Download MCQ 1. A moving electric charge produces A. electric field only. B

ELECTROSTATICS::BHSEC 2009-2014
MCQ
1. A moving electric charge produces
A. electric field only.
B. magnetic field only.
C. both electric field and magnetic field.
D. neither of these two fields.
Ans.: C. both electric field and magnetic field
2. If the electric field at A and B are EA and EB and the distance between them is ‘r’ as shown in the
figure given below, then
Ans.: E A  EB
3. If an electric current of 2mA flows through a wire, then the number of free electrons passing a given
point in a wire per second will be
A. 1.25 x 1013 .
B. 1.25 x 1016.
C. 1.25 x 1018.
D. 1.25 x 1021.
Ans.: B 1.25 x 1016
4. What is the equivalent capacitance of the following circuit?
A.
B.
C.
D.
3 F
2 F
1.5 F
1 F
Ans.: D 1 F
4. The capacitors in the given circuit diagram have equal capacitance of 10 F each
The equivalent capacitance between the points X and Y is
A. 40 F .
B. 30 F .
C. 20 F .
D. 10 F .
Ans.: D 10 F
5. The capacitors in the given circuit diagram have equal capacitance of 10 F each.
The equivalent capacitance between the points X and Y is
A 40 F .
B 30 F .
C 20 F .
D 10 F .
Ans.: D 10 F
6. The equivalent capacitance between the points A and B from the network of capacitors shown below
is
A.
B.
C.
D.
2 μF.
3 μF.
4 μF.
20 μF.
Ans.: A 2 μF
7. The electric field in a region where the electric potential is constant throughout is
A. maximum.
B. minimum.
C. neutral.
D. zero.
Ans.: D zero
Fill in the blanks
1. Guass’ law relates the ………. field at points on a closed Gaussian surface to the ………. charge
enclosed by the surface. (magnetic, electric, net, test)
Ans.: electric, net
2. Positron is a ………… charged particle having the same ………… as an electron. (positively,
negatively, mass, symbol).
Ans.: positively, mass
True or False
1. Dielectric substance placed between two plates of a capacitor increases the potential energy of the
plates.
Ans.: False. Dielectric substance placed between two plates decreases the potential energy of the
plates and increases the capacitance.
2. A dipole in a non-uniform electric field experiences only a torque.
Ans.: False. In a non-uniform electric field, a dipole experiences both torque and force. It
experiences only a torque when placed in a uniform field.


3. If an electric dipole of moment p is placed normal to the lines of force of an electric field E , then the
potential energy of dipole is 2pE.
Ans.: False, the potential energy of dipole placed normal to lines of force is Zero ( ie, work done is
zero )
4. Gauss’ theorem cannot be used to determine the electric field of any surface.
Ans.: True
Answer the following Questions
1. The electric field at a point due to a point charge is 20 N
and the electric potential at that point is
C
10 V. Calculate the distance of the point from the charge and the magnitude of the charge.
1 q
1
q
V 10
0.5 x10
. →r  
. 2 → q  4 0 rV 
 5.6 x1010 C
Ans.: V 
 0.5m → E 
9
4 0 r
4 0 r
E 20
9 x10
2. The tyres of aircraft are not made of ordinary rubber (which is an insulator) but of a special rubber
which is slightly conducting. Why?
Ans.: The tyres of aircrafts are highly charged during landing due to friction between the tyres and
the air strip (earth). If the tyres are slightly conducting, the charge will not accumulate on them and
will leak to the earth.
3. A regular pentagon of side 20 cm has a charge 5 μ C at each of its vertices. Calculate the electric
potential at the centre of the pentagon.
Ans.: V 
5q
4 0 r

9 x109 x5 x5 x106
 1125000  1.1x106 V
.2
4. Name the factors on which capacitance of a parallel plate capacitor depends. Give the corresponding
relation.
Ans.: Factors:
i. Area of the plates
ii. Distance between the plates
iii. Dielectric constant
Relation
C
0 A
d
C
 0 KA
d
CV 2
, for the potential energy of a charged conductor of capacitance ‘C’
2
carrying a charge ‘Q’ at a potential ‘V’.
Ans.:
5. Establish the formula U 
Small Work done:
dw  V dQ 
Q
dQ
C
Total work done:
Q
1
1
1 Q2
1  Q2 




