Question

Consider a circular ring of radius r, uniformly charged with linear charge density λ. 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

Given:
Radius of the ring = r
So, circumference = 2πr
Charge density = λ,
Total charge, q = 2πr × λ
Distance of the point from the centre of the ring = x
Distance of the point from the surface of the ring, $r\text{'}=\sqrt{r^2+x^2}$
Electricity potential, $V=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q}{r\text{'}}$
$\Rightarrow V=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{2\mathrm{\pi }r\lambda }{\sqrt{r^2+x^2}}$
$\Rightarrow V=\frac{1}{2{\epsilon }_0}\frac{r\lambda }{(r^2+x^2{)}^{1/2}}$


Due to symmetry at point P, vertical component of electric field vanishes.
So, net electric field = Ecosθ
$$\begin{gathered} \Rightarrow E=\frac{r\lambda }{2{\epsilon }_0(r^2+x^2{)}^{1/2}}\frac{x}{(r^2+x^2)} \\ \Rightarrow E=\frac{r\lambda x}{2{\epsilon }_0(r^2+x^2{)}^{3/2}} \end{gathered}$$