Question

There are four point charges $+q,-q,+q$ and $-q$ are placed at the corners $A,B,C,$ and $D$ respectively of a square of side a. The potential energy of the system is $\frac{1}{4\pi {\epsilon }_0}$ times

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

Answer 1

Answer: (A)

There will be six pairs,
So, potential energy of the system is
$U=\frac{(q)(-q)}{4\pi {\epsilon }_{0}a}+\frac{(-a)(+q)}{4\pi {\epsilon }_{0}a}+\frac{(+q)(-q)}{4\pi {\epsilon }_{0}a}$
$+\frac{(-q)(q)}{4\pi {\epsilon }_{0}a}+\frac{(q)(q)}{4\pi {\epsilon }_{0}a\sqrt{2}}+\frac{(-a)(-q)}{4\pi {\epsilon }_{0}a\sqrt{2}}$
$U=4\left(\frac{-q^2}{4\pi {\epsilon }_{0}a}\right)+2\frac{q^2}{4\pi {\epsilon }_0\sqrt{2}a}$
$U=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a}(-4+\sqrt{2})$