TPS Physics Papers with Solution
12
200
Electrostatics - Paper with Solution
Section A - (4 Marks)
1.
Soln.:
2.
The capacity of parallel
capacitor increases with the
Section B – (14 Marks)
plate
[1]
a)
Increase of its area
b)
Decrease of its area
c)
Increase of distance between plates
d)
It is independent of area
5.
Soln.:
a) increase of its area
If a 4 μf capacitor is charged to 1 KV
then energy stored in the capacitor is [1]
a) 1 J
b) 4 J
c) 6 J
d) 2 J
E =
Soln.:
3.
6.
E = 2J
d) 2J
[1]
State the principle of van de Graff
generator
[1]
Soln.: If a charge is continuously supplied
to an insulated metallic conductor the
potential of conductor goes on increasing. It
is used to produce very high potential of the
7
order of 10 -V
U =
1
QV
2
Q =
2U
V
2  3.5  10– 4
700
Q = 10– 6 C
Soln.: The amount of induced surface
charge per unit area is called as polarization
of charges.
4.
U = 3.5  10– 4 J
=
1
4  10-6  103  103
2
Define polarization of charges
_4
Electrostatic energy of 3.5  10 J is
stored in capacitor at 700V. What is
the charge on capacitor ?
[2]
V = 700 volts
1
E = CV2
2
∴
Electrostatics
[½]
[½]
[1]
What do you mean by polar molecules
and non polar molecules? Give one
example on each
[2]
Soln.: Polar molecules : A molecules in
which centre of gravity of positive nuclei and
revolving electron are separated. They have
permanent electric dipole moment and
behave as tiny electric dipoles.
[½]
e.g. N2O, HCl
[½]
Non-polar :
Non-polar molecules – molecules in
which centre of gravity of positive nuclei and
revolving electron coincide. They do not have
permanent electric dipole moment.
[½]
e.g. O2, H2, CO2
[½]
TPS Physics Papers with Solution
7.
Obtain an expression for electric field
intensity at a point outside uniformly
charged thin plane sheet
[2]
Soln.: Electric field intensity at a point
outside uniformly charged thin plane sheet.
Consider a uniformly charged thin
infinite plane sheet. Let  be the surface
charge density of sheet. By symmetry electric
field is perpendicular to plane sheet and
directed outwards. Electric field intensity has
same magnitude at a given distance on the
either sides of sheet.
201
Electrostatics

Since E is perpendicular to plane sheet it is
parallel to the curved surface of Gaussian
cylinder. Hence electric flux does not pass
through the curved surface of Gaussian
cylinder.
Total normal electric induction over
Gaussian surface =  E (2dS)
Where dS is surface area of end faces
of the cylinder.
Algebraic sum of charges enclosed by
Gaussian cylinder =  ds
[½]
According to Gauss law
 E = dS =  dS
E =
8.
Diagram - ½ Mark
P-
Point at which electric field intensity is
to be found
E-
Electric field intensity at point P
S-
Gaussian surface
Ps -
Plane sheet
To obtain electric field intensity at a
point P due to uniformly charged cylinder
around P with its axis perpendicular to plane
sheet carrying charge with ends having cross
sectional area ds. The plane sheet passes
through the middle of cylinder’s length so
that the ends of cylinder are equidistant from
the plane sheet carrying charge.
[½]

Electric
field
intensity
E
is
perpendicular to the ends of cylinder, hence
the electric flux through each end is Eds.

