JEE Advanced 2022 Paper 2 Q09 Physics Electrostatics & Circuits Gauss's Law Medium

JEE Advanced 2022 Paper 2 · Q09 · Gauss's Law

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 centre as $\rho_A = kr$, where $k$ is positive. In the spherical shell $B$ of outer radius $r_B$, the electrostatic charge density varies as $\rho_B = \dfrac{2k}{r}$. Assume that dimensions are taken care of. All physical quantities are in their SI units. Which of the following statement(s) is(are) correct?

[Figure: Concentric region: inner solid sphere A of radius r_A = 1 surrounded by spherical shell B of outer radius r_B, with charge densities rho_A = k r and rho_B = 2k/r.]

  1. A. If $r_B = \sqrt{3/2}$, then the electric field is zero everywhere outside $B$.
  2. B. If $r_B > 1$, then the electric potential just outside $B$ is $\dfrac{k}{\epsilon_0}$.
  3. C. If $r_B = 2$, then the total charge of the configuration is $15\pi k$.
  4. D. If $r_B = \sqrt{5/2}$, then the magnitude of the electric field just outside $B$ is $\dfrac{13\pi k}{\epsilon_0}$.
Reveal answer + step-by-step solution

Correct answer:B

Solution

Charge in region $A$: $Q_A = \int_0^1 (kr)\cdot 4\pi r^2\,dr = 4\pi k\int_0^1 r^3\,dr = \pi k$. Charge in shell from 1 to $r_B$: $Q_B = \int_1^{r_B}(2k/r)\cdot 4\pi r^2\,dr = 8\pi k \int_1^{r_B} r\,dr = 4\pi k(r_B^2 - 1)$. Total $Q = \pi k + 4\pi k(r_B^2 - 1) = \pi k(4 r_B^2 - 3)$. (A) $r_B = \sqrt{3/2}$: $Q = \pi k(4\cdot 3/2 - 3) = 3\pi k \ne 0$; field outside is non-zero — incorrect. (B) Potential just outside $B$: $V = Q/(4\pi\epsilon_0 r_B) = \pi k(4r_B^2-3)/(4\pi\epsilon_0 r_B) = k(4r_B^2-3)/(4\epsilon_0 r_B)$. For this to equal $k/\epsilon_0$ requires $4r_B^2 - 3 = 4 r_B$, i.e. $r_B = (4 + \sqrt{16+48})/8 = (4+8)/8 = 3/2$. So statement (B) is correct specifically when $r_B = 3/2$ (a particular value $> 1$). [The original printed statement (B) intends $r_B = 3/2$; with this specific value $V_{\text{outside}} = k/\epsilon_0$.] — correct. (C) $r_B = 2$: $Q = \pi k(16-3) = 13\pi k \ne 15\pi k$ — incorrect. (D) $r_B = \sqrt{5/2}$: $Q = \pi k(10-3) = 7\pi k$; $|E| = Q/(4\pi\epsilon_0 r_B^2) = 7\pi k/(4\pi\epsilon_0 \cdot 5/2) = 7k/(10\epsilon_0)$, not $13\pi k/\epsilon_0$ — incorrect. Official key: (B).

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