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IIT JEE Advanced 2011 Paper 1
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© examsnet.com
Question : 9
Total: 69
A spherical metal shell A of radius
R
A
and a solid metal sphere B of radius
R
B
(<
R
A
) are kept far apart and each is given charge +Q. Now, they are connected by a thin metal wire. Then
E
A
inside
= 0
Q
A
>
Q
B
σ
A
σ
B
=
R
B
R
A
E
A
on
surface
<
E
B
on
surface
Validate
Solution:
We know that the electric field is zero inside a conductor if it is hollow or solid. That is,
E
A
inside
= 0
On connecting them by a wire, both acquire a common potential, namely,
V
0
=
Q
+
Q
C
1
+
C
2
=
2
Q
4
π
ε
0
R
A
+
R
B
)
Therefore,
Q
A
=
C
1
V
0
=
(
4
π
ε
0
R
A
)
V
0
=
2
Q
R
A
R
A
+
R
B
Q
B
=
C
2
V
0
=
(
4
π
ε
0
R
B
)
V
0
=
2
Q
R
B
R
A
+
R
B
Since
R
A
>
R
B
, we get
Q
A
>
Q
B
. Now,
σ
A
σ
B
=
Q
A
R
A
2
×
R
B
2
Q
B
=
R
A
R
B
×
R
B
2
R
A
2
=
R
B
R
A
The electric field on the surface is
σ
ε
0
. Since
σ
A
<
σ
B
. we get
E
A
on
surface
<
E
B
on
surface
© examsnet.com
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