Question

A point charge $q$ is located at the centre of a thin ring of radius $R$ with uniformly distributed charge $-q$. Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance $x$ from its centre, if $x>>R$.

Answer 1

Let the charge at the centre inside a charge on the ring of $dq=\lambda dl$

At P, horizontal components cancel;

So, $E=\frac{1}{4\pi {\epsilon }_0}\left[\frac{\lambda dl}{r^2}\cos \theta \right]\hat{x}$

Here, $r^2=R^2+x^2$

$\cos \theta =\frac{x}{r}$ and

$\int dl=2\pi r$

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}.\frac{\lambda \left(2\pi r\right)x}{(R^2+x^2)\frac{3}{2}}$

for $x>>R$

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}.\frac{\lambda \left(2\pi r\right)x}{x^3}\hat{x}$

$=\frac{1}{4\pi {\epsilon }_0}.\frac{\lambda \left(2\pi r\right)}{x^2}\hat{x}$