of curvature of 2 surface.

Adding (1) + (2) , we get n1 [ 1 – 1 ] = ( n2 – n1 ) [ 1 – 1 ]
v u R1 R2
Dividing both sides by n1 , we get,
1 – 1 = ( n2 – 1 ) [ 1 – 1 ]
v u n1 R1 R2
Let the lens be placed in air , n1 =1 and n2 = n.

1 – 1 = ( n – 1 ) [ 1 – 1 ] ---------- (3)
v u R1 R2
When u = ∞ , then v = f
1 = (n –1) [ 1 – 1 ] ------------------ (4)
f R1 R2
This is the Lens makers formula.
Compairing (3) and (4) , we get 1 – 1 = 1 .This is the lens formula.
v u f
9. Power of a lens
Power of a lens is defined the reciprocal of its focal length expressed in metre..
P= 1 S.I. unit of power is dioptre (D).
f

10. Refraction through a prism.

In the quadrilateral AQNR,

From Δ QNR ,

From the above equations, r1 + r2 = A ---------- (1)
We know, exterior angle = sum of interior angles, thus d = (i - r1 ) + (e - r2 )
d = (i + e – ( r1 + r2 ) ) or d = (i + e - A)
At the minimum deviation, d = D, і = e, r1 = r2 = r therefore

Equation (1) becomes, 2r =A or r = A
2
Equation (2) becomes, A + D = 2i or i =A+ D
2
Refractive index , sin A + D
n = 2
sin A
2
i – d curve

12. Microscopes
I. Simple microscope:

The object to be magnified is placed in between the principle focus and the optic centre of the
lens. A magnified , virtual erect image is formed at the least distance of distinct vision(D).

Magnifications, m =1- V
f
At the least distance of distinct vision , V = - D

Magnification , m= 1+ D
f

II. Compound microscope :

Magnification due to objective , m0 = L where L – distance between objective and
f0 eyepiece.
f0 - focal length of objective.

As the first image is near the focal point of eyepiece , the magnification due to eyepiece is
me = 1 + D
fe
When the image is formed at infinity ,
me = D
fe
Total magnification, m = m0 x me
m = L x D
f0 fe
Important Questions :
1. Consider refraction of light at a spherical surface separating two media of refractive indices n1 and
n2 (n2 > n1 ). With the help of ray diagram show the formation of image of point object placed in
the medium of refractive index n1 .
2. i) Define power of a lens. What is its unit ?
ii) Derive the lens makers formula.
3. i) Draw the diagram showing the path of a monochromatic light through a triangular prism.
ii) What do you mean by angle of minimum deviation ?
Iii) Arrive at the expression sin A + D
n = 2
sin A
2
4. Draw the ray diagram showing the formation of image by a compound microscope. Derive
expression for its magnification.
5. Draw ray diagram of simple microscope that uses a single convex lens. Derive an expression for its

linear magnification.
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CHAPTER 10
WAVE OPTICS
Topics Included
 Wavefront
 Huygen’s Principle
 Superposition Principle
 Coherent Sources of light
 Interference Of Light
 Polarization
 Law of Malus

Wavefront
When a wave propagates through a medium, the particles of the medium vibrate about their mean
position. The locus of all particles of the medium vibrating in the same phase at a given instant constitute a
wave front.
Wave fronts are of three types

Huygen’s Principle
According to this principle
1) “Each point of the given wave front is a source of
secondary wavelets. The secondary wavelets from these
points spread out in all direction with speed of the wave”.
2) “The position of a wave front after a time t is given
by the forward envelop of these secondary wave lets”.

Superposition Principle
When two or more wave propagates through a medium simultaneously; the resultant displacement of a
particular point in the medium at any instant will be the vector sum of displacement due to individual waves
ie Y = Y1 + Y2 + Y3 + Y4
a) Constructive Superposition
Crest of one wave falls on crest of other ( or trough on trough )
the amplitude and intensity of resultant wave increases.
b) Destructive superposition
Here crests of one wave falls on trough of other,
the amplitude and intensity of resultant wave decreases.
Coherent sources of light
Two light sources are said to be coherent if they emit light waves of same frequency, (wave length), and
nearly same amplitude also the waves must be in the same phase or must have constant phase difference.

