Question

Four charges $+q,-q,+q$ and $-q$ are placed in order on the four consecutive corners of a square of side $a$. The work done in interchanging the positions of any two neighbouring charges of the opposite sign is

• A. $\frac{q^2}{4\pi {\epsilon }_{0}a}(-4+\sqrt{2})$
• B. $\frac{q^2}{4\pi {\epsilon }_{0}a}(4+2\sqrt{2})$
• C. $\frac{q^2}{4\pi {\epsilon }_{0}a}(4-2\sqrt{2})$
• D. $\frac{q^2}{4\pi {\epsilon }_{0}a}(4+\sqrt{2})$

Answer 1

The correct option is C $\frac{q^2}{4\pi {\epsilon }_{0}a}(4-2\sqrt{2})$

Let, bottom two charges $+q$ and $-q$ are interchanged.


Work done, $\left(W\right)$ in interchanging the charges will be equal to the change in potential energy, $\left(\Delta U\right)$ of the system due change in the positions of the charges,

$W=\Delta U=U_f-U_i...\left(1\right)$

where,
Initial potential energy of the system,

$U_i=\frac{1}{4\pi {\epsilon }_0}(\frac{-q^2}{a}+\frac{q^2}{\sqrt{2}a}-\frac{q^2}{a}-\frac{q^2}{a}+\frac{q^2}{\sqrt{2}a}-\frac{q^2}{a})$

Final potential energy of the system,

$U_f=\frac{1}{4\pi {\epsilon }_0}(\frac{-q^2}{a}-\frac{q^2}{\sqrt{2}a}+\frac{q^2}{a}+\frac{q^2}{a}-\frac{q^2}{\sqrt{2}a}-\frac{q^2}{a})$

Substituting the values in equation $\left(1\right)$, we get

$W=\frac{1}{4\pi {\epsilon }_0}[(\frac{-q^2}{a}-\frac{q^2}{\sqrt{2}a}+\frac{q^2}{a}+\frac{q^2}{a}-\frac{q^2}{\sqrt{2}a}-\frac{q^2}{a})-(\frac{-q^2}{a}+\frac{q^2}{\sqrt{2}a}-\frac{q^2}{a}-\frac{q^2}{a}+\frac{q^2}{\sqrt{2}a}-\frac{q^2}{a})]$

$\therefore W=\frac{q^2}{4\pi {\epsilon }_{0}a}[4-2\sqrt{2}]$

Hence, option (c) is correct answer.