W

dw

Q
dQ
w

Q
dQ
→
→
→
w




0 C
C 2
C  2 0
C 0
Q
Q
1 Q2
1
1
w .
w  CV 2 [ Q  CV ] U  CV 2
2 C
2
2
6. Calculate the equivalent capacitance of the network shown below between the points ‘A’ and ‘B’,
given C1  C2  12  F , C3  7  F , C4  C5  C6  15 F .
Ans.:
Capacitors in series C4, C5 and C6
𝟏
𝟏
𝟏
𝟏
=
+
+
𝑪 𝑪𝟒 𝑪𝟓 𝑪𝟔
𝟏𝟓
𝑪=
= 𝟓µ𝑭
𝟑
C3 and C are in parallel
𝑪′ = 𝑪𝟑 + 𝑪 ⇒ 𝟕µ𝑭 + 𝟓µ𝑭 = 𝟏𝟐µ𝑭
Now C1, C’ and C2 are in series
𝟏
𝟏
𝟏
𝟏
𝟏
𝟏
𝟏
𝟑
=
+
+
⇒
+
+
=
𝑪" 𝑪𝟏 𝑪′ 𝑪𝟐 𝟏𝟐 𝟏𝟐 𝟏𝟐 𝟏𝟐
∴ 𝑪" = 𝟒µ𝑭
7. An α-particle is accelerated to a potential 106 volt from the position of rest. Calculate its energy in
electron volt.
Ans.:
Energy (W) = q.v
W = (2e) v  α-particle = 2e
W = 2 x 1.6 x 1019 x 106J
2 x1.6 x1019 x106 J
W
eV = 2 x 106 eV = 2MeV
1.6 x1019
8. Obtain an expression for potential energy of a system of three charges.
Ans.:
Work done for moving the charge q2 from infinity to the distance r12 is
1 q1 q2
W12 
4 0 r12
Work done for moving the charge q3 from infinity to distance r13 and
r23 is
1 q1 q3
W13 
4 0 r13
1 q2 q3
W23 
4 0 r23
Total work done to assemble the system is
  W  W12  W13  W 23
=
q1 q2
1 q1 q3
1 q2 q3


4 0 r12
4 0 r13
4 0 r23
1
1  q1 q2 q1 q3 q2 q3 




4 0  r12
r13
r23 
9. An electric charge is uniformly distributed on the surface of a hollow sphere. State how the values of
electric intensity E and potential V vary inside and outside the sphere.
Ans.:
i. Inside the sphere
E = 0 and V remains same
ii. Outside the sphere
E and V decreases as it moves away from the sphere. E decreases at a faster rate.

10. State one essential property of a Gaussian surface and its use.
Ans.: The Gaussian surface should pass through the point at which the electric field is required.It is
used to complete the electric field intensities due to certain charge system or charged bodies.
11. Three capacitors of 20 F ,30 F and 60 F are connected in series to a 100 volt d.c supply.
Calculate the total energy stored in the capacitors.
Ans.:
1
1
1
1



