Question

A point of charge 10×10^-6C has a potential of 4×10^4V from a point 2mm from it. Calculate the potential due to the charge when the distance is now 16mm.

Answer 1

Given:

$V_1=4*10^4\:\rm V$

$r_1=2*10^{-3}\:\rm m$

$r_2=16*10^{-3}\:\rm m$

The potential of electric field of point charge

$$V=k\frac{q}{r}\sim\frac{1}{r}$$

Hence

$$V_2=V_1\frac{r_1}{r_2}=4*10^4*\frac{0.002}{0.016}=5000\:\rm V$$