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JEE Advanced 2017 Paper 2
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© examsnet.com
Question : 9
Total: 54
A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct?
The electric flux passing through the curved surface of the hemisphere is
−
Q
2
ε
0
(
1
−
1
√
2
)
The component of the electric field normal to the flat surface is constant over the surface
Total flux through the curved and the flat surfaces is
Q
0
ε
0
The circumference of the flat surface is an equipotental
Validate
Solution:
Since the charge lies outside the sphere, net flux passing through the sphere is zero.
ϕ
curved surface
+
ϕ
disc
=
0
Option (C) is incorrect
ϕ
curved surface
=
ϕ
disc
c
o
s
θ
=
R
√
R
2
+
r
2
E
=
1
4
π
ε
0
Q
(
R
2
+
r
2
)
ϕ
disc
=
∫
−
E
.
d
−
A
=
∫
(
1
4
π
ε
0
Q
(
R
2
+
r
2
)
×
2
π
r
d
r
×
c
o
s
θ
)
=
Q
.2
π
4
π
ε
0
∫
r
d
r
R
2
+
r
2
×
R
(
R
2
+
r
2
)
1
2
Q
R
2
ε
0
R
∫
0
r
d
r
(
R
2
+
r
2
)
3
2
=
Q
R
2
ε
0
[
1
2
(
R
2
+
r
2
)
−
1
2
−
1
2
]
0
R
=
Q
R
2
ε
0
[
1
R
−
1
√
R
2
+
R
2
]
=
Q
2
ε
0
[
1
−
1
√
2
]
⇒
ϕ
curved surface
=
−
Q
2
ε
0
Option (A) is correct Potential at any point on the circumference of the flat surface is
1
4
π
ε
0
Q
√
R
2
+
R
2
=
Q
4
π
ε
0
(
√
2
R
)
Hence it is equipotential
Option (D) is correct
E
=
1
4
π
ε
0
Q
(
R
/
c
o
s
θ
)
2
=
θ
4
π
ε
0
R
2
c
o
s
2
θ
E
normal
=
E
c
o
s
θ
=
Q
4
π
ε
0
R
2
c
o
s
3
θ
Which is not constant
Option (B) is in correct
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