Question

Show that electric potential at a point P, at a distance 'r' from a fixed point charge Q, is given by $V=\left(\frac{1}{4\pi {ϵ}_0}\right)\frac{Q}{r}$.

Answer 1

Electric field produced at a distance $r$ is given by $E=\frac{1}{4\pi {ϵ}_o{r}^2}$

Electric potential at that point $V=-{\int }_{\infty }^{r}E.dr$

$V=-{\int }_{\infty }^r$ $\frac{1}{4\pi {ϵ}_o{r}^2}dr$

$V=-$ $\frac{1}{4\pi {ϵ}_o}\times \frac{-1}{r}{|}_{\infty }^r$

$V=-$ $\frac{1}{4\pi {ϵ}_o}\times \frac{-1}{r}=\frac{1}{4\pi {ϵ}_{o}r}$