What is the potential difference half way between two charges of magnitude 4.5 x 10^-6 C 1.5 M apart?

Question

I have no Idea how to answer this, if you could explain the method to solving it I would greatly appreciate it!

Answer 1

Assuming positive charges

potential = (K*q / r )+ (K*q/r)

k=9*10^9

q=4.5*10^-6C

r= 1.5 meter/2 = 0.75 meter

potential = 2*(9*10^9)*(4.5*10^-6)/0.75 = 108,000 volts

This potential is also called "potential difference" relative to infinity.