Question

Find the maximum electric field due to a uniformly charged ring of charge $Q$ and radius $R$ along the axis of the ring $[K=\frac{1}{4\pi {ϵ}_0}]$

• A. $\frac{KQ}{\sqrt{3}{R}^2}$
• B. $\frac{KQ}{3\sqrt{3}{R}^2}$
• C. $\frac{2KQ}{3\sqrt{3}{R}^2}$
• D. $\frac{2KQ}{\sqrt{3}{R}^2}$

Answer 1

The correct option is C $\frac{2KQ}{3\sqrt{3}{R}^2}$


We know that the net electric field $\left(E_{\text{net}}\right)$ on the axial point of ring is given by,

$E=\frac{KQx}{(R^2+x^2{)}^{\frac{3}{2}}}...\left(1\right)$

For $E$ to be maximum with respect to axial point at distance $x$,

$\frac{dE}{dx}=0$

$\Rightarrow KQ\left[\frac{d}{dx}\right(x\left)\right(R^2+x^2{)}^{-\frac{3}{2}}]=0$

$\Rightarrow 1(R^2+x^2{)}^{-\frac{3}{2}}+x\times (-\frac{3}{2})(R^2+x^2{)}^{-\frac{5}{2}}\times 2x=0$

$\Rightarrow 1-\frac{3x^2}{R^2+x^2}=0$

$\Rightarrow R^2-2x^2=0$

$\Rightarrow x=\pm \frac{R}{\sqrt{2}}$

For , $x=+\frac{R}{\sqrt{2}}$

Maximum electric field from $\left(1\right)$,

$E_{\text{max}}=\frac{\frac{KQR}{\sqrt{2}}}{{(R^2+\frac{R^2}{2})}^{\frac{3}{2}}}$

$\therefore E_{\text{max}}=\frac{2KQ}{3\sqrt{3}{R}^2}$

Hence, option (c) is correct answer.