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JEE Advanced 2019 Paper 1
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© examsnet.com
Question : 4
Total: 54
A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential at its surface is
V
0
.A hole with a small area
α
4
π
R
2
(
α
<
<
1
)
is made on the shell without affecting the rest of the shell. Which one of the following statement(s) is correct?
The magnitude of electric field at the centre of the shell is reduced by
α
V
0
2
R
The ratio of the potential at the centre of the shell to that of the point at
1
2
R
from centre towards the hole will be
1
−
α
2
−
α
The magnitude of electric field at a point, located on a line passing through the hole and shell’s center, on a distance 2R from the center of the spherical shell will be reduced by
α
V
0
2
R
The potential at the center of the shell is reduced by
2
a
V
0
Validate
Solution:
v
0
=
K
Q
R
=
K
σ
4
π
r
2
R
At center,
v
1
−
v
0
-(potential due to removed element)
=
v
0
−
K
q
R
=
v
0
−
K
R
(
σ
α
4
π
R
2
)
=
v
0
−
v
0
α
=
v
0
(
1
−
α
)
At mid point between center and hole is
V
2
=
V
0
−
k
q
R
/
2
=
V
0
−
2
V
0
α
=
V
0
(
1
−
2
α
)
V
1
V
2
=
V
0
(
1
−
α
)
V
0
(
1
−
2
α
)
=
1
−
α
1
−
2
α
→
(
2
)
is correct
Option A:
E
initial
=
0
E
final
=
K
q
R
z
=
k
σ
α
4
π
R
2
R
2
=
V
0
α
R
E
i
−
E
f
=
V
0
α
R
i.e E will decrease by
V
0
α
R
i.e. (C) is wrong
Option D:
V
initial
=
V
0
V
final
=
=
V
0
−
K
σ
α
4
π
R
2
R
=
V
0
−
V
0
α
∴
V
center
r decreases by
V
0
α
.i.e.(D) is wrong
© examsnet.com
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