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 stored in this system of charges.

• A. $3.6\times {10}^{-2}\text{J}$
• B. $-7.62\times {10}^{-2}\text{J}$
• C. $7.62\times {10}^{-2}\text{J}$
• D. $-3.6\times {10}^{-2}\text{J}$

Answer 1

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

Potential energy of the system of four charges is given by:

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

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

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

Substituting the given values with sign in equation $\text{(1)}$, we get

$U=(9.0\times {10}^9)\left({10}^{-6}\right)\left({10}^{-6}\right)[\frac{\left(4\right)(-3)}{1}+\frac{\left(4\right)\left(2\right)}{\sqrt{2}}+\frac{\left(4\right)\left(1\right)}{1}+\frac{(-3)\left(2\right)}{1}+\frac{(-3)\left(1\right)}{\sqrt{2}}+\frac{\left(2\right)\left(1\right)}{1}]$

$U=(9.0\times {10}^{-3})[-12+\frac{5}{\sqrt{2}}]$

$U=-7.62\times {10}^{-2}\text{J}$

Hence, option (b) is the correct answer.