KENDRIYA VIDYALAYA
SANGATHAN
RAIPUR REGION
STUDY MODULES CLASS-XII
PHYSICS
2012-13
a
A
b
B
c
C
cal
d
dB
e
eV
E
f
F
g
G
h
H
Hz
i
I
J
k
kg
K
l
L
m
M
n
N
NA
o
p
P
Pa
q
Q
r
R
rms
s
S
t
T
Symbols used in Physics
acceleration (m/s2)
Amplitude (m), Ampere (fundamental unit of current)
y-intercept
magnetic field strength (T = 104 G), Bulk Modulus (N/m2)
speed of light (3 x 108 m/s), specific heat (J/(kg•K))
Coulomb (unit of charge = A•s), center of curvature (m), Celcius (unit of Temperature),
Capacitance (F)
calories (unit of heat energy = 4.186 J)
distance (m), deci (prefix = 10-1)
decibels (unit of sound intensity)
symbol for charge of an electron (1.6 x 10-19), efficiency (dimensionless)
electron-Volt (unit of energy = 1.6 x10-19 J)
energy (J), electric field intensity (N/C)
frequency (Hz = s-1), femto (prefix = 10-15), force of friction, focal length (m), final value
(subscript)
force (N), farad (unit of capacitance = A2•s4/(kg•m2), Farenheit (unit of Temperature)
acceleration due to gravity (9.8 m/s2), gram (unit of mass)
giga (prefix = 109) Universal Gravitational Constant (6.67 x 10-11 m3/(kg•s2)
Planck’s constant (6.626 x 10-34 J•s), height (m)
Henry (unit of inductance)
Hertz (unit of frequency)
current (A), initial value (subscript)
moment of inertia (kg•m2), Luminous Intensity (W/m2)
Joule (unit of energy / work), impulse (N•s)
kilo (prefix = 103)spring constant (N/m = kg/s2), thermal conductivity (W/(m•K)), Boltzman
constant (J/K)
kilogram (fundamental unit of mass)
Kelvin (fundamental unit of temperature), kinetic energy (J)
length (m)
liter (unit of volume), latent heat (J/kg), angular momentum (J•s), inductance (H), latent
heat (J/kg)
mass (kg), meter (fundamental unit of distance), slope, milli (prefix = 10-3)
mega (prefix = 106)
nano (prefix = 10-9), index of refraction (dimensionless), number of moles
Newton (unit of force), Normal Force (N), number of particles
Avogadro’s number (6.022 x 1023 particles / mole)
subscript - initial
momentum (N•s), pico (prefix = 10-12), proton
power (W), pressure (N/m2, atm, mmHg, psi, Pa)
Pascals (unit of pressure = N/m2)
charge (C)
charge (C), Heat (J)
radius (m)
ideal gas constant (8.31 J/mol K), resistance (Ω)
root – mean – square
seconds (fundamental unit of time)
entropy (J/K)
time (s)
temperature (°C, K), period (s), tension (F), Tesla (unit of magnetic field)
u
U
v
V
w
W
x
X
y
Y
z
Z
atomic mass unit (also amu, 1.6 x10-27 kg)
potential energy (J)
velocity or speed (m/s)
Volt (unit of potential difference), Voltage or Potential Difference, volume (L or m3)
weight (N)
work (J), Watt (unit of power)
distance – usually horizontal
reactance – capacitive or inductive (Ω)
distance – usually vertical
Young’s Modulus (N/m2)
distance – usually 3rd dimension
impedance (Ω)
Greek Letters
α (alpha) - angular velocity (rad/s2), proportional symbol, temperature coefficient of linear
expansion (K-1)
α – type of radiation
β (beta) – temperature of area expansion (K-1), type of radiation, electron
γ (gamma) – relativistic gamma factor (dimensionless), type of radiation
δ (delta)– instantaneous change
∆ (capital delta) – finite change
ε (epsilon) – electrical permittivity
θ (theta) – angle (rad or °)
κ (kappa) – dielectric constant (dimensionless)
λ (lambda) – wavelength (m), ), linear charge density (C/m)
µ (mu) – coefficient of friction (dimensionless), micro (prefix = 10-6), magnetic permeability
ν (nu) – frequency (Hz, s-1)
π (pi) – ratio of circumference to diameter (3.1416..)
ρ (rho) – density (kg /m3), resistivity (Ω•m)
σ (sigma) – surface charge density (C/m2), Stefan Boltzmann constant (5.6704 × 10-8 kg s-3 K-4)
∑ (capital sigma) – summation or sum of
τ (tau) – torque (N•m), time constant (s)
φ (phi) – angle (rad)
Φ (capital phi) – flux – either magnetic or electric
Ψ (psi) – wave function (dimensionless)
ω (omega) – angular velocity (rad/s)
Ω (capital omega) – Ohms (unit of resistance = V/A)
KENDRIYA VIDYALYA SANGATHAN
RAIPUR REGION
STUDY MATERIAL
FOR LOW ACHIEVERS
PHYSICS
CLASS XII
SESSION : 2012-13
GUIDANCE
: KVS REGIONAL OFFICE
RAIPUR
Prepared By :
NAGENDRA SHARMA
PRINCIPAL
KENDRIYA VIDYALAYA AMBIKAPUR
IN CONSULTATION WITH PGTs
OF
KV AMBIKAPUR, KV KUSMUNDA, KV DURG &KV BILASPUR
CONTENTS
PART - I
Electrostatics*
Unit – 1* Marks – 8* KV Ambikapur
1 (a) Electrostatic Charges
1 (b) Electrostatic Field
1 (c) Electrostatic Potential and Flux
1 (d) Capacitance
Current Electricity*
Unit – 2* Marks – 7*
KV Kusmunda
Unit – 3* Marks – 8*
KV Kusmunda
2 (a) Electric Current and Resistance
2 (b) Electric Measurement
Magnetic Effects of*
Current and Magnetism
3 (a) Magnetic Field Due to Current
3 (b) Forces on Charged Particles in Electric & Magnetic Fields
3 (c) Magnets and Earth’s Magnetism
Electromagnetic Induction* Unit – 4* Marks – 8*
& Alternating Currents
KV Ambikapur
4 (a) Electromagnetic Induction
4 (b) Alternating Currents & Electrical Devices
Electromagnetic Waves*
Unit – 5* Marks – 3*
5. Electromagnetic Waves
KV Ambikapur
142 - 146
PART - II
Optics*
Unit – 6* Marks – 14*
KV Durg
6 (a) Reflection, Refraction & Dispersion of Light
6 (b) Optical Instruments
6 (c) Huyghens Principle and Interference
6 (d) Diffraction and Polarization
Dual Nature of Matter*
Unit – 7* Marks – 4*
KV Durg
Unit – 8* Marks – 6*
KV Bilaspur
Unit – 9* Marks – 7*
KV Bilaspur
7. Dual Nature of Matter and Radiation
Atoms and Nuclei*
8. Atoms and Nuclei
Electronic Devices*
9 (a) Conductors, Insulators and Semi-conductors
9 (b) Semi-conductor Devices
9 (c) Logic Gates
Communication Systems* Unit – 10* Marks – 5*
10 (a) Analog and Digital Communication
10 (b) Space Communication
KV Bilaspur
1(a)
ELECTROSTATC CHARGES
IMPORTANT CONCEPTS
1. Electrostatics (Static Electricity or Frictional Electricity): The branch of Physics,
which deals with the study of electric charges at rest, the forces between the static
charges, fields and potentials due to these charges is called Electrostatics or Static
Electricity or Frictional Electricity.
2. Electrification: When certain pair of substances are rubbed together, they acquire a
property of attracting small bits of paper, light feathers etc. towards them. The substances
are said to have been electrified or charged and the process is known as electrification.
3. Electric Charge: Charge is something possessed by material object that makes it
possible to exert electrical force and to respond to electrical force. Excess or deficiency
of electron on a material object constitutes an electric charge.
4. Polarity of Charge: The property which differentiates the two kinds of charges is called
the polarity of charge. When a body possesses no charge, it is said to be neutral.
Conventionally two kinds of charges are said to be positive & negative.
5. Dielectrics: Insulating materials which transmit electric effects without conducting are
known as dielectrics. Insulators are the materials which can not conduct electricity and
are also called Dielectrics.
6. Conductor: A substance which can be used to carry or conduct electric charge from one
place to the other is called a conductor.
7. Charging by Induction: In charging by induction, a charged body A imparts to another
body B, some charge of opposite sign without any actual contact between A and B.
Obviously, body A shall not lose any charge as it is not in contact with body B.
8. Quantization of Electric Charge: The property by virtue of which all free charges are
integral multiple of a basic unit of charge represented by e is known as the quantization
of electric charge, where e = 1.6×10−19 coulombs.
Thus charge q of a body is always given by
q = ne
where n is any integer, positive or negative.
9. Conservation of Electric Charge: The property by virtue of which total charge of an
isolated system always remains constant is known as the conservation of electric
charge.
10. Additivity of Charge: This is a property by virtue of which total charge of a system is
obtained by adding algebraically all the charges present anywhere on the system.
Charges adds up like real numbers i. e., they are Scalars more clearly if any system has n number of
charges
q1,
q2,
q3,
qn
then
total
charge
of
the
system
is
q = q1 + q2 + q3 + ................ qn
Proper sign have to be used while adding the charges for example if
q1 = +1C
q2 = -2C
q3 = +4C
then total charge of the system is
q = q1 + q2 + q3
q = (+1) + (-2) + (+4) C
q = (+3) C
11. Coulomb’s Law: The force of interaction between any two point charges is directly
proportional to the product of the charges and inversely proportional to the square of the
distance between them. The force acts always along the line joining the two charges.
Consider two point charges q1 and q2 at points with position vector r1 and r2 with respect to the
origin
vector r21= r2 - r1 is the difference between r2 and r1 and the distance of separation r is the
magnitude of vector r21.
Point wise it can be written as
r1 = position vector of charge q1 with respect to origin
r2 = position vector of charge q2 with respect to origin
r21 = vector from 1 to 2 (r2 - r1)
r12 = -r21 = vector from 2 to 1 (r1 - r2)
r = r12 = r21 = distance between 1 and 2.
Coulomb's law can then be expressed as
F21 = force on q2 due to q1
= K [(q1 q2) / r2] (r21/r) (2a)
F12 = force on q1 due to q2
= -F21 = K [(q1 q2) / r2] (r12/r)
(2b)
K = some position constant
From these two equation we see that electric forces exerted by two charges on each other are
equal in magnitude but are opposite in direction
In above equation, we find a positive constant K and experimentally found value of K is
K = 8.98755 × 10 9 Nm2/C2
K ≅ 9 × 10 9 Nm2/C2
sometimes K is written as 1/4π ε0 where ε0 is the permittivity of the vacuum whose value is
K = 1/4πε0
(ε0 = 9 × 10 -12 C2/Nm2)
12. Unit of Charge: Unit charge in SI (i.e. one coulomb) is that charge which when placed in
vacuum at a distance of one metre from an equal and similar charge would repel it with a
force of 9 × 10 9 Newton.
13. Dielectric Constant: Dielectric constant of a medium is defined as the ratio of absolute
electrical permittivity of the medium (ε) to the absolute permittivity of vacuum or free
space (ε0).
Dielectric Constant, also called relative electrical permittivity of the medium and
denoted by εr, is thus given by
εr = ε/ ε0
14. Principle of Superposition (Forces between Multiple Charges): Force on any charge
due to a number of other charges at rest is the vector sum of all the forces on that charge
due to other charges, taken one at a time. The forces due to individual charges are
unaffected due to presence of other charges.
If a system of charges has n number of charges say q 1, q2, ...................., qn, then total force
on
charge
q1
according
to
principle
of
superposition
is
F
=
F12
+
F13
+
..................................
F1n
Where F12 is force on q1 due to q2 and F13 is force on q1 due to q3 and so on.
15. Charge Density: The linear, surface, or volume charge density is the amount of electric
charge in a line, surface, or volume per unit length, per unit area or per unit volume
respectively.. It is measured in coulombs per metre (C/m), square metre (C/m²), or cubic
metre (C/m³), respectively. Since there are positive as well as negative charges, the
charge density can take on negative values. Like any density it can depend on position. It
should not be confused with the charge carrier density. As related to chemistry, it can
refer to the charge distribution over the volume of a particle, molecule, or atom.
Therefore, a lithium cation will carry a higher charge density than a sodium cation due to
its smaller ionic radius.
IMPORTANT FORMULAE, UNITS AND CONSTANTS
1. q = ne, where q is the amount of the charge, n is an integer and e is electronic charge.
The constant e is given by: e = 1.6×10−19 coulombs.
2. F = K [(q1 q2) / r2], where F is the force of interaction between q1 & q2 , the two point
3.
charges separated by a distance r in vacuum or free space. The constant K here is given by: K ≅
9 × 10 9 Nm2/C2 in vacuum or free space.
The constant K above for any medium (other than vacuum or free space) is given by:
K=
1/4πε, where ε is known as the absolute permittivity of that medium
4. εr = ε/ ε0, where εr is the relative permittivity or the dielectric constant of the medium, ε is the
absolute permittivity of the medium and ε0 is the absolute permittivity of free space or vacuum. ε0
= 8.854 × 10 -12 C2/Nm2 ≅9 × 10 -12 C2/Nm2
5. λ = q / l, where λ is the Linear Charge density in homogeneous distribution of charge q
over a length l..
6. σ = q / A, where σ is the Surface Charge density in homogeneous distribution of charge
q over a surface area A.
7. ρ = q / V, where ρ is the Surface Charge density in homogeneous distribution of charge
q over a given volume V.
8. F = F0 / K, where F is force in the medium of dielectric constant K & F0 is force in vacuum or
free space between two point charges.
VERY SHORT ANSWER TYPE QUESTIONS (CARRYING 1 MARK EACH)
1. What is SI unit of electrostatic charge?
Ans. Coulomb (C)
2. What is c.g.s. unit of electrostatic charge?
Ans. Electro Static Unit (e.s.u.) or Stat Coulomb.
3. How are coulomb and stat coulomb units related to each other?
Ans. 1 coulomb = 3 ×109 stat coulomb (or e.s.u.)
4. What is one e.m.u. of electrostatic charge?
Ans. Electro Magnetic Unit of charge also called e.m.u. of charge is given by: 1 e.m.u.
of charge = 10 coulomb = 3 ×1010 stat coulomb (or e.s.u.).
5. How many electrons make up one coulomb of negative charge?
Ans. From q = ne, n = q / e = 1/ 1.6×10−19 = 6.25 ×1018 .
6. Give two points of distinction between charge and mass?
Ans. Charge can be positive, negative or zero. But mass is a positive quantity. Further,
electric charge is always conserved. But mass is not conserved.
7. Does Coulomb’s law of electric force obey Newton’s third law of motion? Ans. Yes,
it obeys. Forces exerted by two charges on each other are always equal and opposite.
8. What is the value of the charge on a body which carries 20 excess electrons?
Ans. Here n = 20, q = ?, e = -1.6×10−19 C. As q = ne, q = 20 (1.6×10−19), hence charge q
= -3.2×10−18 C.
9. Is a charge of 5.8×10−18 C possible?
Ans. From q = ne, n = q/e = 5.8×10−18 / 1.6×10−19 = 36.25; As n is not an integer, this
value of charge is not possible.
10. What is the basic cause of quantization of charge?
Ans. Only integral number of electrons can be transferred from one body to another.
11. What is the cause of charging?
Ans. The cause of charging is actual transfer of electrons from one body to another.
Besides, charging can also be done by induction, when already charged body is brought
very near to the neutral body.
12. Is mass of a body affected on charging?
Ans. Yes, very slightly. The negatively charged body which gains electrons, gains some
mass. It’s however negligible on macroscopic point of view.
13. Does motion of a body affect its charge?
Ans. No, charge on a body does not change with motion of the body.
14. Name any two basic properties of electric charge?
Ans. (i) Quantization of charge (ii) Conservation of charge.
15. How do the charges interact?
Ans. Like charges repel and unlike charges attract each other. However charged body
always slightly attracts neutral body.
16. A glass rod rubbed with silk acquires a charge of +1.6 × 10−12 C. What is the charge
on the silk?
Ans. Charge on silk is equal and opposite to charge on glass rod i.e. q = -1.6×10−12 C.
17. Electrostatic forces are much stronger than gravitational forces. Give one example.
Ans. A charged glass rod can lift a piece of paper against the gravitational pull of earth
on this piece.
18. By what factor is the electric force between two electrons greater than the
gravitational
force
between
them?
Ans.
19. How is the force between two charges affected when dielectric constant of the
medium in which they are held increases?
Ans. As F = F0 / K, therefore force deceases, when K increases.
20. What is the dimensional formula for ε0 (absolute permittivity
Ans. [M-1L-3 T4A2]
of free space) ?
21. What is the relevance of large value of K (=81) for water?
Ans. It makes water a great solvent. This is because binding force of attraction between
oppositely charged ions of the substance in water becomes 1/81 of the force between
these ions in air.
22. Can ever photons have a charge? If not, why?
Ans. No, photons can never have a charge. This is because charge cannot exist without
rest mass.
SHORT ANSWER TYPE QUESTIONS (CARRYING 2 MARKS EACH)
1. Dielectric constant of the medium is unity. What will be its permittivity?
Ans. K = εr = ε/ ε0 , therefore ε = K ε0 = 1 × 8.854 × 10 -12 C2/Nm2
i.e
ε = 8.854 × 10 -12 C2/Nm2
2. A +2.0 μC charge is located on the x-axis at +0.3 m and another at -0.3 m. What is
the force of interaction between the two charges?
Ans. F = K [(q1 q2) / r2] = 9×109 × 2 ×10-6 × 2 ×10-6 / 0.6 × 0.6 = 0.1 N repulsive.
3. An attractive force of 5N is acting between two charges of +2.0 μC & -2.0 μC placed
at some distance. If the charges are mutually touched and placed again at the same
distance, what will be the new force between them?
Ans. On touching, charges neutralize. Therefore, F = 0.
4. Two point charges of +2 μC and +6 μC repel each other with a force of 12N. If each
is given an additional charge of -4 μC, what will be the new force?
Ans. q1 = +2 μC , q2 =+6 μC , F = 12N
q1’ = +2-4 = -2 μC and q2’ = +6-4 =+2 μC, F’ = ?
F’ / F = q1’ q2’ / q1 q2 = (-2) (2) / (2) (6) = - 1/3,
F’ = - F/3 = -12/3 = - 4N (attractive).
1(b)
ELECTROSTATC FIELD
IMPORTANT CONCEPTS, FORMULAE AND DERIVATIONS.
1. Electric Field (as the space property): Electric field due to a given charge is defined as
the space around the charge in which electrostatic force of attraction or repulsion due to
the charge can be experienced by any other charge (the test charge).
2. Electric Field Intensity or Electric Field: The electric field intensity (or in short called
the Electric Field) at any point is defined as the force experienced by unit positive charge
placed at that point. Electric Field is a vector quantity.
3. Relation between Electric Field Intensity and Force: If F is a force acting on a small
test charge + q0 at any point r, then electric field intensity at this point is given by:
E (r) = F / q0,The SI unit of electric field intensity is Newton per Coulomb (N/C).
4. Electric Field due to a group of charges: Electric field at any point due to a group of
point charges is equal to the vector sum of the electric field intensities due to individual
charges at the same point.
Properties (in electrostatics)
Illustration of the electric field surrounding a positive (red) and a negative (green) charge.
Electric field is dependent on position. The electric field due to any single charge falls off as the
square of the distance from that charge.
Electric fields follow the superposition principle. If more than one charge is present, the total
electric field at any point is equal to the vector sum of the respective electric fields that each
object would create in the absence of the others.
Coulomb's law
The electric field surrounding a point charge is given by Coulomb's law:
where
5.
6.
7.
8.
9.
Q is the charge of the particle creating the electric force,
r is the distance from the particle with charge Q to the E-field evaluation point,
is the unit vector pointing from the particle with charge Q to the E-field evaluation
point,
is the electric constant.
Physical significance of electric field: From the knowledge of electric intensity E at any
point r, we can readily calculate the magnitude and direction of force experienced by any
charge q held that point, i.e. F = q E.
Electric Field Lines: Electric field line is a path, straight or curved, such that tangent to
it at any point gives the direction of electric field intensity at that point.
Electric Dipole: An electric dipole consists of a pair of equal and opposite point charges
separated by some small distance.
Electric Dipole Moment: Dipole Moment (p) is a measure of the strength of electric
dipole. It is a vector quantity whose magnitude is equal to the product of the magnitude
of either charge or the distance between them. The direction of dipole moment is from –q
to +q.
Ideal Dipole or Point Dipole: If charge q gets larger, and the distance d (=2a, where a is
distance of either charge from the centre of the dipole) gets smaller and smaller, keeping
the product p = q×d = constant, we get what is known as an ideal dipole or point dipole.
Dipole Moment
The electric dipole moment for a pair of opposite charges of magnitude q is
defined as the magnitude of the charge times the distance between them and the
defined direction is toward the positive charge. It is a useful concept in atoms and
molecules where the effects of charge separation are measurable, but the distances
between the charges are too small to be easily measurable. It is also a useful
concept in dielectrics and other applications in solid and liquid materials
10. Dipole Field: The dipole field is the electric field produced by an electric dipole. It is the
space around the dipole in which the electric effect of the dipole can be experienced.
11. The Field of an Electric Dipole
Electric field produced by a dipole is known as dipole field.
Let +q and -q be equal and opposite point charges separated by a small distance 2l. The strength
of an electric dipole is measured by a vector quantity known as electric dipole moment known as
electric dipole moment
charges, that is
The direction of
Coulomb-meter.
, which is the product of the charge and separation between the
is always from negative to positive. The SI unit of dipole movement is
(a) For points on axial line
The axial line of a dipole is the line passing through the positive and negative charges of the
electric dipole.
Consider a system of charges (-q and +q) separated by a distance 2a. Let 'P' be any point on an
axis where the field intensity is to be determined.
Electric field at P (EB) due to +q
Electric field at P due to -q (EA)
Net field at P is given by
Simplifying, we get
As a special case :
(b) For points on the equatorial line
An equatorial line of a dipole is the line perpendicular to the axial line and passing through a point
mid way between the charges.
Electric Field Intensity due to a Dipole at a Point lying on the Perpendicular
Bisector of a Dipole
Consider a dipole consisting of -q and +q separated by a distance 2a. Let P be a point Consider a
point P on the equatorial line.
The resultant intensity is the vector sum of the intensities along PA and PB. E A and EB can be
resolved into vertical and horizontal components. The vertical components of E A and EB cancel
each other as they are equal and oppositely directed. It is the horizontal components which add
up to give the resultant field.
E = 2EA cos 
As 2qa = p
As a special case,
We find that at very far off points i.e., 2a < r.< r.
Electricity intensity at an axial point is twice the electric intensity on the equatorial line.
The simplest set of sources that can occur
