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 $\frac{1}{4 \pi \varepsilon_{0}}$ 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 \varepsilon_{0} a}+\frac{(-a)(+q)}{4 \pi \varepsilon_{0} a}+\frac{(+q)(-q)}{4 \pi \varepsilon_{0} a} $
$+\frac{(-q)(q)}{4 \pi \varepsilon_{0} a}+\frac{(q)(q)}{4 \pi \varepsilon_{0} a \sqrt{2}}+\frac{(-a)(-q)}{4 \pi \varepsilon_{0} a \sqrt{2}} $
$U=4\left(\frac{-q^{2}}{4 \pi \varepsilon_{0} a}\right)+2 \frac{q^{2}}{4 \pi \varepsilon_{0} \sqrt{2} a} $
$U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{a}(-4+\sqrt{2})$