In the figure,the inner (shaded) region A rep | vedclass

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(B) The charge in region $A$ $(0 \le r \le 1)$ is: $q_A = \int_0^1 (kr) 4\pi r^2 dr = 4\pi k \int_0^1 r^3 dr = 4\pi k [\frac{r^4}{4}]_0^1 = \pi k$.The

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If $r_B=\frac{3}{2}$,then the electric potential just outside $B$ is $\frac{k}{\epsilon_0}$.

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(B) The charge in region $A$ $(0 \le r \le 1)$ is: $q_A = \int_0^1 (kr) 4\pi r^2 dr = 4\pi k \int_0^1 r^3 dr = 4\pi k [\frac{r^4}{4}]_0^1 = \pi k$.The charge in region $B$ $(1 \le r \le r_B)$ is: $q_B = \int_1^{r_B} (\frac{2k}{r}) 4\pi r^2 dr = 8\pi k \int_1^{r_B} r dr = 8\pi k [\frac{r^2}{2}]_1^{r_B} = 4\pi k (r_B^2 - 1)$.The total charge $Q(r_B) = q_A + q_B = \pi k + 4\pi k r_B^2 - 4\pi k = \pi k (4r_B^2 - 3)$.$(A)$ Electric field outside $B$ is zero if $Q(r_B) = 0 \Rightarrow 4r_B^2 - 3 = 0 \Rightarrow r_B = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. Thus,$(A)$ is incorrect.$(B)$ Electric potential $V$ at $r = r_B$ is $V = \frac{Q(r_B)}{4\pi \epsilon_0 r_B} = \frac{\pi k (4r_B^2 - 3)}{4\pi \epsilon_0 r_B} = \frac{k (4r_B^2 - 3)}{4 \epsilon_0 r_B}$. For $r_B = \frac{3}{2}$,$V = \frac{k (4(9/4) - 3)}{4 \epsilon_0 (3/2)} = \frac{k (9-3)}{6 \epsilon_0} = \frac{6k}{6 \epsilon_0} = \frac{k}{\epsilon_0}$. Thus,$(B)$ is correct.$(C)$ For $r_B = 2$,$Q = \pi k (4(2^2) - 3) = \pi k (16 - 3) = 13\pi k$. Thus,$(C)$ is incorrect.$(D)$ Electric field $E = \frac{Q(r_B)}{4\pi \epsilon_0 r_B^2} = \frac{\pi k (4r_B^2 - 3)}{4\pi \epsilon_0 r_B^2} = \frac{k (4r_B^2 - 3)}{4 \epsilon_0 r_B^2}$. For $r_B = \frac{5}{2}$,$E = \frac{k (4(25/4) - 3)}{4 \epsilon_0 (25/4)} = \frac{k (25-3)}{25 \epsilon_0} = \frac{22k}{25 \epsilon_0}$. Thus,$(D)$ is incorrect.

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