with electricity but not with gravity is the
dipole, consisting of a positive charge and
a negative charge with equal magnitudes.
More generally, an electric dipole can be
any object with an imbalance of positive
charge on one side and negative on the
other.
A dipole field. Electric fields diverge from a positive charge
and converge on a negative charge.
13. Electric Dipole in an Electric Field
The net force acting on a neutral object placed in a uniform electric field is zero. However, the
electric field can produce a net torque if the positive and negative charges are concentrated at
different locations on the object. An example is shown in Figure below. The figure shows a
charge Q located on one end of a rod of length L and a charge - Q located on the opposite end of
the rod. The forces acting on the two charges are given by
Figure : Electric Dipole in an Electric Field.
Clearly, the net force acting on the system is equal to zero. The torque of the two forces with
respect to the center of the rod is given by
As a result of this torque the rod will rotate around its center. If [theta] = 0deg. (rod aligned with
the field) the torque will be zero.
The distribution of the charge in a body can be characterized by a parameter called the dipole
moment p. The dipole moment of the rod shown in Figure above is defined as
In general, the dipole moment is a vector which is directed from the negative charge towards the
positive charge. Using the definition of the dipole moment from above equation, the torque of an
object in an electric field is given by
Torque on Electric Dipole
The torque produced on an electric dipole by an electric field can be expressed as a vector product with direction
given by the right hand rule.
The lever arm for each charge with respect to
the center is
Dipole field Dipole potential
Energy of an Electric Dipole
An electric field produces a torque on a dipole
which tends to take it to its low energy configuration. To
rotate it from the low energy state against the field requires
work
where the shorter form employs the scalar product.
VERY SHORT ANSWER TYPE QUESTIONS (CARRYING 1 MARK EACH)
1. How does a free electron when released in an electric field move?
Ans. A free electron released at rest will move in a direction opposite to the electric
field.
2. When is an electric line of force straight?
Ans. Electric line of force is straight in the field of a single charge.
3. When is electric line of force curved?
Ans. Electric line of force is curved in the field of more than one point charges.
4. Why do two electric lines of force not cross each other?
Ans. If two lines of force cross each other then at the point of intersection, there will be
two tangents at that point of intersection, which means that there are two values of the
electric field at that point, which is not possible.
5. Is electric field intensity a scalar or a vector?
Ans. Electric field intensity is a vector
6. What are the units of electric field intensity?
Ans. N/C or V/m
7. Give the unit of electric dipole moment.
Ans. Cm
8. When is the torque on an electric dipole in a field maximum?
Ans. The torque is maximum when dipole is held at 90° to the field.
9. What is the net force on a dipole in a uniform magnetic field?
Ans. Zero.
10. How does a torque affect the dipole in an electric field?
Ans. Torque tries to align the dipole along the field.
11. When is an electric dipole in stable equilibrium in an electric field?
Ans. When p is parallel to E i.e.  = 0.
12. When is an electric dipole in unstable equilibrium in an electric field?
Ans. When p is antiparallel to E i.e.  = 180.
13. At what points, dipole field intensity is parallel to the line joining the charges?
Ans. At any point on axial line or equatorial line of dipole.
14. Two point charges of +3 μC each are 100 cm apart. At what point on the line joining
the charges will the electric intensity be zero?
Ans. At the centre.
PROBLEMS FOR PRACTICE (SHORT ANSWER TYPE QUESTIONS - CARRYING 23 MARKS EACH)
1. A charge of -2 x 10-6 C experiences a force of 0.08 N [left]. What is the
electric field at that point?
2. A charge of +3.0 x 10-6 C is 0.25 m away from a charge of -6.0 x 10-6C.
a. What is the force on the 3.0 x 10-6 C charge?
b. What is the force on the -6.0 x 10-6 C charge?
3. Three charges, q1 = 4 x 10-6 C, q2 = -2 x 10-6 C, and q3 = 5 x 10-6 C are
placed at the corners of a square with sides 0.30 m. What is the field
at the fourth corner?
4. Four point charges form the vertices of a square with sides = L. Two diagonally opposite
charges have a charge of 2√2 μC each. The other two charges are identical to each other
and each has a charge, q. If there is no net force on either of the 2√2 μC points, what is the
value of q?
5. Two point charges lie on the x-axis. A charge of +9 μC is at the origin, and a charge of -4
μC is at x=10cm.
a. At what position x would a third charge q3 be in equilibrium?
b. Does your answer to part a depend on whether q3 is positive or negative? Explain.
6. Four particles each with a positive charge of q are placed on the vertices of a square
having sides L. A fifth particle with a positive charge Q is placed at the center of the
square. What is the force on the particle at the center of the square?
7. A charge of 6.00*10-9 C and a charge of -3.00*10-9 C are separated by a distance of 60.0
cm. Find the position at which a third charge of 12.0*10-9 C can be placed so that the net
electrostatic force on it is zero.
8. Two charges, +q and 4q, are 1 m apart. What are the location, magnitude and sign of a
third charge, Q, placed so that the entire system is at equilibrium?
9. A +2.0 μC charge is located on the x-axis at +0.3 m and another at -0.3 m. A third
charge, +4.0 μC, is located on the y-axis at +0.4m. Find
a. the net force on the third charge
b. the electric field at (0,-0.4m)
1(c) ELECTROSTATC POTENTIAL AND FLUX
IMPORTANT CONCEPTS
1. Electrostatic Potential Energy. Potential energy of a charge q at a point in an
electrostatic field due to any charge configuration is defined as the work done by the
external force (equal and opposite to electric force) in bringing the charge q from infinity
to that point.
2. Electric Potential Difference. Electric potential difference between two points B and A
in an electrostatic field is defined as the amount of work done in carrying unit positive
test charge from A to B (against the electrostatic force of the field) along any path
between the two points.
3. Electrostatic or Electric Potential. Electrostatic potential at any point in a region of
electric field is defined as the minimum work done in carrying a unit positive charge
(without acceleration) from infinity to that point.
4. Units of Potential Energy. The SI unit of potential energy is joule. One joule is the
energy stored in moving a charge of one coulomb through a potential difference of one
volt.
5. One Volt of Potential. Electrostatic potential at any point is a scalar quantity. Its SI unit
is volt. Electric potential at a point is said to be one volt, when one joule of work is done
in moving one coulomb of positive charge from infinity to that point against the
electrostatic force of the field.
Electric Potential Difference
Consider the task of moving a positive test charge within a uniform electric field from location A to
location B as shown in the diagram at the right. In moving the charge against the electric field from
location A to location B, work will have to be done on the charge by an external force. The work done on
the charge changes its potential energy to a higher value; and the amount of work which is done is equal
to the change in the potential energy. As a result of this change in potential energy, there is also a
difference in electric potential between locations A and B. This difference in electric potential is represented by the
symbol V and is formally referred to as the electric potential difference. By definition, the electric potential
difference is the difference in electric potential (V) between the final and the initial location when work is done upon
a charge to change its potential energy. In equation form, the electric potential difference is
Significance of Electrical Potential
Just as the electric field is described as force per unit charge, electric potential at a point can be
described as electrical potential energy per unit charge . This concept is useful in calculations
involving energies of charged particles. Not only that, as force is related to work, electric potential is
closely related to the electric field
Potential being a scalar quantity, it is easy to deal with, than electric field. Hence, when we need to
determine an electric field, it is easy to find the potential first and find the field from it .
The electrostatic potential at any point in an electric field is defined as the work done in bringing a
unit positive charge from infinity to that point against the electric force of the field.
V = W/q
where V is the potential
W is work done
q the charge
S.I. unit is Joule/Coulomb = 1 Volt
Potential Difference
Consider two points in an electric field. If a free test charge +q o was placed in the field, it would
flow from b to a. So we say b is at higher potential. The electric field does negative work in moving
the test charge qo from a to b.
If a and b are two points in an electric field having a potential difference of 12V and if V B>VA, it
means 12 Joules/C of work has to be done in moving a test charge from A to B. Conversely this 12
J/C of energy will be released if it moves from B to A. This process does not depend on the route
traversed and like gravitational electrostatic forces are also conservative.
If a is at infinity, then the potential due to a single point charge
Electric potential at any point in vacuum due to group of point charges q 1,q2,q3……….qn is equal to
the algebraic sum of the potentials due to q 1,q2,q3……….qn at point P. Algebraic sum is one in which,
sign of the physical quantity is taken into account.
6. Equipotential Surface: An equipotential surface is that at every point of which electric
potential is the same.
7.
GENERAL MEANING
OF ELECTRIC FLUX
In common language flux refers to the flow or stream of any thing from one point to
another point. In the similar way electric flux is the total number of lines of force passing
through a surface.
8.
PHYSICAL MEANING
OF ELECTRIC FLUX
In physical sense, electric flux is defined as:
"The total number of lines of force passing through the unit
area of a surface held perpendicularly."
9.
MATHEMATICAL
MEANING OF
ELECTRIC FLUX
Mathematically the electric flux is defined as:
"The dot product of electric field intensity (E) and the vector
area (A) is called electric flux."
Where is the angle between E and A
10.
MAXIMUM
FLUX
If the surface is placed perpendicular to the electric field then maximum
electric lines of force will pass through the surface. Consequently
maximum electric flux will pass through the surface.
11.
ZERO
FLUX
If the surface is placed parallel to the electric field then no electric lines of force
will pass through the surface. Consequently no electric flux will pass through the
surface.
Flux is a scalar quantity .
12.
UNIT
OF
FLUX
13. Gauss's Law
The total of the electric flux out of a closed surface is equal to the charge enclosed divided
by the permittivity.
14. Applications of Gauss' Law
Gauss' law is a powerful tool for the calculation of electric fields when they originate from
charge distributions of sufficient symmetry to apply it.
15. GAUSS’ THEOREM:
We have already learnt to find the electric field intensity due to a charged conductor using
Coulomb's law. Gauss' theorem can also be used to calculate the electric field intensity provided
there is symmetry in the charge distribution.
GAUSS’ THEOREM can be considered as an alternative to Coulomb's law for expressing the
relationship between electric charge and electric field. This theorem was formulated by a
German mathematician Karl Friedrich Gauss.
Gauss' theorem states that the total electric flux through any closed surface is proportional to the
total electric charge inside the surface.
Total electric flux q
But we have already obtained the relation
This means the electric flux is independent of the radius of the surface but only depends on the
charge enclosed by the surface.
Proof
Let case I - when Gaussian surface is spherical in shape, a positive charge q be placed at the
centre of an imaginary spherical surface of radius R as shown in the figure.
By symmetry, the field due to the charge +q is radial and E is perpendicular to the sphere and is
directed along the normal to the surface.
So the angle between the normal and the electric intensity is zero.
That is, df = E dA cos0
Which is nothing but the mathematical representation of Gauss' theorem.
Case II When Gaussian Surface is not Spherical in Shape
The total electric flux is,
where dW is the solid angle subtended by the area dA at the point.
Which is nothing but Gauss' theorem
16. Application of Gauss' Theorem
Gauss' theorem can be used to calculate the electric intensity due to
1. an infinitely long straight charged wire
2. a uniformly charged infinite plane sheet
3. a uniformly charged thin spherical shell
1. Field Due to an Infinitely Long Straight Charged Wire
Consider a thin long charged wire. Let the charge per unit length of the wire be
. To
calculate the field at P we consider a Gaussian surface with wire as axis, radius r and
length l as shown in the figure. This Gaussian surface that is, the cylinder is closed at each
end by planes normal to the axis.
The electric lines of force are parallel to the end faces of the cylinder and hence the
component of the field along the normal to the end faces is zero.
The field is radial everywhere and hence the electric flux crosses only through the curved
surface of the cylinder.
If E is the electric field intensity at P, then the electric flux through the Gaussian surface is
E x 2prl (2prl is the surface area of the curved part)
The
charge
enclosed
by
the
Gaussian
surface
is
l
is directed radially outwards if q is positive and radially inwards if q is negative.
Note:
SI unit of l is C/m.
2. Field Due to a Uniformly Charged Infinite Sheet
Let s be the uniform surface charge density of an infinite plane sheet.
If x-axis is taken normal to the given plane then the electric field will not depend on y and z
axes.
The Gaussian surface will be a parallelepiped of cross sectional area A. only the two faces 1
and 2 will contribute to the flux whereas the other two faces do not contribute to the total
flux as the electric field lines are parallel to them.
Flux through the Gaussian surface = 2EA
(The factor 2 appears as the total flux is due to two faces of parallelepiped which are
normal to electric field
SI unit of s is C/m2
3.Field due to a Uniformly Charged thin Spherical Shell
Consider a hollow conducting sphere of radius R with its centre at O. let s be its surface
density. The field at any point P, outside or inside depends upon the distance from the
centre of the spherical shell. Let the distance between the centre of the spherical shell and
the point be r.
Case 1
Field Inside a Hollow Conducting Sphere
Consider a thin hollow conducting sphere with radius R. Let q be the charge on this
sphere. To find the field at a point P, draw a gaussian surface (dotted circle) of radius r.
Since, this surface does not enclose any charge, we have
Case 2
Field on the surface of the shell
When point P lies on the surface of the shell or sphere, r = R so
Case 3
Field at a Point Outside the Spherical Shell
Let the point P be outside the spherical shell. At points outside the sphere the electric field
is radial every where because of spherical symmetry.
where is the unit vector.
is directed outwards if the charge is positive and is directed
inwards if the charge is negative.
The electric field in terms of surface charge density
Case I
Case II
Case III
16.
Electric Field: Sphere of Uniform Charge
The electric field of a sphere of uniform charge density and
total charge charge Q can be obtained by applying Gauss'
law. Considering a Gaussian surface in the form of a sphere
at radius r > R, the electric field has the same magnitude at
every point of the surface and is directed outward. The
electric flux is then just the electric field times the area of
the spherical surface.
The electric field outside the sphere (r > R)is seen to be identical to
that of a point charge Q at the center of the sphere.
For a radius r < R, a Gaussian surface will enclose less
than the total charge and the electric field will be less.
Inside the sphere of charge, the field is given by:
Inside a Sphere of Charge
The electric field inside a sphere of uniform charge is radially outward (by symmetry), but a
spherical Gaussian surface would enclose less than the total charge Q. The charge inside a radius r
is given by the ratio of the volumes:
The electric flux is then given by
and the electric field is
Note that the limit at r= R agrees with the expression for r >= R. The spherically symmetric charge outside the radius r does not affect
the electric field at r. It follows that inside a spherical shell of charge, you would have zero electric field.
17.
Electric Field: Parallel Plates
If oppositely charges parallel conducting plates are treated like infinite planes
(neglecting fringing), then Gauss' law can be used to calculate the electric field
between the plates. Presuming the plates to be at equilibrium with zero electric
field inside the conductors, then the result from a charged conducting surface can
be used:
This is also consistent with treating the charge layers as two
charge sheets with electric field
in both directions.
UNITS AND DIMENTIONS
S. NO.
QUANTITY
1.
Electric Field
2.
Permittivity
3.
Linear Charge Density
4.
Surface Charge Density
5.
Volume Charge Density
6.
Electric potential
7.
Electric Flux
8.
Potential Gradient
SI UNITS
N/C or V/m
C2 /Nm2
C /m
C /m2
C /m3
V or J/C
Nm2/C or Vm
N/C or V/m
DIMENTIONS
M1L1 T-3 A-1
M -1L-3 T4 A2
M 0L-1 T1 A1
M 0L-2 T1 A1
M 0L-3 T1 A1
M1L2 T-3 A-1
M1L3 T-3 A-1
M1L1 T-3 A-1
VERY SHORT ANSWER TYPE QUESTIONS (CARRYING 1 MARK EACH)
1. What is the amount of work done in moving a 200nC charge between two points 5
cm apart on an equipotential surface?
Ans. On an
equipotential surface, work done is zero.
2. If V equals a constant throughout a given region of space, what can you say about E
in that region?
Ans. When V is
constant, E must be zero in the given region.
3. If E equals zero at a given point, must V equal zero for that point? Give some
example to prove your answer.
Ans. No, it
is not necessary that V equals zero when E equals zero. This is because E = dV / dr. In a
sphere, V is constant at every point inside the sphere, but E = 0.
4. How many electron volts make one joule?
Ans. As I
eV = 1.6×10−19 joule i.e. 1.6×10−19 joule = 1 eV;
therefore, 1 joule = 1 /
1.6×10−19 eV = 0.625×1019 eV
5. What is the relation between electric intensity and flux?
Ans. The
surface integral of electric field intensity over a closed surface in free space is 1/ ε0 times
the total charge q enclosed by the surface i.e.
6. What is the number of electric lines of force that radiate outwards from one
coulomb of charge in vacuum?
Ans. It is
−12
11
1/ ε0 = 1/8.85×10
= 1.13×10 .
7. What is the S.I. unit of the line integral of electric field?
Ans. Volt
or Joule/Coulomb.
8. What is the S.I. unit of the surface integral of electric field?
Ans. Volt –
metre or Nm2/C.
9. What is the dielectric constant of a metal?
Ans.
Infinity, every metal is a good conductor.
10. What is the shape of equipotential surface for a point charge?
Ans. It is a
sphere around the point charge.
11. No work is done in moving a test charge over an equipotential surface, why? Ans.
Work done = charge ×potential difference = zero.
12. 5 J of work is done in moving a point charge of 0.5 C before two points. What is the
potential difference between these points?
Ans. Potential
difference = Work done / Charge = 5 / 0.5 = 10 V.
13. A charge of 2 C is moved between two plates maintained at a potential difference of
1 Volt. What is the energy acquired by the charge?
Ans. Energy acquired =
q×V = 2×1 = 2 Volt.
14. It requires 50 μJ of work to carry a 2 μC of charge from a point A to B. What is the
potential difference between these points? Which point is at higher potential? Ans.
Potential difference = VB - VA =Work done / Charge = 50μJ/2μC = 25 Volt. Clearly, VB>VA
15. Can two equipotential surfaces intersect? Explain.
Ans. No,
because equipotential surfaces are normal to electric lines of force, which cannot
intersect.
16. An electric dipole of dipole moment 20×10−6 Cm is enclosed by a closed surface.
What is the net flux coming out of the surface?
Ans. Net flux
coming out of the closed surface is zero, because net charge on electric dipole is zero.
17. What is the potential energy of two equal negative point charges 2μC each held at 1
m apart in air?
Ans. P.E. = K q1 q2 /
/ r = (9×109) (2×10−6) (2×10−6 )/1 = 0.036 J.
SHORT ANSWER TYPE QUESTIONS (CARRYING 2 MARKS EACH)
1. Give two examples of conservative forces in nature.
Ans. (i)
Electrostatic forces (ii) Gravitational forces.
2. What is the work done by the field of a nucleus in a complete circular orbit of
electron? What if the orbit is elliptical?
Ans. Zero.
However if the orbit is elliptical, work is done in moving an electron from one position to
other in elliptical orbit. However, net work done over a complete orbit is zero, because
electrostatic forces are conservative forces.
3. Two protons A and B placed between two parallel plates having a potential
difference V as shown in figure below. Will these protons experience equal or
unequal force?
+++++++++++++++++++++++
Ans. Electric Field is
Uniform. F=qE,
Therefore forces are equal.
.A
.D
.C
.B
- - - - - -- - - - - - - - - 4. A uniform field E exists between two charged plates as shown in Fig. above. What
would be work done in moving a charge q along a closed rectangular path A C B D
A?
Ans. Work done in
moving a charge along the closed rectangular path would be zero, because the field in the
entire space is uniform and electrostatic forces are conservative forces.
5. The electric field at a point due to a point charge is 30 N/C, and the electric potential
at that point is 15 J/C. Calculate the distance of the point from the charge and the
magnitude of the charge.
Ans. Since E = Kq / r2 = 30
N/C
and V = Kq / r = 15 J/C,
therefore, E = (Kq / r)(1/r) = V / r, i.e. r =
V / E = 15/30 m = 0.5 m.
Further, Kq / r = 15, i.e. q = 15 r / K = 15 ×0.5 /
9×109 = 0.833×109 C.
1(d)
CAPACITANCE
IMPORTANT CONCEPTS
1. PARALLEL-PLATE CAPACITORS:
o = Permittivity of empty space ( air o)
= 8.854x10-12 C 2/N.m2 = 1/(4k)
 = Permittivity of material between plates
A = Surface Area of one of the plates (SI: m 2)
d = Separation of the plates (SI: m)

