Question

A point charge pf -6 x 10^-9 C is 5 m from point A and 3 m from point B.

Find the Potential at point A and point B. What is the potential difference between

point A to B?

Answer 1

The potential due point charge

$$V=k\frac{q}{r}$$

$$V_1=9*10^9*\frac{6*10^{-9}}{5}=11\:\rm V$$

$$V_2=9*10^9*\frac{6*10^{-9}}{3}=18\:\rm V$$

The potential difference between point A to B

$$V=V_2-V_1=7\:\rm V$$