2
[½]
Three capacitors of capacities 8 μf,
8 μf and 4uf are connected in a series
and a potential difference of 120v is
maintained across the combination.
Calculate the charge on capacitor of
capacity 4 μf.
[2]
Soln.:
Given: C1 = 8 F, C2 = 8 F, C3 = 4 F, V = 120 V
For series combination
1
1
1
1
=
+
+
Cs
C1 C2 C3
[½]
1
1 1 1 1 1 21
= + + = + +
Cs
8 8 4 8 8 24
1
4
=
Cs
8
 Cs =
8
= 2 F
4
= 2  10– 6 F
[½]
Q1 = Q2 = Q3 = Qs charge on each is same
Q = Cs  V
[½]
= 2  10– 6  120
= 240  10– 6 C
[½]
TPS Physics Papers with Solution
202
OR
8.
A metal sphere of Radius 1 cm is
charged with 3.14 μc. Find the electric
intensity at a distance of 1 cm from
the centre of metal sphere.
[2]
Electrostatics
capacitor is connected to negative terminal of
battery.
[½]
Soln.:
Given
_
R = 1 cm = 10 2 m
q = 3.14  c = 3.14  10-6 c
r = 1m
To find : E
Fig. Q. 10
Calculation :
E =
1
q
4 π Є0 r2
E = 9  109 
3.14  10-6
(1)2
[½]
[½]
Let Q1, Q2, and Q3 be the charges
induced on C1, C2 and C3 respectively. Let V
be the potential deference across each
capacitor.
E = 9  3.14  103
E = 28.26  103 N/C
9.
[1]
∵
Q = Q1 + Q2 + Q3
∵
Q = CV
Q2 = C2 V
Q3 = C3 V
Soln.: The ability of conductor to hold a
charge is called as capacity of conductor.
1 farad  If a charge of 1 coulomb
increases the potential of conductor by one
volt then the capacity of conductor is said to
be one farad
1F =
10.
1C
1V
[1]
Derive an expression for equivalent
capacitance for three capacitors
connected in parallel
[2]
Soln.: In parallel arrangement one terminal
of each capacitor is connected to positive
terminal of battery and other terminal of
[½]
Q1 = C1 V
What do you mean by capacity of a
conductor? Define 1 farad
[2]
Capacity of conductor is define as
quantity of charge required to raise its
potential be one unit
[1]
[½]
CV = C1 V + C2 V + C3 V
CV = V (C1 + C2 + C3)
∴
∴
C = C1 + C2 + C3
In general if n capacitors are connected
in parallel
CP = C1 + C2 + C3 + … + Cn
[½]
Thus
in
parallel
arrangement
equivalent capacitance is equal to sum of
their equivalent capacitances.
Section C – (9 Marks)
11.
A cube of marble having each side
1 cm is kept in an electric field of
intensity 300v/m. Determine the
energy contained in the cube of
TPS Physics Papers with Solution
dielectric constant 8 (Є0 = 8.85
2
203
 10_12
2
C /Nm F/m)
[3]
Soln.:
Data:
K = 8; E = 300 V/m
–2
l = 1 cm = 10
3
Electrostatics
Consider a charge +q situated at point
‘O’ inside a conductor of any shape consider
a small area as on surface of conductor at
distance r from point O.
N.E.I. for area ds = Є E cos θ ds
m
–6
V = (1 cm) = 10
m3
Energy stored per unit volume
U =
=
1
k0 E2
2
[½]
1
 8  8.85  10– 12  (300)2
2
[½]
U = 4  8.85  g  10– 8
U = 3.185  10– 6 J/m3
[½]
Energy contained in cube
= 3.185  10– 6  (10– 2)3
= 3.185  10– 12 J
[½]
[1]
log 4 = 0.6021
log 8.85 = 0.9469
E =
q
4 π Є r2
N.E.I. for area ds = Є 
∴ N.E.I. for area ds =
log 9 = 0.9542
q
cos θ ds
4 π Є r2
q cos θ ∙ ds