2. Electric Field in an Ideal Parallel Plate Capacitor:

The Electric Field due to a sheet of charged infinite sheet is very simple in that the Efield is constant at any distance from the sheet.

When two plate of different charge are placed near each other, the two E-fields between
the plates add while the E-field outside the plate cancel

When the plates are close to each other to form a capacitor, the E-field between the plates
is constant through out the interior of the capacitor as long as one is not near the edges of
the plates.

Since the electric field is the negative of the gradient of the potential and the E-field is
constant inside a capacitor, the magnitude of the Electric field has a very simple relation
to the voltage between the plates and their separation d.

This relationship is also true in an electrical wire were V is the voltage across the ends of
the wire and d is the length of the wire.

Using the definition of capacitance we can determine the capacitance C of an ideal
capacitor as a function of its structure.

This equation for the capacitance of a parallel capacitor shows that C is a constant
independent of the charge stored in on the plates or the voltage across the capacitor.

By placing a thin insulating material (a dielectric) between the plates the separation d can
be reduced thus increasing the capacitance of the capacitor and prevent the plates from
touching.

It takes more voltage to store the same amount of charge on a capacitor because of the
presence of the dielectric. Typically a dielectric contains polar molecules which partially
line up in the presence of the electric field.

The dielectric creates an E-field in the opposite direction which reduces the overall Efield between the plates.

You could calculate the capacitance C of a parallel plate capacitor by replacing the
permittivity of empty space o by the permittivity  of the dielectric material placed
between the plates.

Except for simple capacitors (like the parallel plate capacitor) we do have an equation
from which we can calculate the capacitance C. In practice we simply measure the value
of a capacitor and assume that it is constant.
3. Energy Density of the Electric Field in a Capacitor:

The electrical energy stored in the Electric Field between the plates of an ideal capacitor
has a simple form when expressed as the electrical energy per unit volume, u = U/Vol

This is a general expression that is valid for the energy density of the Electric Field no
matter how the electric field is generated, i.e. it is true at any point in space where there is
an electric field E.

