Question

A non conducting rod AB of length $\sqrt{3}R$, uniformly distributed charge of linear charge density $\lambda$ and.a non-conducting ring of uniformly distributed charge Q, are placed as shown in the figure. Point A is the centre of ring and line AB is the axis of the ring, perpendicular to plane of ring. The electrostatic interaction energy between ring and rod is

• A. $\frac{Q\lambda }{4\pi {ϵ}_o}ln(2+\sqrt{3})$
• B. $\frac{Q\lambda }{2\pi {ϵ}_o}ln(2+\sqrt{3})$
• C. $\frac{Q\lambda }{4\pi {ϵ}_o}ln(2-\sqrt{3})$
• D. $\frac{Q\lambda }{2\pi {ϵ}_o}ln(2-\sqrt{3})$

Answer 1

The correct option is A $\frac{Q\lambda }{4\pi {ϵ}_o}ln(2+\sqrt{3})$

Given a non-conducting rod of length $\sqrt{3}R$; $\lambda =$ linear charge density;

non-conducting ring having charge $=Q$

To find electrostatic interation energy between ring and rod.

Small change in energy, $dU=$ potential $\times$ charge

$=V_p$ (ring) $\lambda dx$ ...(i)

potential due to ring, $V_p$ (ring) $=\frac{kQ}{\sqrt{R^2+x^2}}$ ...(ii)

where, $k=\frac{1}{4\pi {\epsilon }_0}$

$x=$ distance of point $P$ from center of ring.

So, $dU=\frac{kQ}{\sqrt{R^2+x^2}}\lambda dx$

[Here we have substituted (ii) in (i)]

So, total electrostatic energy can be obtained by integrating $\sqrt{3}R$

$U=\frac{Q\lambda }{4\pi {\epsilon }_0}{\int }_0^{\sqrt{3}R}\frac{dx}{\sqrt{R^2+x^2}}$

To solve this integral, $x=R\tan \theta$

On differentiating both sides, we get

$dx=R{\sec }^2\theta d\theta$

For limits of integration: if $x=0\Rightarrow \theta =0$

and for, $x=\sqrt{3}R\Rightarrow \theta =\frac{\pi }{3}$

So, $U$ becomes, $U=\frac{Q\lambda }{4\pi {\epsilon }_0}{\int }_0^{\pi /3}{\sec }^2\theta d\theta$

$=\frac{Q\lambda }{4\pi {\epsilon }_0}{\left[\ln (\sec \theta +\tan \theta )\right]}_0^{\pi /3}$

$=\frac{Q\lambda }{4\pi {\epsilon }_0}\ln (2+\sqrt{3})$