Two point charges, +3.18 µC and -6.40 µC, are separated by 1.32 m. What is the electric potential midway between them?

1 answer

v=kq/r
r=.5(1.32)=0.66m
k=(8.99*10^9)
q1= 3.18x10^-6 C
q2= -6.40x10^-6 C

v1= (k(3.18x10^-6C)/0.66m)=4.3x10^4v
v2= (k(-6.4x10^-6C)/0.66m)=-8.8x10^4v

v1+v2=v(total)= -4.5x10^4 v