For parallel plate capacitors this can easily be derived since the E-field is constant
through out the interior of the capacitor and equal to V/d. Here,
4. CAPACITOR AND PRINCIPLE OF CAPACITANCE.
A capacitor consists of two isolated conductors with a charge +Q on one conductor (typically
referred to as a “plate” and an equal but opposite charge −Q stored on the other conductor. The
charge “stored” in the capacitor is simply referred to as Q. We find that the charge stored by a
capacitor is proportional to the potential difference between the two conductors. In other
words,
Q = (Constant) V or Q = CV
where the proportionality constant C is known as the capacitance. The capacitance of a given set
of conductors depends only upon their geometry (the shape of the conductors and the distance
between them) and on the “stuff” placed between them. We refer to any insulating material
placed between conductors in a capacitor as a dielectric.
There are three principal capacitor geometries you should become familiar with:
• the parallel-plate capacitor,
• the cylindrical capacitor, and
• the spherical capacitor.
If a capacitor has a capacitance C0 when the space between the conducting plates is
occupied by nothing at all (vacuum), or to a very good approximation, air, we find that the
introduction of any insulating material into that space serves to increase the capacitance of
the capacitor. The multiple by which the capacitance has increased is known as the
dielectric constant κ of the insulating material. We can then write
C = κ C0
where again, C0 is the capacitance with nothing between the conducting plates and C is the
capacitance when a material with dielectric constant κ has been inserted between the plate.
5. Capacitance. Capacitance of a capacitor is defined as the ratio of the charge on the
capacitor to the potential of the capacitor. Capacitance of a capacitor can also be defined
as the amount of the charge required to raise its potential by unity, i.e. in the equation Q =
CV, if V = 1, Q = C.
6. One Farad of Capacitance. S.I. unit of Capacitance (C) is farad. In eq. Q =CV, if Q = 1
coulomb and V = 1 volt, C = 1 coulomb / volt or farad. Hence, Capacitance of a
conductor is said to be one farad, when a charge of one coulomb raises its potential
through one volt.1 microfarad(μF)=10−6 farad(F)
7. Spherical Capacitor
The capacitance for spherical or cylindrical conductors can be obtained by
evaluating the voltage difference between the conductors for a given
charge on each. By applying Gauss' law to an charged conducting sphere,
the electric field outside it is found to be
The voltage between the spheres can be found by integrating the electric field along a radial line:
From the definition of capacitance, the capacitance is
8. Isolated Sphere Capacitor?
An isolated charged conducting sphere has capacitance. Applications for such
a capacitor may not be immediately evident, but it does illustrate that a
charged sphere has stored some energy as a result of being charged. Taking
the concentric sphere capacitance expression:
and taking the limits
gives
Further confirmation of this comes from examining the potential of a charged conducting sphere:
9. Electrostatic shielding. Electrostatic shielding/screening is the phenomenon of
protecting a certain region of space from external electric field.
10.SERIES CONECTION:
11. Series connected Capacitors always have the same Charge. They do not have the
same voltage unless the capacitors have the same Capacitance C.
12. The Charge on the equivalent capacitor Ce is the same as the charge on either capacitor.
13. The Voltage across the equivalent capacitor Ce is the sum of the voltage across both
capacitors.
14. PARALLEL CONNECTION
15. Parallel connected Capacitors always have the same voltage drop across each of
them. They do not have the same charge unless they have the same capacitance C.
16. The Charge on the equivalent capacitor Ce is the sum of the charges on both capacitors.
17. The Voltage on the equivalent capacitor Ce is the same as the voltage across either
capacitor.
18. Polarization of Dielectric
If a material contains polar molecules, they will generally be in random orientations
when no electric field is applied. An applied electric field will polarize the material by
orienting the dipole moments of polar molecules.
This decreases the effective
electric field between the
plates and will increase the
capacitance of the parallel
plate structure. The dielectric
must be a good electric
insulator so as to minimize
any DC leakage current
through a capacitor.
Effect on permittivity and capacitance.
19. Parallel Plate with Dielectric
The capacitance of a set of charged parallel plates is increased by the insertion of a dielectric
material. The capacitance is inversely proportional to the electric field between the plates, and
the presence of the dielectric reduces the effective electric field. The dielectric is characterized
by a dielectric constant k, and the capacitance is multiplied by that factor.
20. Parallel Plate with Dielectric
When a dielectric is placed between charged plates, the
polarization of the medium produces an electric field
opposing the field of the charges on the plate. The
dielectric constant k is defined to reflect the amount of
reduction of effective electric field as shown below.
The permittivity is a characteristic of space, and the
relative permittivity or "dielectric constant" is a way to
characterize the reduction in effective field because of
the polarization of the dielectric. The capacitance of the
parallel plate arrangement is increased by factor k.
Table of dielectric constants
21. Capacitance of Parallel Plate with a Dielectric Slab
When a dielectric slab of thickness t is introduced between the plates of a capacitor having
uniform electric field E0, the slab develops a slight opposite electric field due to polarization of
molecules, resulting a net electric field E inside the dielectric slab over the thickness t So,
V = E0 (d - t) + Et
This clearly implies: C>Co
And is expected because introduction of dielectric slab reduces the electric field
and hence potential difference between the plates. Since, charge on capacitor
plates is same, capacitance increases.
22. Dielectric Constant and Electric Susceptibility.
When a static electric field is applied to a dielectric medium, a current flows. The total current flowing in a
real dielectric is, in general, made up of two parts: a conduction and a displacement current. The
displacement current can be considered the elastic response of the dielectric material to the applied electric
field. As the magnitude of the electric field is increased, the additional displacement is stored as potential
energy within the dielectric. When the electric field is decreased, the dielectric releases some of the stored
energy as a displacement current. The electric displacement can be separated into a vacuum contribution and
one arising from the dielectric by
where P is the polarization of the medium, E is the electric field, D is the electric flux density (or
displacement), and ? its electric susceptibility. It follows that the relative permittivity and susceptibility of a
dielectric are related,
23. Van de Graaff generator
[
Electrostatic generator is capable of producing a voltage of over a million volts. It consists of a continuous
vertical conveyor belt that carries electrostatic charges (resulting from friction) up to a large hollow sphere
supported on an insulated stand. The lower end of the belt is earthed, so that charge accumulates on the
sphere. The size of the voltage built up in air depends on the radius of the sphere, but can be increased by
enclosing the generator in an inert atmosphere, such as nitrogen.
VERY SHORT ANSWER TYPE QUESTIONS (CARRYING 1 MARK EACH)
1. For a given potential difference, does a capacitor store more or less charge
with a dielectric than it does without a dielectric?
Ans. A
capacitor with a dielectric would store more charge, as its capacity increases.
2. An isolated conducting sphere is given a positive charge. Does its mass
increase, decrease or remain the same?
Ans. Its
mass decrease slightly as it loses some electrons.
3. A capacitor is charged by using a battery, which is then disconnected. A
dielectric slab is then slipped between the plates. Describe qualitatively what
happens to the charge, the capacitance, the potential difference, electric field
strength and stored energy?
Ans. Charge remains the same; capacitance
increases; potential difference decreases: electric field strength decreases and
stored energy decreases.
4. Is it possible for a metal sphere of 1 cm radius to hold a charge of 1C? Ans.
No, because V = Q / C = Q / 4πε0 r = 9×109 ×1 / 10-2 =9×1011 V. The potential
will become so high that electric break down of air will occur and the entire
charge will leak out.
5. Is there any conductor which can be given almost unlimited charge? Ans.
Yes, earth can be given almost unlimited charge, because its capacity is very
large.
6. Two spheres of silver of same radii, one solid and the other hollow are
charged to the same potential. Which one has greater charge?
Ans. Both
the spheres will have the same charge, as q = CV = (4πε0 r) V.
7. Two insulated charged spheres of radii 10 cm and 20 cm having same charge
are connected by a conductor and then they are separated. Which of two
spheres will carry more charge?
Ans. Bigger sphere will
carry more charge as its capacity is larger, as q = CV. The potential V becomes
same on connecting them with a wire.
8. The distance between the plates of a parallel plate capacitor is d. A metal
plate of thickness d/2 is placed between the plates, what will be the new
capacity?
Ans. An electric
field inside the metal plate is zero, d becomes d/2. Hence C becomes twice (from
C = ε0 A / d).
9. Why does the electric field inside a dielectric decrease when it is placed in an
external electric field?
Ans. It is because the
dielectric gets polarized.
10. The dielectric constant of a conductor can be taken as infinitely large. Why?
Ans. When a conductor is held inside an electric field, the field inside the
conductor becomes zero. The dielectric constant of conductor is the ratio of
strength of applied electric field to the reduced value of electric field inside, and it
becomes infinity.
11. On inserting a dielectric between the plates of a capacitor, its capacitance is
found to increase 5 times. What is the relative permittivity of the dielectric?
Ans. εr = K = C / Co = 5.
12. What should be the capacitance of a capacitor capable of storing 1 joule
energy, when used with a 100 V d.c. supply?
Ans. Here U = 1
J, V = 100 volt, C =?
As U = ½ C V2 , C = 2U /
-4
2
2
V = 2 ×1 / (100) = 2 × 10 F.
13. A capacitor is charged through a potential difference of 200 V, when 0.1 C
charge is stored in it. How much energy will it release, when it is discharged?
Ans. Here, V = 200 volt, q = 0.1 C
The
Energy released on discharging = energy stored on charging = ½ qV = 1/2 × 0.1
×200 = 10 J.
14. How much work must be done to charge a 24 μF capacitor, when the
potential difference between the plates is 500 V?
Ans. Here,
-6
C = 24 μF = 24 × 10 F.
V = 500 V
Work done = ½ C V2 = ½ (24 × 10-6 ) (500)2 = 3 joule.
SHORT ANSWER TYPE QUESTIONS (CARRYING 2 MARKS EACH)
1. Problems on capacitors
Electron charge = -1.6 x 10-19 C
1.
What is meant by the capacitance of a capacitor?
2.
Define the farad.
3. (a) A capacitor of capacitance 5 μF is connected to a 6 V supply. What charge is stored
in the capacitor?
(b)
A 400 pF capacitor carries a charge of 2.5 x 10-8 C. What is the potential difference
across the plates of the capacitor?
4.
A capacitor is charged such that there is a charge of +20 mC on the positive plate.
What is the charge on the negative plate?
5.
A 4700 μF capacitor is connected as shown in the
circuit diagram. When it is fully charged:
(a) What is the charge on the positive plate of the
capacitor?
(b) What is the potential difference across the capacitor?
(c) How many additional electrons are on the negative
4.5 V
C
plate?
6.
A resistor of 100 Ω is now added to the circuit as
shown in the second diagram.
(a)
What effect does this have on the time to charge up
capacitor?
(b) What is the final charge on the plates?
(c) What is the final potential difference across the
capacitor?
4.5 V
the
C
Solution and worked examples
1.
The ratio of the charge on the plates of the capacitor to the potential difference across
them.
C = Q/V.
2.
The charge required to increase the potential difference across the plates by 1 V.
3.
Use C = Q/V
(a) Q = CV = 5 x 10-6 x 6 = 30 x 10-6 = 30 μC
(b) V = Q/C = 2.5 x 10-8 / 400 x 10-12 = 62.5 V
4.
- 20 mC equal and opposite.
4.5 V
5.
(a) Q = CV = 4700 x 10-6 x 4.5 = 0.021 C
(b) 4.5 V
(c) 0.021 / 1.6 x 10-19 = 1.32 x 1017
C
6.
(a) It will take longer
(b) It will be the same as before
(c) 4.5 V since no current is flowing finally there will be no
potential difference across the resistor whatever its value.
.2.
4.5 V
C
Examples of Capacitor Combinations
We can consider some specific examples:
1. Consider the combinatory of 4 parallel-plate capacitors as shown in figure 2.1. A battery
maintains a potential difference of 20 volts between its poles. The values of the
capacitances are:
C1 = 20 μF, C2
= 5 μF, C3 = 10 μF, C4 = 6 μF
a. Find the equivalent capacitance of the circuit.
b. Find the charge on and potential across each of the capacitors.
1.
Figure 2.1: A combination of 4 capacitors.
Solution:
The equivalent capacitance is gotten by realizing that the definition of a series combination is that the
negative plate of one capacitor is connected only to the positive plate of another capacitor. In this
case, the net charge on the capacitor plates (assuming they were initially uncharged) must have the
same magnitude and opposite sign for both plates. By definition this is a series combination. Looking
at the diagram, we realize that capacitors C1 and C2 are in series. The equivalent capacitor for this
pair (call it C12) is in parallel with C4. Therefore, call this equivalent capacitance C124. Finally, this
equivalent capacitor is in series with C3. Hence we can replace all these by one equivalent capacitor
with equivalent capacitance
1
1
=
C12
1
+
C1
C2
C1C2
C12 =
C1 + C 2
20 x 5 μF2
=
20 + 5 μF
C12 = 4 μF
C124 = C12 + C4
= 4 μF + 6 μF
C124 = 10 μF
1
1
=
Ceq
1
+
C124
C3
C3C124
Ceq =
C3 + C124
10 x 10 μF2
=
10 + 10 μF
Ceq = 5 μF
Figure 4.14: The equivalent circuit for figure 2.1.
To find the charge across each capacitor, first note that the charge across C124 and
C3 must be the same since they are in series. This charge must be the same magnitude as
the charge on the equivalent capacitance for the circuit, so
Q3 = Q124 = Qeq =
=
Q3 = Q124 = Qeq =
CeqVb
(5 μF)(20 V)
100 μC
The voltage across C3 and across C124 can then be derived as follows.
100 μC
Q3
V3 =
=
C3
= 10 V
10 μF
100 μC
Q124
V124 =
=
C124
= 10 V
10 μF
Notice in this case that the voltages happen to be equal, but this is strictly a consequence of the
capacitor values. For any capacitor values it would be necessary for
V3 + V124 = Vb
as it does in this case. So we are done with C3. To get the voltage across C4 note that it
must be the same as the voltage across V124. So,
V4 = 10 V.
The charge on C4 is then
Q4 = C4V4 = (6 μF)(10 V) = 60 μC
The charge on capacitors C1 and C2 must be the same since they are in series. Their
equivalent capacitance, C12, must also have this magnitude of charge and the same
potential difference as C4 since it is in parallel with it. Hence,
Q1 = Q2 = Q12 = C12V124 = (4 μF)(10 V) = 40 μC
Finally, the voltages across C1 and C2 are
40 μC
Q1
V1 =
=
40 μC
Q2
V2 =
=2V
20 μF
C1
=
C2
=8V
5 μF
2.
What is the combined capacitance of a 10 F capacitor, a 20 F capacitor
and a 30 F capacitor connected (a) in parallel (b) in series?
3.
What is the combined capacitance of a 10 F capacitor and a 20 F
capacitor connected in parallel, and then connected in series to a 30 F capacitor
as shown below.
Answers and worked solutions
2. (a) From the equation derived for parallel combination:
(b) From the equation derived for series combination:
3.
Taking the two in parallel first, we have
Combining this with the other capacitor in series, we have
Capacitance is measured in units called farads (F). A
farad is a very big unit, and we are much more likely to
use microfarads (F) or nanofarads (nF).