4 π 
r2

[½]
Let dw = solid angle subtended by
2.5032
area ds on charge q
Antilog 2.5032
N.E.I. for area ds =
3.185  102
3.185  102  10– 8
3.185  10
State and prove Gauss’s theorem
q
(dw)
4π
T.N.E.I. can be find out by integrating
this relation
–6
12.
[½]
Electric intensity at a distance r from
charge q
= UV
= 3.185  10– 6  10– 6
Fig. Q. 12
[3]
Soln.: Statement – For a closed surface of
any shape with any number of charges
situated in any position inside it the Total
Normal Electric Induction (T.N.E.I.) over the
closed surface is equal to the algebraic sum of
electric charges enclosed by that surface. [1]
q
T.N.E.I. =
 4 π (dw)
T.N.E.I. =
q
 (dw)
4π
dw
= Total solid angle subtended at
charge q = 4 π
TPS Physics Papers with Solution
∴
T.N.E.I. =
∴
204
Electrostatics
q
is capacitance of capacitor ? If
charge is 1000 μC. What will be the
P.D. between two plates. How
will the capacity of air capacitor
be affected if the space between
plates is completely filled with Mylar
(Kmylar = 3)
[3]
4π4π
T.N.E.I. = q
If q1, q2, q3 … qn be the charges situated
inside the closed surface then by using the
principle of conservation of charges
T.N.E.I. = q1 + q2 + q3 + … + qn
Soln.:
d = 2 mm = 2  10-3 m
n
T.N.E.I. =  q
i=1
Q = 1000 μc = 10-3 c
Hence, Gauss’s theorem is proved
13.
Soln.:
[1]
A metal plate of area 0.01m2 carries a
charge of 100 uc. Calculate the
outward pull on plate.
[3]
2
_2
Kmylar = 3
To find :
1) Cair
1)
Cair =
A = 0.01 M = 10 M
F =
2
σ2
2 Є0 K
2) V
3) Cmylar
q = 100 μc = 100  10-6 = 10-4 c
Cair =
A Є0 K
d
10
_2
[½]
2)
F =
Q
2 Є0 KA
F =
10-8
2  8.85  10-12  1  10-2
2
1
F =
 106
17.70
V =
[½]
V =
1
F =
 106
1.770  101
1
F =
 105
1.770
3)
F = 0.5650  105
F = 56500 N
[1]
OR
A capacitor consist of two parallel
metal plates each of area 100 cm2
separated by a distance 2 mm. What
_11
_12
f
1
[½]
Q = CV
V =
[1]
[½]
 8.85  10
_
2  10 3
Cair = 4.425  10
σ2A
F =
2 Є0 K
13.
A = 100 cm2 = 10-2 m2
[½]
Q
C
10
_3
4.425  10
_11
1
 108
4.425
V = 0.2259  108 V
[½]
C ∝K
[½]
Cmylar Kmylar
=
Cair
Kair
Cmylar
3
=
4.425  10-11
1
Cmylar = 4.425  3  10
Cmylar = 13.275  10
_11
_11
f
[½]
TPS Physics Papers with Solution
Section D – (10 Marks)
14.
With the help of near diagram
describe
the
construction
and
working of van de Graff generator. [3]
Soln.: Van de Graff generator :
Van de Graff generator is an
electrostatic generator that can produce high
potential of the order of millions of volt.
205
Electrostatics
D = Gas discharge tube
T = Target
Van de Graff generator consists of a
hollow metal polished sphere S, called dome
which is supported on an insulating column.
An endless insulated fabric belt is made to
run over two pulleys P1 and P2. Pulley P1 is
rotated using an electric motor. C1 is metal
comb with sharp metal spikes and connected
to high voltage supply comb C2 is called as
collecting comb which is connected to dome.
A discharge tube D, acts as a source of
positive ions such as protons, deuterons.
Target nucleus is kept at other end of
discharge tube. The generator is enclosed in a
steel chamber filled with nitrogen or methane
at high pressure.
[½]
Working:
When an electric motor is switched on,
endless belt wound over pulley P1 and P2
starts rotating. A spray comb C1 is given
Fig. Q. 14
Diagram - 1 Mark
Construction:
P1 = Pulley rotated with motor
P2 = Pulley
