Answer:
The electrical potential energy is 54.9J
Step-by-step explanation:
Given that charge q1=q2=q3=1.15*10-6c
The side length is 0.650mm
We are required to find the electrical potential U of the system
Note before the U of the system is the summation of potential energy interaction of the system it is given by
U= 1/4πe[q1q2/r12+q2q3/r23+q1q3/r13]
But since the charges are the same and the distance apart are also the same hence U reduce to
U= 1/4πe(3q^2/r)
We're 1/4πe=9*10^9N.M^2/C^2
U= 9*10^9(3*(1.15*10^-6)^2/0.65*10^-3)
U=54.9J