1 F = 1 × 10-6 F
1 nF = 1 × 10-9 F
Unit-2
CURRENT ELECTRICITY
Weightage-07
Question Packet-1
1. Define electric current. Name its S I unit. Define this S I unit.
2. Define Electric current density. Is it vector? If yes state its direction.
3. A steady current flows in a metallic conductor of non-uniform cross section. Which of
the following quantities is/are not constant along the conductor?
Charge/time. Current density,
drift speed.
4. State Ohm’s law.
5. Define resistance of a wire in terms of potential difference across its ends.
(Hint V=IR, if I=1 then V=I)
6. On what factors does resistance of a metallic conductor depend?
7. What is relaxation time? What affects relaxation time of electrons in a metallic
conductor?
8. Name two physical conditions on which the resistivity of a metal depends.
9. Two wires of equal length one of copper and the other of manganin, have the same
resistance. Which wire is thicker?
𝒍
𝑹
𝝆
(Hint-R= ρ𝑨 or 𝒍 = 𝑨 = constant. Therefore greater ρ means greater A, means
thicker wire so Manganin)
10. Define resistivity in terms of resistance. Write its SI unit. Deduce dimensional formula of
it.
11. How is electrical conductivity of an electrolyte affected by increase of temperature?
12. A wire is stretched to increase its length to three times of original .What will be effect on
its i) Resistance ii) Resistivity.
13. Manganin is used for making standard resistors. Why?
14. Why is the electrical conductance of an electrolyte less than that of metals?
15. What is the basic principle of Potentiometer?
16. State the principle of balanced Wheat Stone Bridge.
17. Define EMF of a cell.
18. A cell is discharging. Which will be greater it’s EMF or it’s terminal potential difference?
(Hint- If emf is E, Terminal potential difference is V and internal resistance is r then during
charging V=E-I r and during discharging V=E+ I r)
19. What is sensitivity of a potentiometer?
(A potentiometer is said to be more sensitive if it is capable to detect even in small change in
potential. For high sensitivity the potential gradient should be low.)
20. What is superconductivity?
Question Packet-2
1. Write the mathematical expression for a. Drift velocity of electron in terms of electric field intensity E relaxation time
b. Drift velocity as per definition
c. Electric current in terms of number density of free electrons, drift velocity of free
electrons and other factors.
d. Current density in terms of conductivity and electric field intensity.
e. Resistance in terms of resistivity area of cross section and length of conductor
f. Electrical Resistivity in terms of relaxation time and number density of free electrons
g. Electrical resistivity an terms of electron mobility
2. Write the formulae fora. Equivalent resistance when two resistors are in series.
b. Equivalent resistance when two resistors are in parallel.
c. Equivalent resistance when two resistors are in series and one is parallel to them
together.
d. Equivalent resistance when two resistors are in parallel and one is in series to them.
e. Current through each resistor r1 & r2 when they are in parallel and a net current I is
supplied to this combination.
f. Power consumed by each resistor r1 & r2 when they are in parallel and a net current I
is supplied to this combination.
g. Power consumed by each resistor r1 & r2 when they are in series and a current I
passes to this combination.
h. Power consumed by each resistor r1 & r2 when they are in series and a potential
difference V exists across this combination.
i. Terminal potential difference of a cell when it discharges.
j. Terminal potential difference of a cell when it is being charged by other source.
k. Current supplied to an external resistance R by parallel combination of n cells each of
Emf E and internal resistance r.
l. Current supplied to an external resistance R by series combination of m cells each of
Emf E and internal resistance r.
m. Efficiency of delivering power by a cell of Emf E, internal resistance r to an external
resistance R.
n. Experimental formula for comparison of EMF of two primary cells by potentiometer.
o. Experimental formula to find internal resistance of a primary cell through
potentiometer.
Question Packet-3
1. The variation of Potential difference with length in case of two potentiometers A and B is
given below on left. Which of the two is more sensitive and why?
2. V-I graph for a metallic wire at two different temperatures T1 and T2 is shown above right
side. Which is higher T1 or T2?
3. V-I graph for two metallic wires X and Y at constant temperature is shown below left
side. Assuming that the wires have the same length and diameter. Which will have higher
resistivity
4. V-I graph for a conductor at temperatures T1 and T2 is shown in figure above right side.
Find the proportionality of (T1-T2) in terms of angle θ.
Hint: - In graph,
Therefore at T1,
Slope = Resistance
R1= Tan (90- θ) = Cot θ = R0 (1+αT1) ………………..(1)
And at T2
R2=Tan θ = R0 (1+αT2) ……………………………….(2)
Now (1)-(2) gives
Or on solving
Or
Or
Cot θ -Tan θ = R0 (1+αT1) - R0 (1+αT2)
= R0 (T1-T2) α
2 Cot 2 θ
= R0 (T1-T2) α
R0 (T1-T2) α = 2 Cot 2 θ
(T1-T2) œ Cot 2 θ
5. V-I graph for parallel and series combinations of two metallic resistors are shown in
figure. Which graph out of the two is for parallel combination? Justify your answer.
6. Solve the following(a)
Find resistance between points (a) A&B (b) A&C (c) A&D (d) C&D (e) B&C if
each resistor shown as r in figure below is 1 ohm.
(b)
If a battery of Emf 2 volt and negligible internal resistance is connected between
points A & B in above question then find the potential drops between points(i)
AD
(ii)
AC
(iii)
DC
and
(iv)
DB
(c)
Find current through each branch of above network if a battery of Emf 2 volt and
internal resistance 1 ohm is connected between points A&C.
(d)
Calculate the heat produced in the resistor r connected between points C&B in
above figure in the situation of question (b) & (c) both.
7. Use Kirchhoff’s Laws to calculate current in each branch. Given that E1 = 2V, E2 = 1V
and E3 = 4V in the figure given below
E1
E2
E3
8. If a resistor of value 2 ohm is connected between any of the points on left side and the
point on right side then
(i)
Find the current through that resistor
(ii)
And the current through all the branches.
9. Two bulbs are marked 60 W, 220 and 100W, 220V. Which one out of the two will glow
brighter if –
(i)
These are connected in parallel to 220 V mains.
(ii)
These are connected in parallel to 110 V mains.
10. Two bulbs are marked 60 W, 220 and 100W, 220V. These are connected in series. Which
one out of the two will glow brighter if(i)
This combination is connected to 220 V mains.
(ii)
This combination is connected to 110 V mains.
11. Find resistance from the coloured rings as shown below.
Question Packet- 4
1. A resistance of R Ω draws current from a potentiometer. The potentiometer has a total
resistance R0 Ω. A voltage V is supplied to the potentiometer. Derive an expression for
the voltage across R when the sliding contact is in the middle of the potentiometer.
2. Derive a relation between drift velocity of electrons in a conductor and current density.
3. Derive a relation between E I r and R. where symbols have their usual meaning.
4. Draw a diagram of Wheatstone Bridge having resistances R1, R2, R3, R4 connected in sides of a
rectangle in clock wise order and R5 along any diagonal. A voltage source of emf E is connected
along other diagonal. Derive the condition of balanced Bridge.
5. Draw the circuit diagram of a potentiometer set to compare the Emfs of two primary cells using
it. Derive the experimental formula for the same. Why is it necessary for the driver cell to have
emf more than the emf of experimental cells?
6. Draw the circuit diagram of a potentiometer set to find internal resistance of a primary cell.
Derive the experimental formula for the same.
Unit-3
Magnetic effects of current and Magnetism
Weightage-08 Marks
Question Packet-1
1. What is the name given to the curve, the tangent to which at any point gives the direction of
magnetic field at that point?
2. State any rule which relates the direction of electric current and the direction of accompanying
magnetic field?
3. What will be magnetic field at the Centre ‘C ‘of circular current carrying loop?
4. Define the term magnetic moment.
⃗⃗.
5. What is the magnitude of force on a charge q moving with velocity 𝑣⃗ in a magnetic field 𝐵
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
What is the resistance of an ideal voltmeter?
What is the resistance of an ideal ammeter?
How is a voltmeter connected in the circuit?
How is an ammeter connected in the circuit?
Why should an ammeter always be connected in series in the circuit and a voltmeter in parallel
with the circuit element across which voltage difference has to be measured?
What is the angle of dip at a place –
(i)
Where horizontal and vertical components of earth’s magnetic field are equal in
magnitude. Where are such points located on earth’s surface?
(ii)
Where horizontal component of earth’s magnetic field is zero. Where are such points
located on earth’s surface?
(iii)
Where vertical component of earth’s magnetic field is zero. Where are such points
located on earth’s surface?
Give one example of ferromagnetic, paramagnetic and diamagnetic material.
Name the elements of earth’s magnetic field.
An iron bar magnet is heated to 10000C and then cooled in a magnetic field free space. Will it
retain its magnetism?
Hint:-The Curie point for Fe is about 7700C.At 10000C, the magnet losses its magnetism.
How does the magnetic induction of a paramagnetic material vary with temperature?
Hint:-Paramagnetic obeys Curie’s law of magnetism. Intensity decreases with increase in
temperature.
Question Packet-2
1. Write SI unit of following(i)
µ0 (ii)
µ0/4π (iii) µ0 I /4π
⃗⃗) (vi) Magnetic flux(
(𝐵
⃗⃗⃗ . 𝐵
⃗⃗⃗⃗ (viii) 𝑀
⃗⃗⃗ × 𝐵
⃗⃗⃗⃗ (ix)
(vii) 𝑀
(iv)
µ0 𝑖
4𝜋𝑟 2
(v)
Magnetic flux density
⃗⃗⃗⃗
⃗⃗⃗ )
𝐵. ⃗⃗⃗⃗
𝐴 ) (vi)Magnetic dipole moment( 𝑀
Magnetic Intensity(H) (x) Intensity of
Magnetization
(xi)
Magnetic Susceptibility (𝑋𝑚 ) (xii) Current sensitivity
(xii) Voltage sensitivity
2. Write the dimensional formulae of the following(i) µ0
(ii)
Magnetic flux(
µ0/4π (iii) µ0 I /4π
(iv)
µ0 𝑖
4𝜋𝑟 2
⃗⃗) (vi)
(v) Magnetic flux density ( 𝐵
⃗⃗⃗⃗) (vi)Magnetic dipole moment( 𝑀
⃗⃗⃗⃗ 𝐴
⃗⃗⃗ )
𝐵.
⃗⃗⃗ . 𝐵
⃗⃗⃗⃗ (viii) 𝑀
⃗⃗⃗ × 𝐵
⃗⃗⃗⃗ (ix)
(vii) 𝑀
Magnetic Intensity(H) (x) Intensity of
Magnetization
(xi)
Magnetic Susceptibility (𝑋𝑚 ) (xii) Current sensitivity
(xii) Voltage sensitivity
3. Define the following(i)
Magnetic Field intensity in terms of force on a charged particle moving the
field.
(Use B=F/qv, take v=1 and q=1 then F=B)
(ii)
Magnetic flux in terms of magnetic field.
⃗⃗. ⃗𝑨⃗ take Area=1 then ɸ=B)
(Use ɸ = ⃗𝑩
(iii)
(iv)
(v)
Tesla using formula of Biot Savert’s law.
Weber
Ampere in terms of force between two parallel current carrying wires.
(Use 𝐅 =
(vi)
(vii)
𝛍𝟎𝐢𝟐 𝐢𝟏
𝟐𝛑𝐫
takes both currents unity)
A light magnetic dipole is placed in non-uniform magnetic field such that
line joining the two poles is not parallel with magnetic field lines. Explain why
the magnetic dipole –
(a) Will experience net force.
(b) Will experience a net torque.
(Hint- It’s two magnetic poles are under different forces therefore net force
on it will be non-zero. Also forces on the two poles are along different lines
so there will be net torque.)
Figure below shows a small magnetized needle P placed at a point O. The
arrow shows the direction of its magnetic moment. The other arrows show
different positions (and orientations of the magnetic moment) of another
identical magnetized needle Q.
(a) In which configuration the system is not in equilibrium?
(b) In which configuration is the system in
(i) Stable, and
(ii) Unstable equilibrium?
(c) Which configuration corresponds to the lowest potential energy among all
the configurations shown?
Question Packet-3
1. A closely wound solenoid of 800 turns and area of cross section 2.5 × 10–4 m2
carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar
magnet. What is its associated magnetic moment?
2. A bar magnet of magnetic moment 1.5 J T–1 lies aligned with the direction of a
uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the
magnet so as to align its magnetic moment?
(i) Normal to the field direction,
(ii) Opposite to the field direction?
(c) What is the torque on the magnet in cases (i) and (ii)?
3. A telephone cable at a place has four long straight horizontal wires carrying a
current of 1.0 A in the direction east to west. The earth’s magnetic field at the
place is 0.39 G, and the angle of dip is 35º. The magnetic declination is nearly
zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Question Packet-4
1. Explain with the help of a leveled diagram the underlying principle construction and
working of cyclotron.
2. State Ampere’s circuital law. Derive the mathematical expression for force acting on
a current carrying straight conductor kept in a magnetic field using it. State the rule
used to determine the direction of this force.
3. Derive a mathematical expression for the force per unit length acting on each of the
two straight parallel metallic conductors carrying current in the same direction and
kept near each other. Why do such current carrying conductors attract each other?
4. Derive an expression for the torque acting on a loop of N turns, area A, carrying
current I when held in uniform magnetic field B. Hence define current and voltage
sensitivities.
5. With the help of a diagram show how to convert a Galvano meter into (i) An
ammeter (ii) A voltmeter.
Also determine the expression for net resistance of these converted devices
provided their Range is (0-I ammeter) in case of ammeter and (0-V volts) in case of
voltmeter. Take the resistance of Galvanometer coil as G.
6. Draw a typical B-H curve. Explain the graph. Define the terms(i)
Coercivity
(ii)
Retentivity
Explain the significance of the graph. What properties will you see in a material to
make a(a) permanent magnet
(b) an electro magnet
UNIT - IV
ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT
VERY SHORT ANSWER QUESTION S (1 marks)
Q1.
An electric lamp, connected in series with a capacitor and an a.c. source is glowing with
certain brightness. How does the brightness of bulb changes on reducing the capacitance?
Ans. On reducing the capacitance, capacitive reactance increases which reduce brightness of
lamp.
Q2.
Ans.
Q3.
A choke coil and a bulb are connected in series to a d.c. source. The bulb shines brightly.
How does the brightness changes when an iron core is inserted in the choke coil?
Brightness of bulb will not change because at steady d.c. , the choke coil has no
inductive reactance.
The current in the wire PQ is increasing. In which direction does the induced current
flows in the current loop.
P
Q
Ans.
Clockwise
Q4.
Give the direction in which the induced current flows in the wire loop, when the magnet
moves towards it as shown in figure.
N
S
Ans.
Clockwise when looked from magnet side of the loop.
Q5.
Why a transformer cannot be used to step up d.c. voltage?
Ans.
D.C. cannot produce varying field for secondary winding, therefore induced emf cannot
be produced in it.
Q6.
What is the phase difference between the the voltage across an inductor and a capacitor in
an a.c. circuit.
Ans.
1800
Q7.
Give the phase difference between applied a.c. voltage and current in a LCR circuit at
resonance.
Ans.
00 i.e. in phase
Q8.
What is the power consumed in (i) purely inductive and (II) purely capacitive a.c.
circuits?
Ans.
Zero
Q9.
:
What is the power dissipation of an a.c. circuit in which voltage and current are given by
V = 300 sin (ωt –π/2) and I = 10 sin ωt ?
Ans.
Power in a.c. circuit is ,P = VI cos φ
Here φ = π/2 and cos π/2 = 0
Thus P = 0
Q10.
In series LCR circuit, when voltage and current are in same phase?
Ans.
At resonance
Q11.
What is the power factor of an LCR series circuit at resonance.
Ans.
Unity.
Q12.
If number of turns of a solenoid is doubled, keeping the other factors constant, how does
self inductance of the solenoid changes?
Ans.
As L α N2 thus L ˈ α 4 N2
Hence self inductance increases to four times.
Q13.
Ans.
A plot of magnetic flux (φ) versus current is shown in figure for two inductors A & B.
which of the two has larger value of self inductance?
A
L = φ/I
B
Φ
I
For given I, A has larger value of φ, so A has larger self inductance.
Q14.
Why is a.c. more dangerous than d.c. for same voltage?
Ans. A.c. of same r.m.s. voltage as that of d.c. will have higher value of maximum voltage
given as
Vmax = √2 Vrms
This increases the value of a.c. which makes it more dangerous.
SHORT ANSWER QUESTION (2 or 3 marks)
Q1.
State faraday's laws of electromagnetic induction(EMI).
Ans.
1st law : When magnetic flux linking with a coil changes, an e.m.f. is induced in the coil.
This induced e.m.f. lasts so long as the change in magnetic flux continues.
2nd law: the magnitude of the induced e.m.f. produced in a coil is directly proportional to
the rate of change of magnetic flux dφ/dt linked with it.
I.e. I e I = dφ /dt
Q2.
State Lenz’s law. Show that this law follows the principle of conservation energy.
