JEE Advanced 2017 Paper 2

Since the charge lies outside the sphere, net flux passing through the sphere is zero.
ϕcurved surface+ϕdisc=0
Option (C) is incorrect
ϕcurved surface=ϕdisc
cosθ=

R

√R2+r2

   E=

1

4πε0

Q

(R2+r2)

ϕdisc=∫

−

E

.d

−

A

=∫(

1

4πε0

Q

(R2+r2)

×2πrdr×cosθ) =

Q.2π

4πε0

‌∫

rdr

R2+r2

×

R

(R2+r2) 1 2

QR

2ε0

‌

R

∫

0

rdr

(R2+r2) 3 2

=

QR

2ε0

[

1

2

(R2+r2)− 1 2

− 1 2

]0R =

QR

2ε0

[

1

R

−

1

√R2+R2

] =

Q

2ε0

[1−

1

√2

]
⇒ϕcurved surface=−

Q

2ε0

Option (A) is correct Potential at any point on the circumference of the flat surface is

1

4πε0

Q

√R2+R2

=

Q

4πε0(√2R)

Hence it is equipotential
Option (D) is correct
E=

1

4πε0

Q

(R/cosθ)2

=

θ

4πε0R2

cos2θ
Enormal=Ecosθ=

Q

4πε0R2

cos3θ
Which is not constant
Option (B) is in correct