Question

The electric field strength due to ring of radius R at a distance x from its centre on the axis of ring carrying charge Q is given by $E=\frac{1}{4\pi {\epsilon }_0}\frac{Qx}{(R^2+x^2{)}^{3/2}}$. At what distance from the centre will the electric field be maximum?

• A. $x=R$
• B. $x=\frac{R}{2}$
• C. $x=\frac{R}{\sqrt{2}}$
• D. $x=\frac{2}{\sqrt{R}}$

Answer 1

Answer: (C)

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

E is maximum at a point where $\frac{dE}{dx}=0$

$\frac{dE}{dx}=\frac{Q}{4\pi {ϵ}_0}\frac{(x^2+R^2{)}^{\frac{3}{2}}\cdot (1)-x\cdot \frac{3}{2}(x^2+R^2{)}^{\frac{1}{2}}\cdot 2x}{(x^2+R^2{)}^3}=0$

$⟹(x^2+R^2)-3x^2=0$

Thus, $R^2=2x^2$

$⟹x=\pm \frac{R}{\sqrt{2}}$