Four identical charges (+2.0ìC each)

Question

are brought in from infinity and fixed to a straight line. The charges are located 0.40 meters
apart. Determine the electric potential energy of this group.

Answer 1

label the points 1,2,3,4

V1=V2+V3+V4
V1=kq/x+kq/2x + kq/3x
V1=kq/6x(6+3+2)=11kq/6x
where x is .4m
Now, due to symettry, V4=V1

V2=V1+V3+V4=kq/x+kq/x+kq/2x=kq/6x (15)
and due to symettry, V3=V3

Electric PE= sum of V * q

Electric PE=kq^2/6x (15+15+11+11)
= 7kq^2/x

check my thinking.

Answer 2

thank you! It seems to all make sense. I'm guessing you meant V2=V3 though? Due to symmetry that is.