Ans.
Lenz’s law states that induced e.m.f. opposes the cause that produces this e.m.f.
In the arrangement shown in the figure, direction of the induced current is such that it
produces magnetic field which opposes the movement of magnet towards the coil.
Deflection of Galvanometer indicates the presence of electrical energy. Some work has to
be done to move the magnet which results into electrical energy. Electrical energy
produced in the coil is basically due to the mechanical energy applied to move the
magnet towards the coil. Hence Lenz’s law follows from the principle of energy
conservation.
N
S
G
S
N
Q3.
How are eddy currents produced? Give two applications of eddy currents.
Ans.
Eddy current are circulating currents produced in a metal itself due to EMI when it is
placed in changing magnetic flux in accordance with faradays laws of EMI. Eddy
currents are useful in induction furnaces and dead beat galvanometer.
Non-uniform magnetic
field
Eddy current
Metal
Q4.
Define self inductance. Write its unit. Give expression for self inductance of a long
solenoid having N turns.
Ans.
It is defined as the induced as the induced e.m.f. produced in the coil through which the
rate of decrease of current is unity.
OR
It is defined as the magnetic flux linked with a coil when unit current flows through it.
Its S.I. unit is henry. Self inductance of long solenoid is given by L= μ0μrN2A/l .
Q.5
Define mutual inductance. Write its S.I. unit. Give two factors on which the coefficient
of mutual inductance between a pair of coils depends.
Ans. Mutual induction of the two coils or circuits can be defined as the magnetic flux linked
with the secondary coil due to the flow of unit current in the primary coil. Its S.I. unit is
henry. It depends upon number of turns of both the coils, area of cross-section of primary
coil and length of coil.
Q.6 Draw a labeled diagram of a step down transformer . Mention two sources of energy loss
in a transformer.
Ans. Losses in a transformer are mainly because of (i) iron loss in the core of the transformer
and (ii) copper loss i.e. I2R loss in windings of the transformer.
Q..7 For given a.c. circuit, distinguish among resistance, reactance and impedance.
Ans.
S.No.
Resistance (R)
Inductive
Reactance
Capacitive
Reactance (XC)
Impedance (Z)
It opposes
direct current.
It is the total
opposition offered to
current (Due to
resistance, inductive
reactance and
capacitive reactance.)
(XL)
1.
2.
3.
It is opposition to
the flow of any
type of current.
It opposes the
flow of variable
current.
It is independent of It depends
frequency of source directly on the
of supply.
frequency of
source.
It depends
inversely on the
frequency of
source.
It is given by,
It is given by
XC = 1/2 πνC
It is given by
XL =2 πνL
R = ρ I/a
It depends on the
frequency of the
source.
It is given by, Z =
√[R2+(XL-XC)2]
Long Answer Questions ( 5 Marks)
Q.1.(a) State he condition for resonance to occur in a series LCR a.c. circuit and derive an
expression for the resonance frequency.
(b) Draw a plot showing the variation of the peak current (i0) with frequency of the a.c.
source used. Define the quality factor, Q of the circuit.
Ans.(a) Electrical resonance in Series LCR Circuit.
Electrical resonance takes place in a series LCR circuit when the circuit allows
maximum alternating current for a frequency at which capacitive reactance becomes
equal to the inductive reactance.
Impedance of a series LCR circuit,
Z = √[R2+(XL-XC)2]
And current in circuit,
I = E/Z = E/√[R2+(XL-XC)2]
Clearly I will be maximum when Z is minimum.
i.e. for electrical resonance
(XL - XC) = 0
Or XL = XC
the circuit.
i.e. ωL = 1/ωC i.e. ω = 1/√(LC), where ω is the angular frequency of
i.e. νm= 1/2π√(LC),
(b)
Refer to NCERT text book part 1 fig. 7.16 page no 248.
Quality Factor (Q-factor of Resonance Circuit is defined as 2π times the ratio of the energy
stored in the circuit to the energy dissipated in resistance per cycle of a.c. supply.
Q = 2π x energy stored in the circuit per cycle/energy dissipated per cycle.
Q.2.
Ans.
Draw a schematic diagram of a step-up transformer. Explain its working principle.
Deduce the expression for the secondary to primary voltage in terms of the number of
turns in the two coils. In an ideal transformer, how is this ratio related to the currents in
the two coils?
For figure refer to NCERT text book part 1 fig. 7.20. page no. 260.
Principle : It is based on the principle of mutual induction. It is a phenomena of inducing
e.m.f. in coil due to rate of change of current in near by coil.
Primary e.m.f.
Ep = -Npdφ/dt
Secondary e.m.f.
Es = -Nsdφ/dt
Thus
Es/Ep = Vs/Vp = Ns/Np
In ideal transformer, input power = output power
i.e.
ipVp = isVs i.e. Us/Up=Ns/Np=is/ip
NUMERICAL PROBLEMS
Q1. The electric mains in a house are marked 220 V, 50 Hz. Write down the equation for
instantaneous voltage.
Ans. Given Vrms = 220 V and ν = 50 Hz
V0 = √2 Vrms
= 1.414 × 220 = 311 V
And ω = 2πν = 2× 3.14 × 50 = 314 rad-1
Equation for instantaneous voltage, e = 311 sin 314 t
Q2. An alternating voltage E = 200 sin 300 t is applied across a series combination of R = 10 Ω
and an
Inductor of 800 mH. Calculate (i) impedance of circuit (ii) peak value of current in circuit (iii)
power
factor of the circuit.
Ans. (i) Impedance, Z = √(R2 + ω2 L2 )
Here ω = 300 rad-1
Z = √[102 + (300 × 0.8)2]
On solving Z = 240.2 Ω
(ii) peak value of current,
I0 = E0/Z = 200/240.2 = 0.83 A
(iii) Power factor, cos φ = R/Z = 10/ 240.2 = 0.042
Q3. A series LCR circuit with L = 5 H, C = 80 μF, R = 40 Ω is connected to a variable frequency
source of 230 V (i) determine resonance frequency of the circuit (ii) obtain impedance of
circuit and amplitude of current at resonance.
Ans. Given : Vrms = 230 V
(i) Resonance frequency, ω = 1/√(LC) = 1/ √(5 × 80 ×10-6) = 50 rad-1
(ii) At resonance, impedance ,Z = R = 40 Ω
And Amplitude, I0 = V0/Z = (230 × √2)/ 40
{ V0 = Vrms √2}
= 8.13 A
Q4.The output voltage of an ideal transformer connected to a 240 V a.c. mains is 24 V. When
this transformer is used to light a bulb with rating 24 V, 24 W. calculate current in
primary coil of the circuit.
Ans. Since V1/V2 = I2/I1
I1 = I2 × V2/ V1
I2 = W/V2 = 24/24 = 1A
Thus
I1 = 1 × 24/240 = 0.1 A
Q.8
zero.
Prove that the average power over a complete cycle of a.c. through an ideal inductor is
Ans.
Average power over a full cycle, PL = average of PL
Average of sin 2ωt for full cycle =
𝑇
= ∫ sin 2𝜔𝑡 𝑑𝑡 =
0
1
𝑇
[cos 2𝜔𝑡/2𝜔]
0
𝑇
1
= 2𝜔𝑇 [cos 2𝜔𝑇 − cos 0 O
= 1/2𝜔𝑇[cos
2𝜋
𝑇
𝑋 𝑇 − cos 0O]
𝐼
[1 − 1] = 0
2𝜔𝑡
Thus, average power = 0
=
UNIT – V
ELECTROMAGNETIC WAVES
Q1. Write expression for speed of e.m. wave in free space.
Ans. c = 1/√(μ0ε0) = E0/B0 where c = speed of light = 3 × 108 m/s
ε0 = absolute permittivity = 8.8564 × 10-12 C2N-1m-2
μ0 = absolute permeability = 4π × 10-7 TmA-1
Q2.State four properties of e.m. waves.
Ans. (i) they are transverse in nature.
(ii) they travel in free space with the velocity of light in vacuum i.e. 3 × 108 m/s
(iii) they exert radiation pressure
(iv) they do not require matter medium for propagation.
Q3.Identify the constituent radiation of e.m. spectrum which
(i)
(ii)
(iii)
Is used for studying crystal structure.
Is absorbed by the ozone layer in the atmosphere
Produce intense heating effect
Mention one more application for each of these radiation.
Ans. (i ) X – rays : these are used for medical diagnostic
(ii) Ultraviolet rays : these are used for sterilization purpose.
(iii)Infrared rays : these are used for physiotherapy
Q4. What is meant by transverse nature of e.m. waves? Draw a diagram showing the propagation
of an e.m. wave along the X- direction ,indicating clearly the directions of oscillating electric
and magnetic fields associated with it.
Ans. Transverse nature of e.m. waves: it means that variation of electric and magnetic fields are
perpendicular to each other and also to the direction of propagation of e.m. waves.
Refer to NCERT text book fig. 8.4 on page 275. Part 1
Q5. Give names of e.m. radiations in ascending order of wavelength.
Ans. Gamma rays (10-11 to 10-14 m), X – rays (10-8 to 10-12m), ultraviolet rays (4 × 10-7 to 6 × 1010
m),
visible rays (700 – 400 nm), infrared rays (1mm - 700 nm), microwaves (0.1 – 1
mm), radio waves
(> 0.1 m).
Q6. Explain e.m. spectrum.
Ans. Properties of e.m. waves : refer to Ans. 2.
S.No. Name
1.
Gamma
rays
Frequency Wavelength Production
Range
(Hz)
Range
1019 –
1023
10-11 to 1014m
Emitted by radioactive
nuclei,
Uses
In medicine, to
destroy cancer cells.
Produced in nuclear
reaction
2.
X – rays
1016 - 1020
10-8 to 1012m
Generated by
bombarding a metal
target by high energy
electron
Used as diagnostic
tool in medicine, to
study crystal
structures
3.
Ultraviolet
rays
1015 –
1017
(4 × 10-7 to
6 × 10-10) m
Produced by special
lamps & very hot
bodies (sun).
For eye surgery, to
kill germs in water
purifiers.
4.
Visible rays
4 × 1014 –
7 × 1014
700 – 400
nm
Jumping of electrons
in higher orbits
Provide us
information about
the world.
5.
Infrared
rays (heat
waves)
1012 - 1014
1mm - 700
nm
Produced by hot
bodies and molecules.
Infrared detectors
used in earth
satellite, used in
green house to keep
plants warm.
6.
microwaves 1010 - 1012
0.1 – 1 mm
Produced by special
vacuum tubes
(klystrons, gun diode
& magnetrons)
Microwave oven, for
radar system in
aircraft navigation.
7.
Radio
waves
> 0.1 m
Produced by
accelerated motion of
charges in conducting
wires.
In radio & television
communication
system, in cellular
phones to transmit
voice
communication.
10 - 109
UNIT VI
OPTICS
1 mark question
1. equal focal lengths. What is the focal length of the combination?
Ans. Infinite
2. When light travels from a rarer to denser medium, the speed decreases. Does
this decrease in speed imply a decrease in energy carried by the light wave?
Justify your answer.
Ans. Energy remains unchanged because E = hν where h and ν both
remains unchanged
3. How would the angular separation between the interference fringes in
young’s double slit experiment change when the distance between the slits
and screen is doubled.
Ans .Unchanged because, ѳ = λ/d i.e. independent of D.
4. How would the angular separation of in A converging lens is kept coaxially
in contact with a diverging lens - both the lenses being of terference fringe in
Young’s double slit experiment change when the distance between slit and
screen is halved?
Ans. Unchanged because, ѳ = λ/d i.e. independent of D.
5. How does the angle of minimum deviation of a glass prism vary ,if the
incident violet light is replaced with red light ?
Ans.δ= (μ-1) Since μv > μr So δ v > δr
6. Why does the bluish colour predominate in a clear sky?
7. You have learnt that plane and convex mirrors produce virtual images of
objects. Can they produce real images under some circumstances?
Ans. Yes
Plane and convex mirrors can produce real images as well. If the object is
virtual, i.e., if the light rays converging at a point behind a plane mirror (or
a convex mirror) are reflected to a point on a screen placed in front of the
mirror, then a real image will be formed.
8. Does the apparent depth of a tank of water change if viewed obliquely? If so,
does the apparent depth increase or decrease?
Ans The diver is in the water and the fisherman is on land (i.e., in air).
Water is a denser medium than air. It is given that the diver is viewing the
fisherman. This indicates that the light rays are travelling from a denser
medium to a rarer medium. Hence, the refracted rays will move away from
the normal. As a result, the fisherman will appear to be taller. Yes; Decrease.
9. The refractive index of glass is 1.5. What is the speed of light in glass?
Speed of light in vacuum is 3.0 × 108 m s−1)
Ans.
Hence, the speed of light in glass is 2 × 108 m/s.
10. Is the speed of light in glass independent of the colour of light?
Ans. The speed of light in glass is not independent of the colour of light.
11.What is the Brewster angle for air to glass transition? (Refractive index of
glass = 1.5.)
Ans.
Therefore, the Brewster angle for air to glass transition is 56.31°.
12.In what way is diffraction from each slit related to the interference pattern in
a double-slit experiment?
Ans. The interference pattern in a double-slit experiment is modulated by
diffraction from each slit. The pattern is the result of the interference of the
diffracted wave from each slit.
13.Define resolving power of microscope.
Ans. Resolving power of a microscope is defined as te reciprocal of the
least separation between close objects, so that they appear just separate when
seen through a microscope.
14. How does the resolving power of a telescope change when the aperture
when the aperture of its objective is increased?
Ans. If aperture is increased then resolving power is also increased
15.What will happen when a lens is immersed in a liquid of refractive index
more than that of glass?
Ans. The nature of the lens will change. In other words, a convex lens will
behave as a concave lens and a concave lens as a convex lens. The focal
length of the lens will also change.
16.In Young's double slit experiment, one slit is covered. What will be the effect
?
Ans. The interference pattern will be replaced by diffraction pattern due to a
single slit.
17.Why does sky appear blue ?
Ans. The scattering of light by air molecules is inversely proportional to
fourth power of the wavelength
of light (i.e. scattering 1/4). Thus, the sky appears blue because the
scattered light has a large mixture of shorter wavelengths (violet and blue)
than of longer wavelengths (yellow and red).
18.Two thin lens of power 6 D and – 2 D are in contact. What is the focal
length of the combination? Ans.P= 6 D + (-2D) = 4D
F = 100cm/4
= 25 cm
19. What is the convincing proof of wave nature of light ?
Ans. The interference of light waves is a convincing proof of wave nature
of light.
20. Why does sky appear blue ?
Ans. The scattering of light by air molecules is inversely proportional to
fourth power of the wavelength of light (i.e. scattering α1/λ4). Thus, the
sky appears blue because the scattered light has a large mixture of shorter
wavelengths (violet and blue) than of longer wavelengths (yellow and red).
2 Marks Question
1. Define refractive index of a transparent medium.
Ans. A ray of light passes through a triangular prism, plot a graph showing
the variation of the angle of deviation with the angle of incidence.
2. Define refractive index of a transparent medium.
Ans. A ray of light passes through a triangular prism, plot a graph showing
the variation of the angle of deviation with the angle of incidence.
3.(i) State the principle on which working of optical fibre is based.
(ii) What are the conditions for this phenomenon to occur?
4.(i) What is the relation between critical angle and refractive index of a
material? (ii) Does the critical angle depend on the colour of the light?
5. Define the term linearly polarised light.
When does the intensity of transparent light become maximum, when a
Polaroid sheet is rotated between two crossed Polaroids?
6. Draw a labeled ray diagram of an astronomical telescope in the near
point position. Write the expression for its magnifying power.
7. State one feature by which the phenomenon of interference can be
distinguished from that of diffraction a parallel beam of light of
wavelength 600nm is incident normally on a slit of width ‘a’. If the
distance between the slits and the screen is 0.8 m and the distance of 2 nd
order maximum from the center of screen is 15mm, calculate the width
of the slit.
8. Define resolving power of a compound microscope. How does the R.P.
of a compound microscope change when (i) refractive index of the
medium between object and objective lens increases? (ii) Wavelength
of radiation used is increased?
9. A right angled (angle B = 900) crown glass prism with critical angle 410 is
placed before an object, PQ, in two positions as shown. Trace the paths of
the rays from P and Q passing through the prisms in the two cases.
(i)
P
Q
P
Q
10. Draw a ray diagram of an astronomical telescope in near point position.
Write down the expression for its magnifying power.
11. Draw a ray diagram of an astronomical telescope in the normal
adjustment. Write down the expression for length of telescope in normal
adjustment.
12.(i) State the principle on which working of optical fibre is based.
(ii) What are the conditions for this phenomenon to occur?
13.State one feature by which the phenomenon of interference can be
distinguished from that of diffraction a parallel beam of light of
wavelength 600nm is incident normally on a slit of width ‘a’. If the
distance between the slits and the screen is 0.8 m and the distance of 2nd
order maximum from the center of screen is 15mm, calculate the width
of the slit.
14.Define resolving power of a compound microscope. How does the R.P.
of a compound microscope change when (i) refractive index of the
medium between object and objective lens increases? (ii) Wavelength
of radiation used is increased?
15.The refractive index of the material of an equi-double convex lens is 1.5.
What is its focal length?
1/f = (μ-1){1/R1 - 1/R2)
= (1.5 -1){1/R+1/R}
= 1/ R
Therefore f = R
