CBSE Class 12 Physics 2015 Delhi Set 1 Paper

© examsnet.com
Question : 26
Total: 26
An electric dipole of dipole moment
p
consists of point charges +q and q separated by a distance 2a apart. Deduce the expression for the electric field
E
due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment
p
. Hence show that in the limit x>>a,
E
2
p
4πε0x3
.
(b) Given the electric field in the region
E
2
x
^
i
, find the net electric flux through the cube and the charge enclosed by it.
OR
(a) Explain, using suitable diagrams, the difference in the behaviour of a (i) conductor and (ii) dielectric in the presence of external electric field. Define the terms polarization of a dielectric and write its relation with susceptibility.
(b) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge
Q
2
is placed at its centre C and an other charge +2Q is placed outside the shell at a distance x from the centre as shown inthe figure. Find (i) the force on the charge at the centre of shell and at the point A , (ii) the electric flux through the shell.
Solution:  
(a) Derivation of the expression for the Electric field E and its limiting value
(b) Finding the net electric flux

Electric field intensity at point p due to charge q
E
a
=
1
4πε0
q
(x+a)2
(
^
x
)

Due to charge +q
E
q
=
1
4πε0
q
(xa)2
(
^
x
)

Net Electric field at point p
E
=
E
q
+
E
+q

=
q
4πε0
×[
1
(xa)2
1
(x+a)2
]
(
^
x
)

=
q
4πε0
[
4aqx
(x2a2)2
]
(
^
x
)

=
1
4πε0
(q×2a)2x
(x2a2)2
(
^
x
)

E
=
1
4πε0
2px
(x2a2)2
^
x

For x>>a
(x2a2)2=x4
E
=
1
4πε0
2p
x3
^
x

(b) Only the faces perpendicular to the direction of x-axis, contribute to the Electric flux. The remaining faces of the cube give zero contribution.

Total flux ϕ=ϕI+ϕII
=I
E
ds
+II
E
ds

=0+2(a)a2
ϕ=2a3
ϕ=
qenclosed
ε0

or, 2a3=
qenclosed
ε0

qenclosed =2a3ε0
OR
(a) Explanation of difference in behaviour of
(i) conductor
(ii) dielectric
Definition of polarization and its relation with susceptibility
(b) (i) Finding the force on the charge at centre and the charge at point A
(ii) Finding Electric flux through the shell
(a)

In the presence of Electric field, the free charge carriers, in a conductor, the charge distribution in the conductor readjusts itself so that the net Electric field within the conductor becomes zero.
In a dielectric, the external Electric field induces a net dipole moment, by stretching/reorienting the molecules. The Electric field, due to this induced dipole moment,opposes ,but does not exactly cancel, the external Electric field.
Polarisation: Induced Dipole moment, per unit volume, is called the polarization. For Linear isotropic dielectrics having a susceptibility χc , we have
P=χeE
(b) (i) Net Force on the charge
Q
2
, placed at the centre of the shell, is zero.
Force on charge ' 2Q ' kept at point A
F=E×2Q
=
(
3Q
2
)
2
Q
4πε0r2
=
(K)3Q2
r2
(ii) Electric flux through the shell
ϕ=
Q
2ε0
© examsnet.com
Go to Question: