Question

A thin wire ring of radius $r$ carries a charge $q$. Find the magnitude of the electric field strength on the axis of the ring as a function of distance $l$ from its centre. Investigate the obtained function at $l>>r$. Find the maximum strength magnitude and the corresponding distance $l$. Draw the approximate plot of the function $E\left(l\right)$.

Answer 1

From the symmetry of the condition, it is clear that, the field along the normal will be zero.
i.e. $E_n=0$ and $E=E_1$
Now $d{E}_1=\frac{dq}{4\pi {ϵ}_0(R^2+l^2)}\cos \theta$
But $dq=\frac{q}{2\pi R}dx$ and $\cos \theta =\frac{l}{(R^2+l^2{)}^{1/2}}$
Hence
$E=\int d{E}_l={\int }_0^{2\pi R}\frac{ql}{2\pi R}\cdot \frac{dx}{4\pi {ϵ}_0(R^2+l^2{)}^{3/2}}$
or $E=\frac{1}{4\pi {ϵ}_0}\frac{ql}{(l^2+R^2{)}^{3/2}}$
and for $l>>R$, the ring behaves like a point charge, reducing the field to the value,
$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{l^2}$
For $E_{max}$, we should have $\frac{dE}{dl}=0$
So, $(l^2+R^2{)}^{3/2}-\frac{3}{2}l(l^2+R^2{)}^{1/2}2l=0$ or $l^2+R^2-3l^2=0$
Thus $l=\frac{R}{\sqrt{2}}$ and $E_{max}=\frac{q}{6\sqrt{3}\pi {ϵ}_0{R}^2}$.