B = An endless belt made up of
insulating material
C1 = Comb connected to high
voltage supply and is used to
spray positive charges
C2 = Comb connected to dome
S = Metallic dome
positive potential of the high voltage rectifier
(H.V.R) supply. Charge from metal comb is
transferred to belt. The charge is
continuously carried upto large hollow
sphere. A negative charges are induced on
the sharp ends of comb C2 and an equal
positive charges are induced on farther end
of C2 where comb C2 collects these charges,
which are spread on the outer surface of
dome. The potential of dome continuously
increases.
The charges are crowded on outer
surface of dome which leaks to the
surrounding in the form of spark. To avoid
such a leakage, Van de Graff generator is
enclosed in earth connected steel tank filled
with air under pressure.
[1]
The positive ions are introduced from
ion source in the upper part of the evacuated
TPS Physics Papers with Solution
206
E1 = E2
accelerator tube. These ions are accelerated in
downward direction. Due to very high fall of
potential the positive ions acquire high
energy. These charged particles are directed
towards target (T).
[½]
15.
Electrostatics
Obtain an expression for mechanical
force per unit area of charged
conductor
[3]
E1 + E2 =
σ
Є0 K
2 E1 =
σ
Є0 K
E1 =
σ
= E2
2 Є0 K
[½]
Force acting on area ds due to the
charged conductor
Soln.:
Force = (charge on area ds) 
(Electric intensity due to the
rest of conductor)
F = σ ds 
F =
σ
2Є0 K
σ2 ds
2 Є0 K
[½]
Mechanical force per unit area
Fig. Q. 15
[½]
f =
Consider a charged conductor of any
shape situated in a medium of dielectric
constant K. Let ds be the small area of surface
charge density.
f =
σ = E Є0 K
2
…(1)
∴
Let E1 be the electric intensity due to
the charge on small area ds and E2 be the
electric intensity due to charges an rest of the
conductor
σ
E = E1 + E2 =
Є0 K
[½]
Resultant intensity acting inside the
conductor is given by E1 – E2 but intensity
inside the conductor is O
∴
E1 – E2 = 0
σ2
2 Є0 K
From equation (1)
Electric intensity at a point just outside
the conductor is given by
σ
E =
Є0 K
F
σ2 ds
=
ds 2 Є0 K
ds
f =
f =
E2 E0 K2
2 E0 K
1
Є0 K E2
2
[1]
OR
15.
Derive an expression for
stored in a charged capacitor
energy
[3]
Soln.: When capacitor is charged some
work is required for charging. This work is
stored in the form of electrostatic energy
TPS Physics Papers with Solution
207
Electrostatics
Let Q be the charge and V be the
potential difference
∴
Q = CV
∴
Q
V =
C
Let q be the charge and v be the
potential difference at any time t
∴
V =
q
C
…(1)
1 C2 V2
2 C
∴
E =
1
CV2
2
∵
C =
Q
V
∴
E =
1 Q
2
V
2 V
∴
E =
1
QV
2
[½]
Let dw be the work done in displacing
addition charge dq
∵
E =
B)
Total work done can be find out by
integrating this relation within the limit O
to Q.
[1]
Draw a neat labelled diagram of a
parallel plate capacitor completely
filled with dielectric
[2]
Work done = P.D.  charge
Ds = v  dq
[½]
Soln.:
Q
∫ dw =
 v dq
0
Q
W =
Fig. 15(a)
q
 C dq
[Diagram - 1 Mark]
0
Labelling :
Q
W =
1
C
W =
1  q2Q
C  2  0
W =
1
C
P1, P2 = Parallel plates
 q dq
A = Area of each plate
0
d = Distance between the plate
–
E = Intensity of electric field
 Q2 _ 0
 2 2


1 Q2
W =
2 C
K = dielectric constant

[1]
But, work done can be stored in the
form of energy
1 Q2
2 C
∴
E =
∵
Q = CV
[1]
TPS