16.What will be the effect on the interference fringes obtained in Young's
double slit experiment if the monochromatic source of light is replaced by
white light source?
Ans. There is a central white fringe with a few coloured fringes on both
sides. (i) The path difference to the central fringe is zero for all colours.
Consequently, each colour produces a bright fringe here. As they overlap,
a white fringe is formed.
(ii) In white light, there are different colours (wavelengths). The fringe
of one colour is slightly displaced from the fringes of the other colours of
the same order (β= λD/d).
Thus, the fringes of different colours do not exactly overlap. As
a result, the edges of the fringes are seen coloured.
17.In a single slit diffraction experiment, the width of the slit is made double
the original width. How does this affect the size and intensity of the
central diffraction band?
Ans.
In a single slit diffraction experiment, if the width of the slit is
made double the original width, then the size of the central diffraction
band reduces to half and the intensity of the central diffraction band
increases up to four times.
18.What should be the path difference and phase difference for constructive
and destructive interference?
Ans. Interference
Path difference
Phase difference
Constructive
nλ
n (2π)
Destructive
(2n + 1)λ/2
(2n + 1)π
Here n = 0, 1, 2, 3 ....
3 Marks Question
1. A compound microscope uses an objective lens of focal length 4 cm and
eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the
objective lens. Calculate the magnifying power of the compound microscope.
Also calculate the length of the microscope.
2. In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by
monochromatic light of wavelength 450 nm. The screen is 1 m away from the slits
(a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central
maximum.
(b) How will the fringe pattern change if the screen is moved away from the slits?
3. In young’s double slit experiment the monochromatic light of wavelength 630 nm
illuminates the pair of slits and produces an interference pattern in which two consecutive
bright fringes are separated by 7.2 mm. find the wavelength of the light from the second
source.
What is the effect on the interference fringes if the monochromatic source is replaced by
a source of white light?
4. In young’s double slit experiment, monochromatic light of wavelength 600 nm
illuminates the pair of slits and produces an interference pattern in which two consecutive
bright fringes are separated by 10 mm. another source of monochromatic light produces
the interference pattern in which bright fringes are separated by 8 mm. find the
wavelength
of
light
from
the
second
source.
What is the effect on the interference fringe sift the monochromatic source is replaced
with a source of white light.
5. Draw a labeled ray diagram to show the image formation by a compound
microscope. Write the expression for its magnifying power. How does the resolving
power of a compound microscope change , when (i) refractive index of the medium
Between the object and the objective lens increases; and (ii) wavelength of the
radiation used is increased ?
6. Two narrow slits are illuminated by a single monochromatic source. Name the pattern
obtained on the screen. One of the slits is now completely covered. What is the name of
the pattern now obtained on the screen? Draw intensity pattern obtained in the two cases.
Also write two differences between the patterns obtained in the above two cases.
7. How does an unpolarised light gets polarised when passed though a polaroid? Two
polaroids are set in the crossed positions. A third polaroid is placed between the two
making an angle θ with the pass axis of the first polaroid. Write the expression for the
intensity of light transmitted from the second polaroid. In what orientation the intensity of
the transmitted light will be (i) minimum and (ii) maximum.
.
5 marks questions
1. State the importance of coherent sources in the phenomenon of interference.
In Young's double slit experiment to produce interference pattern, obtain the conditions for constructive
and destructive interference. Hence deduce the expression for the fringe width.
How does the fringe width get affected, if the entire experimental apparatus
of Young is
immersed in water?
2.
State Huygens Principle. Using this principle explain how a .-diffraction pattern is obtained on a screen due
to a narrow slit on which a narrow beam coming from a monochromatic source of light is incident
normally.
Show that the angular width of the first diffraction fringe is half of that of the central fringe.
If a monochromatic source of light is replaced by white light, what change would you observe in the
diffraction pattern?
3. (a) (i) Draw a labelled ray diagram to show the formation of image in an astronomical
telescope for a distant object.
(ii) Write three distinct advantages of reflecting type of telescope over a refracting
type of telescope.
(b) A convex lens of focal length 10 cm is placed coaxially 5 cm away from a
concave lens of focal length 10 cm if the object is placed 30 cm in front of the
convex lens. Find the position of the final image formed by the combined system. /4
a. With the help of a suitable ray diagram, derive the mirror formula for the concave
mirror.
b. The near found of a hypermetropic person is 50 cm from the eye. What is the
power of the lens required to enable the person to read clearly a book held at 25 cm
from the eye.
4. Derive the lens formula, 1/f = 1/v – 1/u for a concave lens, using the necessary ray
diagram. Two lenses of power 10 D and -5 D are placed in contact (i) calculate the
power of the new lens. (ii) Where should an object be held from the lens, so as to
obtain a virtual image of magnification
5. What are coherent sources of light? Two slits in Young’s double slit exp are
illuminated by two different sodium lamps emitting light of the same Wavelength.
Why is no interference pattern observed? (b) Obtain the condition for getting dark
and bright fringes in Young’s exp. Hence write the expression for fringe width. (c) If
s is the size of the source and d its distance from the plane of the two slits, what
should be the criterion for the interference fringes to be seen?
6. State the essential condition for diffraction of light to take place. Use Huygens’
principle to explain diffraction of light due to a narrow single slit and the formation of
a pattern of fringes obtained on the Screen. Sketch the pattern of fringes formed due
to diffraction at a single slit showing variation of intensity with angle θ.
7. What are coherent sources of light? Why are coherent sources required to obtain
sustained interference pattern? State three characteristics features which distinguish
the interference pattern due to two coherently illuminated sources as compared to that
observed in a diffraction pattern due to a single slit.
What are coherent sources of light? State two conditions for two light
sources to be coherent. Derive a mathematical expression for the width of
interference fringes obtained in Young’s double slit experiment with the
help of a suitable diagram.
9. Using Huygens’ principle, draw a diagram to show propagation of a wave
front originating from a monochromatic point source. Describe diffraction
of light due to a single slit. Explain formation of a pattern of fringes
obtained on the screen and plot showing variation of intensity with angle θ
in single slit diffraction.
10. What is meant by a linearly polarized light? Which type of waves can be polarized?
Briefly explain a method for producing polarized light. Two Polaroids are placed at
90o to each other and the intensity of transmitted light is zero. What will be the
intensity of transmitted light when one more Polaroid is placed between these two
bisecting the angle between them? Take intensity of unpolarised light as Io.
8.
11.
12.
Important points
1.
The mirror formula for concave mirror as well as convex mirror is
1/u + 1/v = 1/f = 2/R
Here u = distance of the object from the pole of the mirror
v = distance of the image from the pole of the mirror
f = focal length of mirror; R = radius of curvature of the mirror
2. Linear magnification, m =
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑚𝑎𝑔𝑒
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡
𝑣
= hi/ho = - 
𝑢
The arbitrary minus sign given to linear magnification helps to tell us whether the
image is erect or inverted w.r.t. object
3.
4.
UNIT 7
DUAL NATURE OF MATTER AND RADIATION:
IMPORTANT CONCEPTS
1.
Photon
The basic unit ("quantum") of electromagnetic radiation (and therefore light), usually denoted
by . Photons were first postulated by Planck, whose measurements of the blackbody
spectrum showed that electromagnetic radiation had to come in discrete units, which were
dubbed "photons" by the chemist Gilbert Lewis in 1926.
The energy of a photon of frequency
is given by
where h is Planck's constant. Because the energy of photons is directly proportional to their
frequency, low-energy photons have low frequencies, while high-energy photons have high
frequencies. Low-energy photons are called radio waves or microwaves, medium-energy
photons are called light (or light waves, or visible light), high-energy photons are called Xrays, while those having higher energy still are called gamma rays.
2. Electron Emission
Electron emission is defined as liberation of free electron from a surface of a
substance caused by the external energy transferred to the electrons.
3. Work Function
Electron emission tends to occur on metal, because metal is a
substance with much free electron in between its molecules.
Nucleus attracting force does not strong enough to put the
electron standstill. Every time the free electrons move around
from one molecule to another but it can't leave out
from metal surface. In order to emit from the metal surface
these free electrons require additional external energy. The
amount of outside energy require by electron to emit from the metal surface is
known as work function. The work function usually defined in electron volt (eV)
unit.
4. Kinds of electron emission:
The additional external energy
required by the electron to emit from the metal surface could come from few
sources such as
Contd…172
171
Heat energy, energy stored in the electron field, light energy or kinetic energy.
Accordingly there are following four method of obtaining electron emission from
the metal surface.
4.1 Thermionic emission
4.2 Field emission
4.3 Secondary emission
4.4 Photoelectric emission
1.1
1.2
1.3
1.4
When additional energy comes to the electron in the form of heat
energy, the kinetic energy of electron increases and its movement
becomes uncertain. Finally there will be electrons that leave out from the
metal surface. This is called as thermionic emission.
When a conductor put in a place very close to high voltage conductor, the
electric field from the conductor will exert attractive force on the
free electron in metal. If the positive field is big enough the free
electron will succeed in overcoming restraining of the metal surface and
it will emit from the metal surface. This is called as field emission.
Electron emission from a metallic surface by the bombardment of high
speed electrons or other particles is known as secondary emission. When
high speed electrons suddenly strike a metallic surface, they may give
some or all of their kinetic energy to the free electrons in metal. If the
energy of the striking electrons is sufficient, the free electron
will escape from the metal surface and this phenomenon is called as
Secondary Emission.
When a beam of light strike the surface of cathode the energy from
photons will be transferred from the photons to free electron within the
cathode. If the energy from photons is greater than the metal work
function the free electron will knock out from the cathode surface. The
emitted electron called as photo electron. The amount of photo electron
depends of the light intensity. This phenomenon is called as Photoelectric
Emission.
5. The Photoelectric Effect
Albert Einstein gave the explanation in 1905: Light consists of particles (photons), and the
energy of such a particle is proportional to the frequency of the light. There is a certain minimum
amount of energy (dependent on the material) which is necessary to remove an electron from the
surface of a zinc plate or another solid body (work function). If the energy of a photon is bigger
than this value, the electron can be emitted. From this explanation the following equation results:
Ekin = h ν – W
Ekin ... maximal kinetic energy of an emitted electron
h ..... Planck constant (6.626 x 10-34 Js)
ν ..... frequency
W ..... work function .The above is known as Einstein’s photoelectric
equation.
172
Contd…173
6. The Laws of Photo-Electric Emission
i) The number of electrons emitted per second is directly proportional to the
intensity of the radiation.
ii) The maximum kinetic energy of the electrons emitted increases with the
frequency of the radiation.
iii) There is a minimum frequency below which no emission occurs.
A graph of K.E.max of photo-electrons against frequency of radiation has the
following form.
The minimum frequency for photo-electric emission is called the threshold
frequency, fo or νo .
7. Hertz's spark gaps
In 1887, Heinrich Hertz observed the photoelectric effect and the production and reception of electromagnetic (EM) waves. He published
these observations in the journal Annalen der Physik. His receiver consisted of a coil with a spark gap, where a spark would be seen
upon detection of EM waves. He placed the apparatus in a darkened box to see the spark better. However, he noticed that the maximum
spark length was reduced when in the box. A glass panel placed between the source of EM waves and the receiver absorbed ultraviolet
radiation that assisted the electrons in jumping across the gap. When removed, the spark length would increase. He observed no decrease
in spark length when he substituted quartz for glass, as quartz does not absorb UV radiation. Hertz concluded his months of
investigation and reported the results obtained. He did not further pursue investigation of this effect, nor did he make any attempt at
explaining how this phenomenon was brought about.
8. Von Lenard's observations
In 1902, Philipp Lenard observed the variation in electron energy with light frequency. He used a powerful electric arc lamp which
enabled him to investigate large changes in intensity, and had sufficient power to enable him to investigate the variation of potential with
light frequency. His experiment directly measured potentials, not electron kinetic energy: he found the electron energy by relating it to
the maximum stopping potential (voltage) in a phototube. He found that the calculated maximum electron kinetic energy is determined
by the frequency of the light. For example, an increase in frequency results in an increase in the maximum kinetic energy calculated for
an electron upon liberation - ultraviolet radiation would require a higher applied stopping potential to stop current in a phototube than
blue light. However Lenard's results were qualitative rather than quantitative because of the difficulty in performing the experiments:
the experiments needed to be done on freshly cut metal so that the pure metal was observed, but it oxidized in a matter of minutes even
in the partial vacuums he used. The current emitted by the surface was determined by the light's intensity, or brightness: doubling the
intensity of the light doubled the number of electrons emitted from the surface. Lenard did not know of photons.
9. Matter wave
In quantum mechanics, a matter wave or de Broglie wave is the wave (wave-particle duality) of
matter. The de Broglie relations show that the wavelength is inversely proportional to the
momentum of a particle and that the frequency is directly proportional to the particle's kinetic
energy. The wavelength of matter is also called de Broglie wavelength. The theory was
advanced by Louis de Broglie in 1924 in his PhD thesis; he was awarded the Nobel Prize for
Physics in 1929 for this work, which made him the first person to receive a Nobel Prize on a PhD
thesis.
173
Contd…174
10. DeBroglie Wavelength
Louis de Broglie was the first person to establish an equation for the relationship between an
electron’s momentum and its wavelength. He concluded that:
where h is a number called Planck’s constant (named after Max Planck) and is equal to 6.63 x
10-34 J × s or 4.14 x 10-15 eV × s.
When we observe electron diffraction, the electrons’ kinetic energy is easier to measure than
their momentum, so we write the de Broglie wavelength as
or
This equation is consistent with our results in the previous activity — as the energy increases the
wavelength decreases.
11. Davisson and Germer's Experiment
INTRODUCTION
In the year 1927, Davisson and Germer conducted their famous experiment which was
the experimental verification of De-Broglie's hypothesis i.e.  = h/p.
THE EXPERIMENT
Their experimental setup was enclosed in a vacuum chamber as shown below:
