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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 $\rho_A=k r$, where $k$ is positive. In the spherical shell $B$ of outer radius $r_B$, the electrostatic charge density varies as $\rho_{\bar{B}}=\frac{2 k}{r}$. Assume that dimensions are taken care of. All physical quantities are in their $SI$ units.
Which of the following statement($s$) is(are) correct?
If $r_B=\sqrt{\frac{3}{2}}$, then the electric field is zero everywhere outside $B$.
If $r_B=\frac{3}{2}$, then the electric potential just outside $B$ is $\frac{k}{\epsilon_0}$.
If $r_B=2$, then the total charge of the configuration is $15 \pi k$.
If $r_B=\frac{5}{2}$, then the magnitude of the electric field just outside $B$ is $\frac{13 \pi k}{\epsilon_0}$.
Solution
$q _1=\int_0^1 k r 4 \pi r^2 d c=\frac{4 \pi k}{4}=\pi k$
$q_2=\int_1^2 \frac{2 k}{r} 4 \pi r^2 d r=\frac{8 \pi k\left(r^2-1^2\right)}{2}$
$q_2=4 \pi k\left[r^2-1\right]=4 \pi k r^2-4 \pi k$
$q_{\text {net }}=q_1+q_2$
$=4 \pi r^2-3 \pi k$
$q_{p e t}=\pi k\left[4 r^2-3\right]$
$(A)$ $E_{2 x t}=0 \Rightarrow q_{\infty}=0 \Rightarrow r=\frac{\sqrt{3}}{2}$
$(B)$ $V =\frac{ kQ Q _{ R }}{ r }=\frac{1}{4 \pi \varepsilon_0} \frac{\pi k\left(4 r^2-3\right)}{r}$
$V =\frac{ k }{4 \varepsilon_0}\left[4 r-\frac{3}{r}\right]$
$=\frac{k}{4 \varepsilon_0}\left[4 \times \frac{3}{2}-\frac{3 \times 2}{3}\right]=\frac{k}{\varepsilon_0}$
$\text { (C) } q _{2 e }=\pi k \left[4(2)^2-3\right]$
$=13 \pi k$
$(C)$
$q _{2 m}=\pi k \left[4(2)^2-3\right]$
$=13 \pi k$
$(D)$
$E_z =\frac{k Q}{r^2}$
$=\frac{1}{4 \pi \varepsilon_0} \frac{\pi k\left(4 r^2-3\right)}{r^2}$
$=\frac{k}{4 \varepsilon_0}\left[\frac{4\left(\frac{5}{2}\right)^2-3}{(5 / 2)^2}\right]$
$=\frac{k}{25 \varepsilon_0}[25-3]=\frac{22}{25} \frac{k}{\varepsilon_0}$