Interference
The phenomenon of re distribution of energy due to the super position of light waves from two coherent
sources is called interference.
Condition for constructive interference
Phase difference Φ = 2 n π
Path difference δ = n λ

Where n = 0,1,2,3, ……
Amplitude of resultant wave
Expression for fringe width
β = λ D /d
R max = (a + b )
Intensity of Resultant wave Expression for fringe
I max = (a + b ) 2 width
Condition for destructive interference
Phase difference Φ = (2 n + 1) π
β = λ D /d
Path difference δ = (2n + 1) λ
Where n = ,1,2,3, …… 2
Amplitude of resultant wave
R mini = (a - b )
Intensity of Resultant wave
I mini = (a - b ) 2
Polarization
When un Polarised light is allowed to pass through certain crystals like Tourmaline, calcite etc.. the light
emerging from the crystal contains electric field vectors vibrating in a single plane (Polarised light) this
phenomenon is called Polarizations.

Law of Malus
“The intensity of polarized light transmitted through the analyzer varies as the squire of cosine of angle
between the pass axis of polarizer and analyzer”. Let
I → intensity of light transmitted through the analyzer
Io → the intensity of light falling on analyzer
θ → angle between the pass axis of polarizer and analyzer
I ά cos2θ
I = Io cos2θ

Important Previous Questions
1. State Huygen’s Principle
2. Define Interference of light. Give the expression of fringe width.

3. Define polarization.
4. State Malu’s Law
5. State Brewster’s Law
6. Explain Polarization by scattering.

Chapter 11
DUAL NATURE OF RADIATION AND MATTER

Different methods to emit electrons from metal surface
1.Thermionic emission 2.Field emission 3.Secondary emission 4.Photo-electric emission
Photo electric effect
The phenomenon of ejection of electrons from metal surface when light of suitable
frequency falls on it is called photoelectric effect
Threshold frequency (υo)
The minimum frequency of incident radiation to emits electron from metal surface.
Work Function (ϕo)
The minimum energy required to eject an electron from the metal surface is called
work function. ϕo = hυo
Effect of intensity of light on photoelectric current

The photoelectriccurrent
increases with intensity of
incident light

Effect of potential on photoelectric current
Potential –photo current graph (for different intensities I1,I2&I3 with fixed frequency)

The photoelectric current increases with increase
in the potential applied to the collector.
Stopping potential doesn’t change with intensity
of incident radiation

Potential-photo electric current graph (for fixed intensity with different frequenciesυ1,υ2 &υ3)
Stopping potential changes with change in frequency
of incident radiation
Stopping potential is directly proportional to fre quency

EINSTEIN’S PHOTOELECTRIC EQUATION
Kmax=Maximum Kinetic
h 0 + KmaxKmax = h– 0 Energy of ejected
electron
0= h= work function

CHAPTER 3
ATOMS

BOHR MODEL OF THE HYDROGEN ATOM
Postulates:
1. An electron in an atom could revolve in certain stable orbits without the emission of
radiant energy.

2. The electrons can orbit only those orbits for which the angular momentum is an integral
multiple h/2π.
Ie, angular momentum, L = mvr = nh / 2π
Here n is called the principal quantum numb er and it has the integral values 1,2,3....

3. When an electron jumps from higher energy orbit to lower energy orbit, a photon is
emitted having energy equal to the energy difference between the initial and final states.

If Ei and Ef are the energies associated with the orbits of principal quantam number ni and
nf respectively (n i > n f ), then the amount of energy radiated in the foam of photon is

hν = Ei -Ef
Here ν is the frequency of photon

rn α n2

The radii of the stationary orbits are in the ratio, 1 2 :2 2 :3 2 :..... or 1: 4: 9: .....

is Bohr radius. ( radius of the lowest orbit )

a0 =5.29 X 10 -11 m

CHAPTER 13
NUCLEI

Mass Defect

The difference in the actual mass of a nucleus and its constituents is called the mass defect
(Δ M)
Mass defect , Δ M = [ Zmp + ( A – Z ) mn - M ]

mp – Mass of proton, mn - mass of neutron and M – actual mass of nucleus.

