Question

Consider a circular ring of radius r,uniformly charged with linear charge density $\lambda$. Find the electric potential at a point on the axis at a distance x from the centre of the ring. Using this expression for the potential, find the electric field at this point.

Answer 1

The relation between the electric field and the potential is:

$E=\frac{-dV}{dx}$

Linear charge density $\lambda$

Charge on ring $=\lambda \cdot 2\pi r$

Electric potential $V=\frac{kq}{d}$

The distance of the point from the circumference of the ring.

$d=\sqrt{r^2+x^2}$

So, the electric potential can be obtained as:

$V=\frac{k.2\pi r\lambda }{\sqrt{r^2+x^2}}\frac{dV}{dx}=\frac{-k.2\pi r\lambda .\frac{1}{2}\frac{2x}{\sqrt{r^2+x^2}}}{(r^2+x^2)}E=-\frac{dV}{dx}=\frac{k.2\pi rx\lambda }{{(r^2+x^2)}^{3/2}}E=\frac{rx\lambda }{2{\epsilon }_0{(r^2+x^2)}^{3/2}}$