Question

Four charges $q_1=1\mu \text{C},q_2=2\mu \text{C},q_3=-3\mu \text{C}$ and $q_4=4\mu \text{C}$ are kept on the vertices of a square of side $1\text{m}$. Find the electric potential energy of this system of charges.

• A. $+7.62\times {10}^{-2}\text{J}$
• B. $-7.62\times {10}^{-2}\text{J}$
• C. $2\times {10}^{-2}\text{J}$
• D. $-0.1\text{J}$

Answer 1

The correct option is B $-7.62\times {10}^{-2}\text{J}$

Given

$q_1=1\mu \text{C},q_2=2\mu \text{C},q_3=-3\mu \text{C},q_4=4\mu \text{C}$

To find the electrical potential energy of a system of charges, we make pairs. If there are total $n$ charges then total number of pairs are $\frac{n(n-1)}{2}$.

Electric potential energy for four point charges $q_1,q_2,q_3$ and $q_4$ would be given by,

$U=\frac{1}{4\pi {\epsilon }_0}[\frac{q_4{q}_3}{r_{43}}+\frac{q_4{q}_2}{r_{42}}+\frac{q_4{q}_1}{r_{41}}+\frac{q_3{q}_2}{r_{32}}+\frac{q_3{q}_1}{r_{31}}+\frac{q_2{q}_1}{r_{21}}].......\left(1\right)$

From the diagram we can deduce that,

$r_{41}=r_{43}=r_{32}=r_{21}=1\text{m}$

and $r_{42}=r_{31}=\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{2}\text{m}$

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

$U=9\times {10}^9\times {10}^{-6}\times {10}^{-6}[\frac{4\times (-3)}{1}+\frac{4\times 2}{\sqrt{2}}+\frac{4\times 1}{1}+\frac{(-3)\times 2}{1}+\frac{(-3)\times 1}{\sqrt{2}}+\frac{2\times 1}{1}]$

$\Rightarrow U=9\times {10}^{-3}[-12+\frac{5}{\sqrt{2}}]=-7.62\times {10}^{-2}\text{J}$

Hence, option (b) is the correct answer.