Question

Suppose we have a uniformly charged circular ring of radius $R$ and charge $Q$. The electric field at a distance $x$ from the centre such that $x>>R$ is given by :

Take $k=\frac{1}{4\pi {ϵ}_0}$.

• A. $E=\frac{kQ}{R^2}$
• B. $E=\frac{kQ}{x^2}$
• C. $E=\frac{kQ}{R}$
• D. $E=\frac{kQ}{x}$

Answer 1

The correct option is B $E=\frac{kQ}{x^2}$

Let us consider a small charge element of charge $dq$

And, $dq=\frac{Q}{2\pi R}$.

The field at point P due to this element is $=E=\frac{Kdq}{r^2}$

$E=\frac{Kdq}{(R^2+x^2)}$

Now, from figure we see that component of field normal to axis is cancelled by two diametrically opposite points.

Hence, only component of field along axis is left which add up for all such elements.

$E_{net}=\int E\cos \theta$ where $\theta$ is same for all elements means $\theta =constant$

$⟹E_{net}=\int \frac{K\cos \theta }{(R^2+x^2)}dq$

$⟹E_{net}=\frac{KQ}{(R^2+x^2)}\cos \theta$

$⟹E_{net}=\frac{KQ}{(R^2+x^2)}\frac{x}{\sqrt{R^2+x^2}}$

$⟹E_{net}=\frac{KQx}{(R^2+x^2{)}^{3/2}}$

For $x>>R$,we can say $x^2+R^2\approx x^2$

$E_{net}=\frac{KQx}{(x^2{)}^{3/2}}=\frac{KQx}{x^3}$

$⟹E_{net}=\frac{KQ}{x^2}$

Answer-(B)