JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Electrostatics

In the figure, the inner (shaded) region

A

represents a sphere of radius

r_{A} = 1

, within which the electrostatic charge density varies with the radial distance

r

from the center as

ρ_{A} = k r

, where

k

is positive. In the spherical shell

B

of outer radius

r_{B}

, the electrostatic charge density varies as

ρ_{B} = \frac{2 k}{r}

. Assume that dimensions are taken care of. All physical quantities are in their SI units.

Which of the following statement(s) is(are) correct?

(A)If $r_{B} = \frac{3}{2}$ , then the electric field is zero everywhere outside $B$ .
(B)If $r_{B} = \frac{3}{2}$ , then the electric potential just outside $B$ is $\frac{k}{ϵ _{0}}$ .✓ Correct
(C)If $r_{B} = 2$ , then the total charge of the configuration is $15 π k$ .
(D)If $r_{B} = \frac{5}{2}$ , then the magnitude of the electric field just outside $B$ is $\frac{13 π k}{ϵ _{0}}$ .

Answer: (B)

Solution

Calculate the charges in regions A and B:

q_{1} = \int_{0}^{1} k r \cdot 4 π r^{2} d r = π k

q_{2} = \int_{1}^{r_{B}} \frac{2 k}{r} \cdot 4 π r^{2} d r = 4 π k (r_{B}^{2} - 1)

Total charge:

q_{net} = π k (4 r_{B}^{2} - 3)

(A) For

E = 0

outside,

q_{net} = 0

, giving

r_{B} = \frac{3}{2}

, not

\frac{3}{2}

. Incorrect.

(B) Potential:

V = \frac{1}{4 π ϵ _{0}} \frac{q _{net}}{r _{B}} = \frac{k}{4 ϵ _{0}} (4 r_{B} - \frac{3}{r _{B}})

For

r_{B} = \frac{3}{2}

V = \frac{k}{4 ϵ _{0}} (6 - 2) = \frac{k}{ϵ _{0}}

. Correct.

(C) For

r_{B} = 2

q_{net} = π k (16 - 3) = 13 π k

, not

15 π k

. Incorrect.

(D) For

r_{B} = \frac{5}{2}

E = \frac{k}{4 ϵ _{0}} \frac{4 r _{B}^{2} - 3}{r _{B}^{2}} = \frac{22 k}{25 ϵ _{0}}

, not

\frac{13 π k}{ϵ _{0}}

. Incorrect.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · One or more correct options