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JEE Main 10 April 2016 Paper
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© examsnet.com
Question : 12
Total: 90
Within a spherical charge distribution of charge density
ρ
(
r
)
,N equipotential surfaces of potential
V
0
,
V
0
+
∆
V
,
∆
V
0
+
2
∆
V
,
.
.
.
.
.
V
0
+
N
∆
V
(
∆
V
>
0
)
, are drawn and have increasing radii
r
0
,
r
1
,
r
2.
.
.
.
r
N
, respectively. If the difference in the radii of the surfaces is constant for all values of
V
0
and
∆
V
then :
ρ
(
r
)
α
r
ρ
(
r
)
=
constant
ρ
(
r
)
α
1
r
ρ
(
r
)
α
1
r
2
Validate
Solution:
∆
V
∆
r
→
constant
⇒ unifor E.field
(
E
)
(
4
π
r
2
)
=
1
ε
0
∫
ρ
d
V
(
E
)
(
4
π
r
2
)
=
1
ε
0
r
∫
0
ρ
4
π
r
2
d
r
(
E
)
(
4
π
r
2
)
=
1
ε
0
4
π
r
∫
0
ρ
r
2
d
r
after integral on RHS
We must obtain
r
2
⇒
ρ
∝
1
r
© examsnet.com
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