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, 

\[r' = \sqrt{r^2 + x^2}\]

Electricity potential,

\[V = \frac{1}{4\pi \epsilon_0}\frac{q}{r'}\]

\[\Rightarrow V = \frac{1}{4\pi \epsilon_0}\frac{2\pi r\lambda}{\sqrt{r^2 + x^2}}\]

\[\Rightarrow V = \frac{1}{2 \epsilon_0}\frac{r\lambda}{( r^2 + x^2 )^{1/2}}\]

Due to symmetry at point P, vertical component of electric field vanishes.
So, net electric field = Ecosθ

\[\Rightarrow E = \frac{r\lambda}{2 \epsilon_0 ( r^2 + x^2 )^{1/2}}\frac{x}{( r^2 + x^2 )}\]
\[ \Rightarrow E = \frac{r\lambda x}{2 \epsilon_0 ( r^2 + x^2 )^{3/2}}\]