Question

In each situation of column-$I$, some charge distributions are given with all details explained. The electrostatic potential energy and its nature is given situation in column -$II$. Match the proper entries
from column-$II$ to column-$I$ using the codes given below the columns,

	Column-$I$		Column-$II$
p.	A thin shell of radius a and having a charge – Q uniformly distributed over its surface as shown	1.	$\frac{1}{8{\mathrm{\pi \epsilon }}_o}\times \frac{Q^2}{a}$in magnitude
q.	A thin shell of radius $\frac{5a}{2}$having a charge $-Q$ uniformly distributed over its surface and a point charge $-Q$ placed at its centre as shown	2.	$\frac{3}{20{\mathrm{\pi \epsilon }}_o}\times \frac{Q^2}{a}$in magnitude
r.	A solid sphere of radius $a$ and having a charge $-Q$uniformly distributed throughout its volume as shown	3.	$\frac{27Q}{80\pi {\epsilon }_o}\times \frac{Q^2}{a}$in magnitute
s.	A solid sphere of radius a and having a charge – Q uniformly distributed throughout its volume. The solid sphere is surrounded by a concentric thin uniformly charged spherical shell of radius $2a$ and carrying charge $-Q$ as shown	4.	$\frac{27Q}{60\pi {\epsilon }_o}\times \frac{Q^2}{a}$in magnitude

• A. p-1; q-3; r-2; s-3
• B. p-1; q-2; r-3; s-3
• C. p-1; q-2; r-2; s-3
• D. p-1; q-3; r-2; s-4

Answer 1

The correct option is C p-1; q-2; r-2; s-3

p. Electrostatic potential energy = $\frac{1}{4{\mathrm{\pi \epsilon }}_o}\frac{{\left(-Q\right)}^2}{2a}=\frac{Q^2}{8{\mathrm{\pi \epsilon }}_{o}a}$
q. Electrostatic potential energy = $\frac{1}{4{\mathrm{\pi \epsilon }}_o}\left(\frac{\left(-Q\right)\times \left(-Q\right)}{\frac{5a}{2}}+\frac{{\left(-Q\right)}^2}{2\times \left(\frac{5a}{2}\right)}\right]=\frac{3}{20}\times \frac{Q^2}{{\mathrm{\pi \epsilon }}_{o}a}$r. Electrostatic potential energy = $\frac{1}{4{\mathrm{\pi \epsilon }}_o}\frac{3Q^2}{5a}=\frac{3}{20}\times \frac{Q^2}{{\mathrm{\pi \epsilon }}_{o}a}$
s. Electrostatic potential energy = $\frac{1}{4{\mathrm{\pi \epsilon }}_o}\left(\frac{3Q^2}{5a}+\frac{-Q^2}{2\left(2a\right)}+\frac{\left(-Q\right)\times \left(-Q\right)}{2a}\right]=\frac{27}{80}\times \frac{Q^2}{{\mathrm{\pi \epsilon }}_{o}a}$