Question

A thin circular ring of radius “a” has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring at a distance x from its center.

Answer 1

`E=\frac{KQx}{(x^2+a^2)^{(3)/(2)}}`

Explanation:

Given that

Radius of ring=a

Distance of point P from center=x

Total charge=Q

We know that electric field is given by

`E=(KQcos\theta )/(r^3)`

Here electric field in two direction will be cancel out and only in one direction electric field will exits.

`cos\theta =(x)/(r)`

`r^2=x^2+a^2`

`E=\frac{KQx}{(x^2+a^2)^{(3)/(2)}}`