Binding energy ( Eb )

Binding energy is the energy required to hold the nucleons together.
Binding Energy, Eb = ∆M c2
If the mass defect is in atomic mass unit, then binding energy in MeV is
Binding Energy, Eb = ∆M × 931 MeV.

Nuclear fusion
The process of combining two lighter nuclei into a stable and heavier nuclei.
The energy liberated by the sun and other stars is due to the nuclear fusion reactions

occurring at very high stellar temperatures .
Hydrogen bomb is a fusion bomb.

Atom bomb is a fission bomb.

Isotopes
Isotopes are variants of a particular element with different number of neutrons.
For example , the two isotopes of uranium are 23592U and 23992U .

Isobars
Isobars are elements that have the same number of nucleons ( sum of protons and
neutrons ).
For example 4016S and 4017Cl
Isotones
Isotones are atoms that have the same neutron number but different proton
number.
For example 3818 Ar and 3919 K

Important Questions
1. What is the source of solar energy ?
2. Define binding energy of a nucleus.
3. What do you meant by mass defect of a nucleus?

Chapter 14
Semiconductor Electronics:
Materials, Devices and Simple Circuits Introduction
Classification of Metals, Conductors and Semiconductors
On the basis of conductivity: On the basis of the relative values of electrical conductivity (σ) or
resistivity (ρ = 1/σ ), the solids are broadly classified as:

Classification of Metals, Conductors and Semiconductors On the basis of energy bands
i. metals

In some metals, the conduction band is partially filled and the valence band is partially empty
with small energy gap and in some others the conduction and valance bands overlap. When there
is overlap electrons from valence band can easily move into the conduction band. Therefore, the
resistance of such materials is low or the conductivity is high.
ii. Insulators

In insulators a large band gap, Eg > 3 eV. There are no electrons in the conduction band, and
therefore no electrical conduction is possible. The energy gap is so large that electrons cannot be
excited from the valence band to the conduction band by thermal excitation.

iii. Semiconductors

In semiconductors a finite but small band gap (Eg < 3 eV) exists. Because of the small band gap,
at room temperature some electrons from valence band can acquire enough energy to cross the
energy gap and enter the conduction band. These electrons (though small in numbers) can move
in the conduction band. Hence, the resistance of semiconductors is lower than that of insulators.

When the electrons from valence band move to the conduction band vacant energy levels will be
created in the valence band. This vacancy of electrons is called hole. Other valence electrons can
move to this hole thereby producing hole current.
Intrinsic Semiconductor
Pure semiconductors are called ‘intrinsic semiconductors’.
Si and Ge have four valence electrons. In a pure Si or Ge crystal, each atom makes covalent
bond with four neighbouring atoms and share the four valence electrons.
In intrinsic semiconductors, the number of free electrons, ne is equal to the number of holes, nh.

where ni is called intrinsic carrier concentration
The total current, I is thus the sum of the electron current Ie and the hole current Ih:
I = Ie + Ih

Energy-Band Diagram of an Intrinsic Semiconductor at T=0K
An intrinsic semiconductor will behave like an insulator at T = 0 K .

Energy-Band Diagram of an Intrinsic Semiconductor at T > 0K
At temperatures (T > 0K), some electrons are excited from the valence band to the conduction
band, leaving equal number of holes there.

Extrinsic Semiconductor

When a small amount of a suitable impurity is added to the pure semiconductor, the conductivity
of the semiconductor is increased . Such materials are known as extrinsic semiconductors or
impurity semiconductors.
n-type semiconductor
n-type semiconductor is obtained by doping Si or Ge with pentavalent atoms (donors) like As,
Sb, P, etc.
For n-type semiconductors, ne >> nh
Here electrons become the majority carriers and holes the minority carriers. The electron
and hole concentration in a semiconductor in thermal equilibrium is given by

Energy bands of n-type semiconductor at T > 0K

p-type semiconductor
p-type semiconductor is obtained when Si or Ge is doped with a trivalent impurity like A𝑙, B, In,
etc.
For p-type semiconductors nh >> ne
Here holes become the majority carriers and electrons the minority carriers. The electron and
hole concentration in a semiconductor in thermal equilibrium is given by

