Question 826: Electrostatics & Capacitance

Question

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 $\rho_{B}=$ $\frac{2 k}{r}$. Assume that dimensions are taken care of. All physical quantities are in their SI units.
Which of the following statement(s) is(are) correct?

Accepted Answer

$\begin{aligned} Q_{	ext {Total }} & =\int_0^{r_A} k r\left(4 \pi r^2ight) d r+\int_{r_A}^{r_B} \frac{2 k}{r}\left(4 \pi r^2ight) d r \\ & =\frac{4 \pi k}{4} r_A^4+\frac{8 \pi k}{2}\left(r_B^2-r_A^2ight) \\ & =\pi k+4 \pi k\left(r_B^2-r_A^2ight)\end{aligned}$
(A) If $\mathrm{r}_{\mathrm{B}}  =\sqrt{\frac{3}{2}}, \ r_A=1 $
$
\mathrm{Q}_{	ext {total }}  =\pi k r_A^4+4 \pi k\left(\frac{3}{2}-r_A^2ight) $
$
 =\pi k+4 \pi k\left(\frac{3}{2}-1ight) $
$ =3 \pi k$
(B) If $

r_B=\frac{3}{2} \& r_A=1 $
$
 \mathrm{Q}_{	ext {total }}=\pi k r_A^4+4 \pi k\left(\frac{9}{4}-1ight) $
$
 =6 \pi k $$
 \mathrm{~V}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{6 \pi k}{r_B}ight)=\frac{k}{\varepsilon_0} $
$
 	ext { Hence, } \quad \mathrm{V}=\frac{k}{\varepsilon_0} $
(C) If
$
r_B  =2 $
$
\mathrm{Q}_{	ext {total }}  =13 \pi K$
(D) If $
r_B  =5 / 2 $
$
Q_{	ext {total }}  =22 \pi k $
$
E  =\frac{1}{4 \pi \varepsilon_0} \frac{(22 \pi k)}{25} 	imes 4 $
$$ \Rightarrow $$ $E  =\frac{22 k}{25 \varepsilon_0}$

Answer

If $r_{B}=\sqrt{\frac{3}{2}}$, then the electric field is zero everywhere outside $B$.

Answer

If $r_{B}=\frac{3}{2}$, then the electric potential just outside $B$ is $\frac{k}{\epsilon_{0}}$.

Answer

If $r_{B}=2$, then the total charge of the configuration is $15 \pi k$.

Answer

If $r_{B}=\frac{5}{2}$, then the magnitude of the electric field just outside $B$ is $\frac{13 \pi k}{\epsilon_{0}}$.

Electrostatics & Capacitance

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