# Four charges are brought from infinity and placed at one meter intervals as shown. Determine the electric potential energy of this group?

## Answer is E, why? A) 0.068 J B) 0.032 J C) 0 J D) - 0.014 J E) - 0.021 J

Mar 15, 2018

suppose,the charge placed at origin is ${q}_{1}$ and next to it we give name as ${q}_{2} , {q}_{3} , {q}_{4}$

Now,potential energy due to two charges of ${q}_{1}$ and ${q}_{2}$ separated by distance $x$ is $\frac{1}{4 \pi \epsilon} \left({q}_{1}\right) \frac{{q}_{2}}{x}$

So,here potential energy of the system will be,

$9 \cdot {10}^{9} \left(\frac{{q}_{1} {q}_{2}}{1} + \frac{{q}_{1} {q}_{3}}{2} + \frac{{q}_{1} {q}_{4}}{3} + \frac{{q}_{2} {q}_{3}}{1} + \frac{{q}_{2} {q}_{4}}{2} + \frac{{q}_{3} {q}_{4}}{1}\right)$ (i.e sum of potential energy due to all possible charge combination)

$= 9 \cdot {10}^{9} \left(- \frac{1}{1} + \frac{1}{2} + \frac{- 1}{3} + \frac{- 1}{1} + \frac{1}{2} + \frac{- 1}{1}\right) \cdot {10}^{-} 6 \cdot {10}^{-} 6$

$= 9 \cdot {10}^{-} 3 \cdot \left(- \frac{7}{3}\right) = - 0.021 J$