Energy bands of p-type semiconductor at T > 0K
For p-type semiconductor, the acceptor energy level EA is slightly above the top EV of the
valence band. With very small supply of energy an electron from the valence band can jump to
the level EA and ionize the acceptor.
Negatively.

p-n junction

A p-n junction can be formed by adding a small quantity of pentavalent impurity to a p-type
semiconductor or by adding a small quantity of trivalent impurity to an n-type semiconductor.
Two important processes occur during the formation of a p-n junction: diffusion and drift

Barrier Potential
The loss of electrons from the n-region and the gain of electron by the p- region causes a
difference of potential across the junction of the two regions. Since this potential tends to
prevent the movement of electron from the n region into the p region, it is often called a barrier
potential.
The barrier potential of a Ge diode is 0.2Vand that of a Si diode is 0.7V.

Semiconductor Diode

A semiconductor diode is basically a p-n junction with metallic contacts provided at the ends for
the application of an external voltage. It is a two terminal device.
Symbol of a p-n junction Diode

p-n junction diode under forward bias

If p-side of the diode is connected to the positive terminal and n-side to the negative terminal of
the battery, it is said to be forward biased.
• Conducts current and it is due to majority charge carriers
• Width of depletion region decreases

• Potential barrier decreases.
p-n junction diode under reverse bias

If n-side of the diode is connected to the positive terminal and p-side to the negative terminal of
the battery, it is said to be reverse biased.

• Doesn’t Conduct current
• Width of depletion region increases
• There will be a very small current due to minority charge carriers.

V-I characteristics of a silicon diode.

In forward bias, the current first increases very slowly, till the voltage across the diode crosses a
certain value. . This voltage is called the threshold voltage or cut-in voltage (0.2V for
germanium diode and 0.7 V for silicon diode).
▪ After threshold voltage, the diode current increases significantly , even for a very small
increase in the diode bias voltage.
▪ For the diode in reverse bias, the current is very small (~μA) and almost remains constant
with change in bias. It is called reverse saturation current. However, at very high reverse bias
called break down voltage Vbr, the current suddenly increases. The general purpose diode are
not used beyond the reverse saturation current region.
Threshold Voltage
The forward voltage beyond which the diode current increases significantly is called threshold
voltage or cut-in voltage.
Break down Voltage
The reverse voltage at which the reverse current increases suddenly is called break down
voltage.
Dynamic Resistance(rd)
Dynamic resistance is defined as the ratio of small change in voltage ΔV to a small change in
current ΔI

Application of Junction Diode as a Rectifier
The diode allows current to pass only when it is forward biased.

Rectifier
The process of conversion of ac voltage to dc voltage is called rectification and the circuit used
for rectification is called rectifier.
Half wave Rectifier

In the positive half-cycle of ac there is a current through the load resistor 𝑅𝐿 and we get an
output voltage, whereas there is no current in the negative half cycle. Since the rectified output
of this circuit is only for half of the input ac wave it is called as half-wave rectifier.
In the positive half-cycle of ac there is a current through the load resistor 𝑅𝐿 and we get an
output voltage, whereas there is no current in the negative half cycle. Since the rectified output
of this circuit is only for half of the input ac wave it is called as half-wave rectifier.
Input ac voltage and output voltage waveforms from the rectifier circuit

Full wave rectifier

For a full-wave rectifier the secondary of the transformer is provided with a centre tapping and
so it is called centre-tap transformer.
• During this positive half cycle, diode 𝐷1 gets forward biased and conducts ,while
𝐷2 being reverse biased is not conducting. Hence we get an output current and a
output voltage across the load resistor 𝑅𝐿.
• During negative half cycle, diode 𝐷1 would not conduct but diode 𝐷2 conducts,
giving an output current and output voltage across 𝑅𝐿 in the same directionas in
positive half.
• Thus, we get output voltage during both the positive as well as the negative half of
the cycle. This is a more efficient circuit for getting rectified voltage or current than
the halfwave rectifier.
Input ac voltage and output voltage waveforms from the rectifier circuit.