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

Answer:

$E=\dfrac{KQx}{(x^2+a^2)^{\frac{3}{2}}}$

Step-by-step 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

$E=\dfrac{KQcos\theta }{r^3}$

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

 $cos\theta =\dfrac{x}{r}$

$r^2=x^2+a^2$

$E=\dfrac{KQx}{(x^2+a^2)^{\frac{3}{2}}}$