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) $\frac{1}{8\pi {ϵ}_0}\frac{Q^2}{a}$ in magnitude
(Q) A thin shell of radius $\frac{5a}{2}$ and having a charge $-Q$ uniformly distributed over its surface and a point charge $-Q$ is brought from infinity and placed at its centre as shown.	(2) $\frac{3}{20\pi {ϵ}_0}\frac{Q^2}{a}$ in magnitude
(R) A solid sphere of radius $a$ and having a charge $-Q$ uniformly distributed throughout its volume as shown.	(3) $\frac{27Q}{80\pi {ϵ}_0}\frac{Q^2}{a}$ in magnitude
(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) $\frac{5}{27\pi {ϵ}_0}\frac{Q^2}{a}$ in magnitude

• A. $p\to 1,Q\to 2,R\to 2,S\to 3$
• B. $p\to 4,Q\to 1,R\to 2,S\to 3$
• C. $p\to 1,Q\to 2,R\to 3,S\to 1$
• D. $p\to 2,Q\to 4,R\to 3,S\to 1$

Answer 1

The correct option is A $p\to 1,Q\to 2,R\to 2,S\to 3$

(A) Electrostatic potential energy=$\frac{1}{4\pi {ϵ}_0}\frac{-Q^2}{2a}=\frac{1}{4\pi {ϵ}_0}\frac{(-Q{)}^2}{2a}=\frac{Q^2}{8\pi {ϵ}_{0}a}$

(B) Electrostatic potential energy=$\frac{1}{4\pi {ϵ}_0}[\frac{-Q\times (-Q)}{5a/2}+\frac{(-Q{)}^2}{2\left(5a/2\right)}]=\frac{3}{20}\frac{Q^2}{\pi {ϵ}_{0}a}$

(C) Electrostatic potential energy=$\frac{1}{4\pi {ϵ}_0}\frac{3Q^2}{5a}=\frac{3}{20}\frac{Q^2}{\pi {ϵ}_{0}a}$

(D) Electrostatic potential energy=$\frac{1}{4\pi {ϵ}_0}[\frac{3Q^2}{5a}+\frac{(-Q{)}^2}{2\left(2a\right)}+\frac{(-Q)\times (-Q)}{2a}]=\frac{27Q^2}{80\pi {ϵ}_{0}a}$