Question

Uniformly charged ring is shown is figure, $E$ due to ring is maximum at

• A. Centre
• B. $\infty$
• C. $\frac{R}{\sqrt{2}}$ distance from centre
• D. $R$ distance from centre

Answer 1

The correct option is C $\frac{R}{\sqrt{2}}$ distance from centre

By graph (theoretical)

OR
$E=\frac{kQx}{(R^2+x^2{)}^{\frac{3}{2}}}$
But $\frac{dE}{dx}=0$
$\frac{dE}{dx}=\frac{d}{dx}\left(\frac{kQx}{(R^2+x^2{)}^{\frac{3}{2}}}\right)=0$
$\frac{x.\frac{d}{dx}(R^2+x^2{)}^{\frac{3}{2}}-(R^2+x^2{)}^{\frac{3}{2}}\frac{d}{dx}\left(x\right)}{(R^2+x^2{)}^3}=0$
Solve $\Rightarrow x=\frac{R}{\sqrt{2}}$