Total capacitance,
C C1 C2 C3
1
1
1
3  2 1 6
1
=
=




20 30 60
60
60 10
C  10 F
1
Energy E = Cv 2
2
1
 10 106  (100) 2  5 102 OR 0.05Joules
2
12. With the help of a sketch diagram, deduce Coulomb’s law from Gauss’ theorem
Ans.:
By Gauss’ theorem,
q
 EdS   0
For complete Gaussian surface,
2
 dS  4 r
 E (4 r 2 ) 
E
q
0
1
q
4 0 r 2
If = q0 is the test charge placed at P, the force between the charges =q and =q0 will be
F = q0 E
1 qq0
1 q
F
F  q0
2
4 0 r 2
4 0 r
13. On moving away from a point charge, the electric field due to the charge decreases. This is also true
for a small electric dipole. Does the electric field decrease at the same rate in both the cases? Give
reasons.
1
Ans.: No. The field due to electric dipole decreases more rapidly as E 3 but the field due to a
r
1
point charge decreases less rapidly as E 2 .
r
14. X and Y are two hollow concentric spheres enclosing charge 2Q and 3Q respectively as shown in the
diagram given below. What is the ratio of the electric flux through X and Y?
Ans.: Electric flux through X, φ x =
Ratio =
φ x 2Q ε 0 2
=
=
φ y 5Q ε0
5
2Q
5Q
Electric flux through Y, φ y =
ε0
εo
15. A square of sides 2m has charge of 5 109 C , 2 109 C , 3 109 C and  6 10 9 C at its corners. Find
the potential at the centre of the square.
Ans.: The distance of the centre from each corner of the square of sides 2m is r = 1m. The
potential at the centre due to the group of charges q1, q2, q3 and q4
q q 
1
q q
V=
× 1 + 2 + 3 + 4 
4πε 0  r
r
r
r 
2
1
9
-9 Nm
=9.0×10 ×
5+2+3-6×10 2 ×C
1m
C
Nm
= 36
= 36 J = 36V
C
C
16. Define relative permittivity. What is its unit?
Ans.: Relative permittivity of a medium is defined as the ratio between absolute permittivity (  ) of
the medium to the absolute permittivity (  0 ) of free space.
It is a dimensionless constant so it has no unit.
17. The area of each plate of a parallel plate capacitor is 104 cm2 and the electric field between
the plates is 106 N/C. Compute the charge on each plate.
Ans.:
εA
C= 0
d
Q = CV
εA
= 0 ×V
d
Q=ε 0 AE
=8.85×10-12C2 N-1m-2 ×104 ×10-4 m2 ×106 NC-1 = 8.85  10-6C
18. An electric dipole is free to move in a uniform electric field. Explain its motion when it is placed
parallel and perpendicular to the field.
Ans.:
i. Parallel: The net force as well as torque is zero, so the dipole will not move.
ii. Perpendicular: A maximum torque will act on the dipole which will rotate to become parallel
to the electric field.
19. Derive an expression for the electric field intensity at any point
uniform charge density σ Cm-2 .
Ans.:
Electric field at a certain distance from the sheet is equal in
directions as well.
1
E =  EdA +  EdA From Gauss’ theorem E = q- (ii)
ε0
A
A
E = E  dA + E  dA Equating (i) and (ii) : 1 q=2EA E = 1 q
2ε 0 A
ε0
A
A
 E = EA + EA
E = 2EA -(i)
due to a plane sheet of charge of
other
 q = σ
 A 
20. Obtain an expression for the electric potential at a point distant ‘r’ from a point charge ‘q’. What will
be the potential at a point as a combined effect of several charges?
Ans.: Electric potential at point p, distance r from 0 is the work done in bringing a test charge ‘q’
from infinity. Let it come through ‘B’ to ‘A’ then to ‘p’ where OB = x.
The work done in moving the test charge from B to A, a distance dx is,
dw   Fdx
1 qq0
w
dw  
V=
2
4 0 x
q0
Integrating both side
1 qq 0 1
V=
×
w
r
1 qq 0
4πε 0 r q 0
dw
=
dx
o
 4πε0 x 2
1 q

V=
r
4πε 0 r
1
1
w=
qq 0  2 dx
4πε 0
x

The potential at a point as a combined effect of several
-1 r
x 
-1
charges is the algebraic sum of the potential due to
=
qq 0  
4πε 0
individual charges.
 -1  
r
1 n qi
1
1
ie,
V
=

=
qq 0  
4πε0 i=1 ri
4πε 0
x
 
1
1 1 
=
qq 0  - 
4πε 0
r 
1 qq 0
4πε 0 r
21. When a medium of dielectric constant K is introduced between the plates of a parallel plate capacitor,
how do the
a) capacitance and
b) p.d across the plate change?
w=
1
Ans.: i) Capacitance increase K times. ii) P.D falls to 𝐾 times

22. Define potential energy of a point charge in a uniform electric field E . Write the expression for the


potential energy of a dipole of moment p in a uniform electric field E .
Ans.: The potential energy of a point charge in an electric field is defined as the work done in
bringing the point charge from from infinity to that point. W= -pE Cosθ
23. Calculate the electric intensity due to a point charge 20 μC at a point distant 40cm from the charge.
Ans.:
9 109  20  106
1 q
E=
E=
(0.04) 2
4πε 0 r 2
E = 1.125×106 Nc-1
24. A parallel plate capacitor of plate area 600cm2 and plate separation 2mm is connected to a d.c source
of 200V. Calculate in S.I. unit the:

a) magnitude of the uniform electric field E between the plates,
b) charge density σ on any plate.
Answer
V
200V
a) E  
 105 V
3
m
d 2 10 m
b)   E 0  105  8.85  1012  8.85 107 C 2
m
25. There are two identical metal spheres of exactly equal masses. One is given a positive charge and the
other an equal negative charge. After charging, are their masses equal? Explain why.
Ans.: No. Positive charge has fewer electrons while negative charge has more electrons. So
negatively charged metal sphere have more mass than positively charged metal sphere.