Question

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

Answer 1

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 `1/(4 pi epsilon) (q_1)(q_2)/x`

So,here potential energy of the system will be,

`9*10^9((q_1 q_2)/1 +(q_1 q_3)/2 +(q_1 q_4)/3 +(q_2 q_3)/1 +(q_2 q_4)/2 +(q_3 q_4)/1)` (i.e sum of potential energy due to all possible charge combination)

`=9*10^9(-1/1 +1/2 +(-1)/3+(-1)/1 +1/2 +(-1)/1)*10^-6*10^-6`

`=9*10^-3*(-7/3)=-0.021J`