A beam of electrons accelerated through the potential V was allowed to strike a nickel crystal.
Measurements were made to count the number of electrons scattered by the crystal.
OBSERVATIONS
Davisson and Germer further investigated properly oriented crystals to observe if could be possible
to interpret that electron behave as waves of all wave lengths () as given by De-Broglie's
hypothesis. They calculated the wave length of electron from the known accelerating potential V by
applying the relation:
174
Contd…175
The wave length associated with the above equation agreed with the De-Broglie's
prediction. Thus it is confirmed that electron has a wave like nature because only a wave
has wave length.
VERY SHORT ANSWER TYPE QUESTIONS (CARRYING 1-2 MARKS EACH)
1. What are photo-electrons?
Ans. The electrons ejected from metals in photo-electric effect are called photoelectrons.
2. What is the rest mass of a photon?
Ans. Zero.
3. Which photon is more energetic, violet one or red one?
Ans. Violet: Since E = h ν, the frequency (ν) of violet is more than that of red.
4. If h is Planck’s constant, find the momentum of a photon of wavelength 0.01 A°?
Ans. Momentum of a photon = h / λ = h / 0.01 ×10-10 =1012 h.
5. How will the photoelectric current change on decreasing the wavelength of incident
radiation for a given photo-sensitive material?
Ans. Photoelectric current is not affected on decreasing the wavelength of incident
radiation.
6. The work function of cesium is 2 eV. Explain this statement.
Ans. For the emission of photoelectrons from the cesium metal, the minimum energy of
the incident light on the metal surface should have the photons of energy 2 eV.
7. What is the main aim of Davisson and Germer’s experiment?
Ans. To verify the wave nature of electron.
8. Calculate the energy of a photon of frequency 6 ×1014 Hz in electron volt. Given that
h = 6.62 ×10-34 Js.
Ans. E = h ν = 6.62 ×10-34 ×6 ×1014 J = (6.62 ×6 ×10-20) / (1.6 ×10-19) eV =2.49 eV.
9. What is the value of stopping potential between the cathode and the anode of a
photo cell, if the maximum kinetic energy of the electrons emitted is 5 eV?
Ans. If Vo is the stopping potential, then e Vo = Max. KE = 5 eV or Vo = 5 V.
10. Two metals A and B have work functions 2 eV and 5 eV respectively. Which metal
has lower threshold wavelength? Ans. As work function, W = hc / λo or λo = hc/ W i.e.
λo is proportional to 1 / W, so metal B of greater work function has lower threshold
wavelength.
UNIT – VIII
Atom and Nuclei
Q 1.
radii.
1
Ans:
Q 2.
Ans :
Q 3.
1
Ans:
Q 4.
Ans:
Q 5.
Ans:
Q 6.
Ans:
Q 7.
Ans:
Two nuclei have mars numbers in the ratio 1:8 what is the ratio of their nuclear
R1/R2 = (A1/A2)1/3 we have get R1/R2 = 1/2
Two nuclei have means numbers in the ratio 1:2 What is the ratio of their
numclear densities?
1
D1/D2 = 1/1 , it is independent of mass number.
Name the series of hydrogen spectrum lying in the visible region.
Balmer.
What is impact parameter? What is the value of impact parameter for a head on
collision?
1
Perpendicular distance between the velocity vector of α-particle and center of the
nucleus and for head an collision, β = 0.
Bohr’s theory of module of a Hydrogen atom, name the physical quantity which
ℎ
equal to an integral multiple of 2𝜋 ?
1
Angular momentum.
What is the relation between n and radius ‘r’ of the orbit of electron in a
Hydrogen atom according to Bohr’s theory?
1
2
rαn
How is the half life of a radio active substance related to its decay constant?
1
T1/2 = 0.693/λ where λ is decay constant.
Q 8.
Ans:
Q 9.
Ans:
Q 10.
Ans:
Q 11.
Write the nuclear decay process for β – decay of 32
15𝑃 .
1
32
32
−
−
𝑃
→
(Refer NCERT book page no. 450)
16𝑆 + 𝑒 + 𝑣
15
Define activity of a radioactive nuclide. Write its S I unit.
1
It is define as the rate of disintegration of the substance.
Activity A= Number of atoms disintegrated per second.
S I unit is Becquerel (Bq).
If ground state energy of hydrogen atom is -13.6 ev. What is the K.E. and K.E. of
the electron in this state.
2
K.E. = - E = 13.6ev.
P.E. = -2 K.E. = -27.2 ev.
Sketch the energy level diagram for Hydrogen atom. Mark the transitions
corresponding to Lyman and Bulmer Series.
2
Ans:
ev
1
𝐸 ≈ −.5
2
3
4
𝐸 ≈ −.7
𝐸 ≈ −1
𝐸 ≈ −3
5
Bulmer
𝐸 ≈ −13
Q 12.
1 (n)
Lyman
Draw a labeled diagram of the experimental set up of Rutherford’s alpha particle
scattering experiment. Write two observations draw from this experiment. 2
ev
Ans:
Gold foil about
10-8m thick
vacume
Beam of α –
particles
Source of α –
particles
Most pass
through
𝜃
ZnS screen
Microscope
About 1 in 8000
is reflected
book
Some are deviated through a large
angle 𝜃
Observation
1: Most of the α- particle puss through gold fail without any appreciable deflection.
2: A very small no of α- particle 1 in 8000) suffer deflection of nearly 180o.
Q 13.
What is nuclear force? Mention any two important properties of this force.
2
Ans:
Nuclear force is the strongest force in nature, which acts between the nucleons in
nucleus.
Properties :
1. It is very short range force.
2. It is charge independent force.
Q 14.
2
Ans:
Define muss defect and Binding energy.
Q 15.
Draw the graph showing the variation of binding energy per nucleon with the
muss number.
2
Refer NCERT text book, fig 13.1 on page no 444.
Ans:
Q 16.
Ans:
Muss defect is the difference between the sum of musses of the nucleons of a
nucleus and the rest muss of the nucleus.
∆m = [𝑍𝑀𝑝 + (A − Z)𝑚𝑛 ] − M
Binding energy of a nucleus is the energy equivalent to the muss defect of the
nucleus.
Design the relation N(t) = No e-λt for radio active decay. Obtain the relation
between decay constant and half life.
Refer to NCERT textbook page no. 446
2
Q 17.
3
Ans:
Explain α-decay, β-decay and 𝜗-decay.
Q 18.
3
Define decay constant of radioactive sample. Which of the following radiation.
Ans:
Refer to NCERT Text book page no. 449 and 457.
α-rays, β- rays, 𝜗-rays.
i)
are similar to X-rays?
ii)
are easily absorbed by matter?
iii)
travel with greatest speed?
iv)
are similar in nature to cathode rays?
Decay constant of a radioactive substance is the reciprocal of time at the end of
which, the undecayed atoms reduce to 1/e times the original number of atoms No.
(i)
γ-rays are electromagnetic waves like x-rays.
(ii)
(iii)
(iv)
α-particles are easily absorbed by matter.
𝜗-rays travel with greatest speed.
𝛽-rays are also negatively charged like cathode rays.
Q 19.
A radioactive nucleus ‘A’ undergoes a series of decay according to following
scheme :
𝛼
𝛽
𝛼
𝜗
A → 𝐴1 → 𝐴2 → 𝐴3 → 𝐴4
The muss number and atomic number of A are 180 and 72 respectively. What are
these numbers for A4?
3
Q 20.
Ans:
Q 21
Ans:
Draw a diagram showing the variation of B.E. / nucleon with the muss number.
Mark the region where the nuclei are (i) most stable (ii) Prone to fusion (iii)
Prone to fission. On the basis of this cure explain nuclear fission and fusion.
5
Refer to NCERT text book, fig 13.1 on page no 444 and explain paint III and IV
State Bohr’s Postulates of atomic structure. Using these postulates derive an
expression for total energy and radius of nth orbit of an electron in a hydrogen
atom.
5
Refer NCERT text book. Bohr’s postulate paint (i) (ii) and (iii) on page no 424.
And derivation from NCERT text book on page no 425.
UNIT IX
Electronic Devices
Q 1. Draw the logic circuit of NOT gate and write its truth table.
Ans:
A
Y
A
Y
Q 2. Name the type of biasing of p-n junction diode. So that the junction offers very high
0
1
resistance.
Ans: Reverse biasing
1 which0when added to pure Si, produces (i) n- type and (ii) pQ 3. Name one imparity each,
type semiconductor.
Ans: (i) for n- type – arsenic.
for p- type – indium
Q 4. Draw energy band diagram of a p- type semiconductor.
Ans:
C.B.
Eg. o o o o
Acceptor energy level
o o V.
0.01 – 0.05 ev
B.
(T > 0 K)
Refer to NCERT book fig no 14.9 page no . 477
Q 5.
Name two factors on which electrical conductivity of a pure semiconductor at a given
temperature depends.
Ans: (i)
Band gap
(ii)
Biasing
Q 6.
Give the ratio of the number of holes and the number of electrons in an intrinsic semiconductor.
Ans: ηe = ηn So
1:1
Q 7. How does the width of depletion layer change in reverse li of p-n. junction diode?
Ans: Increases.
Q 8.
Draw the Circuits Symbol for npn transition.
Ans:
C
B
E
Q 9. Draw the circuit symbol for LED.
Ans:
Q 10. Which of the diodes is (i) forward biased and (ii) reverse biased in the following circuits?
Justify your answer.
Ans:
+3 V
-10 V
A
B
-10 V
+4 V
-8 V
C
D
Ans: Fig (a) diode is reverse biased because p-type of the diode is at lower potential.
Fig (b) diode is reverse biased, p-type of the diodes is at lower potential.
Fig (c) diode is forward biased, because p- type of the diode is at higher potential.
Fig (d) diode is forward biased because p- type of the diode is at higher potential.
Q 11. Write two character feature to distinguish between n-type and p-type semiconductor.
Ans: Two distinguish feature are –
(i)
(ii)
n-type
Majority charge carriers are electrons
Mobility is high
(i)
(iii)
p-type
Majority charge carriers are holes.
Mobility is low.
Q 12. How does a light emitting diode (LED) works? Give two advantages of LED’s over the
conventional incandescent lamps.
Ans: Working of LED :- LED works in forward bias at the junction when majority charge
carrier recombine with minority charge carriers, which grow in number due to diffusion
of charges across the junctions, energy is released in the form of photons.
Advanteges:(i)
Low operational voltage and
(ii)
Fast on-off switches capability
Q 13. Draw a circuit diagram to slow how a photo diode is viewed. Draw its characteristics
curres for two different illumination intensities.
N𝜗
Ans:
NA
n side
Refer to NCERT text book fig no. 14.23 Page no. 487
mA
Reverse line
I
1I
1
Volt
p side
Q 14. Which special type of device can act as a voltage regulator? Give the symbol of this
divide and draw the general shape of its V-I characteristics.
Ans: Zener divide
I mA
Reverse
Line
V2
Forward
Line
V
I 𝜇A
Q 15. Identify the logic gates marked P and Q in the given logic circuit. Write down the output
at X for the inputs
(i)
A = 0, B = 0 and
(ii) A = 1, B = 1.
A
Ans: P →
Q→
Q
P
B
X
NOR gate
AND gate
Q 16. How is an n-type semiconductor formed? Name the major charge carriers in it. Draw the
energy band diagram of an n-type semiconductor.
Ans: When and intrinsic semiconductor (like Si, Ge) is doped with pentavalent impurity atoms
(e.g. arsenic , phosphorus etc) n-type semiconductor is formed. Each impurity atom
donates an electron majority charge carriers in n-type semiconductor are electrons.
C. B.
Eg
.
levelO
.
.
.
.
.
V. B.
.
Donor energy
O
Q 17. Define the depletion layer and the potential barrier in a p-n junction diode.
Ans: Refer to NCERT text book.
Q 18. If the output of a two input NOR gate is fed as the input to a NOT gate (i) Name the new
logic gate abstained and (ii) write down its truth table.
Ans. (i) OR gate
A
0
1
0
1
B
0
0
1
1
Y
0
1
1
1
Q 19. In half wave rectification, What is the output frequency if the input frequency is 50 Hz.
What is the output frequency of a full wave rectifier for the same input frequency.
Ans: 50 Hz for half wave
100 Hz for full wave.
Q 20. What is a solar cell? How does it works? Give its one use.
Ans: Solar cell is device for converting solar energy into electricity. It is basically a p-n
junction operating in a photovoltaic mode without external bias.
Working: When light photons fall at the junction electron-hole pairs are generated. those
more in opposite direction due to junction field. These charges accumulate at the two
sides of the junction and photo voltage is developed.
Use: It is used in calculators etc.
Q 21. Draw the circuit diagram showing use of a transistor as an oscillator.
Ans: Refer to NCERT book fig no 14.33 (a)and (b) on page no 500.
Q 22. What is a solar cell. Draw its circuit diagram. Draw I- V characteristics of a solar cell.
Ans: Definition given in Q no 19.
For Circuit and I-V graph.
Refer NCERT text book fig no 14.25 (a) and (b) on page no 489.
Q 23. What is LED with the help of a diagram, show the biasing of a Light Emitting Diode
(LED). Give its two uses.
Ans: It is a heavily doped p-n junction which under forward bias emits spontaneous radiation.
h𝜗
n
n
p
Metalized contact
Q 24. With the help of energy band diagram, distinguish between conductors, semiconductors
and insulators.
Ans: Refer NCERT text book fig. 14.2 on page no. 471.
Q 25. What is a rectifier. With the help of a labeled circuit diagram, explain the working of a
full wave rectifier. Draw the input and output waveforms.
Ans: Refer NCERT text book, fig 14.19 on page no 484.
Q 26. Give the symbol of a n-p-n transistor, show the biasing of a n-p-n transistor and explain
transistor action.
Ans: Refer NCERT text book fig 14.27 and 14.28 on page no 492.
Q 27 What is Zener diode with the help of a circuit diagram explain the use of Zener diode as a
DC voltage regulator.
Ans: Refer to NCERT text book fig. no 14.21 and fig. 14.22
Q 28. Draw a circuit diagram of a common emitter amplifier, Deduce an expression for its
voltage gain. Explain briefly how the output voltage is out of phase by 180o with the
input voltage.
Ans: Refer to NCERT text book fig. no 14.32 on page no 498.
UNIT X
Communication System
Q 1.
Name the type of communication in which the signal is a discrete and binary coded
version of the message of information.
1
Ans: Digital communication.
Q 2.
Ans:
Name the types of communication systems according to the mode of the transmission.
1. Analog communication
2. Digital communication
Q 3.
Ans:
Name the device which can represent digital data by analog signal and vece-versa. 1
Modem.
Q 4.
Ans:
How does the effective power radiated by an antenna vary with wavelength?
Power radiated by an antenna is inversely proportional to the wavelength.
Q 5.
Ans:
What is transducer.
Any device that convert one form of energy into another can be termed as a transducer.
Q 6.
Ans:
Bolck diagram of a receiver is as shown.
output
Recive
Signal
Ans: X = I F Stage
Y = Amplifier
Q 7.
Ans:
Amplifier
X
Detector
Y
Identify X and Y
What is the requirement of transmitting microwaves from one to another on the earth?
The transmitting and receiving antenna must be in line of sight.
Q 8.
Ans:
What type of modulation is required for television broadcast?
Frequency modulation.
Q 9.
Ans:
What type of modulation is required for television broadcast of voice signal.
Amplifier Modulation.
Q 10. What do you mean by modulation.
Ans: Superimposition of baseband signals over carrier wave is called modulation.
Q 11. What is ground wave? Why is ground wave transmission of signals restricted to a
frequency of 1500 KHz?
Ans: Low frequency radio waves propagating along ground is called ground wave. Energy
losses due to absorption of energy by the earth are frequency dependent. For frequencies
above 1500 KHz, it is quite high.
Q 12. What is sky wave proportion of waves? Explain why sky wave transmission of em wave
can not be used for TV transmission.
Ans: In Sky wave propagation, transmitted wave goes up in the sky and is reflected back from
the ionosphere. Waves transmitted towards ionosphere suffer total internal reflection. The
radio waves up to frequency 40 MHz can be reflected back by ionosphere .
TV transmission takes place at frequencies higher than 40 MHz, so sky wave
propagation can not be used for this.
Q 13. What is space wave propagation? Give two examples of communication system which
use space wave mode. A TV tower is 80 m tall, calculate the maximum distance upto
which the signal transmitting from the tower can be received.
Ans: It is the mode of propagation in which waves travel in space in a straight line from the
transmitting antenna to the receiving antenna.
Example: Television broadcast, Microwave links , satellite communication.
(c)
given
h = Som.
Formula
d = √2Rh
Q 14. Draw a schematics diagram showing the (i) Ground wave (ii) Sky wave and (iii) Space
wave propagation modes for e m waves.
Write the frequency range for each of the following.
(i)
Standard AM broadcast
(ii)
Television
(iii) Satellite Commubication
Ans: Refer NCERT text book fig 15.6 on page no 522.
Q 15. Write any two factors which justify the need for modulating a signal. What do you mean
by amplitude modulation? Draw a diagram showing an amplitude modulation wave by
superposing a modulating signal over a sinusoidal carrier wave.
Ans: (i)
Antenna size become practically viable.
(ii)
It avoids mixing up of signals from different transmitters.
Refer NCERT text book fig 15.8.
Q 16. Define modulation index. Give its physical significance.
Ans:
What do you mean by amplitude modulation with the help of block diagram explain how
AM signal is achieved.
Refer to NCERT book fig no 15.10 on page 525.
Q 17. What do you mean by demodulation with the help of block diagram explain detection of
AM signal is carried out.
Ans: Refer NCERT text book fig no 15.13 on page 527.
Q 18. Draw a plot of the variation of amplitude versus W for an amplitude modulated wave.
Define modulation index, State its importance for effective amplitude modulation. What
is the role of a band pass filter in Amplitude Modulation.
Ans: Refer NCERT text book, fig no 15.9 on page 525.
*******************************