JEE Advanced 2022 — Paper 2 — PHY-II solution

JEE Advanced2022Paper 2PHY-II

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 $\frac{k}{ε _{0}}$ .
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 $\frac{13 π k}{ε _{0}}$ .

Solution

The total enclosed charge is $Q_{tot} = \int_{0}^{r_{A}} (k r) (4 π r^{2}) d r + \int_{r_{A}}^{r_{B}} \frac{2 k}{r} (4 π r^{2}) d r = π k r_{A}^{4} + 4 π k (r_{B}^{2} - r_{A}^{2})$ .

With $r_{A} = 1$ : $Q_{tot} = π k + 4 π k (r_{B}^{2} - 1)$ .

Option (A): $r_{B} = 3/2$ gives $Q_{tot} = π k + 4 π k (1/2) = 3 π k \neq = 0$ , so the external field is nonzero. Incorrect.

Option (B): $r_{B} = 3/2$ gives $Q_{tot} = π k + 4 π k (5/4) = 6 π k$ . Potential just outside $B$ : $V = \frac{1}{4 π ε _{0}} \cdot \frac{6 π k}{3/2} = \frac{k}{ε _{0}}$ . Correct.

Option (C): $r_{B} = 2$ gives $Q_{tot} = π k + 12 π k = 13 π k \neq = 15 π k$ . Incorrect.

Option (D): $r_{B} = 5/2$ gives $Q_{tot} = π k + 4 π k (21/4) = 22 π k$ . Field just outside: $E = \frac{1}{4 π ε _{0}} \cdot \frac{22 π k}{( 5/2 ) ^{2}} = \frac{22 k}{25 ε _{0}} \neq = \frac{13 π k}{ε _{0}}$ . Incorrect.

$∴$ Only (B) is correct.

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