Question

A non-conducting ring of radius $0.5\text{m}$ carries a total charge of $1.11\times {10}^{-10}\text{C}$ distributed non-uniformly on its circumference producing on electric field $E$ everywhere in space. The value of the intergal ${\int }_{l=\infty }^{l=0}-E.dl$ ($l=0$ being centre of the ring) in $\text{volt}$ is

• A. $+2$
• B. $-1$
• C. $-2$
• D. zero

Answer 1

The correct option is A $+2$

Given,
Radius of ring, $R=0.5\text{m}$,
Charge on the ring, $Q=1.11\times {10}^{-10}\text{C}$

We know that, the relation between electric potential difference and electric field is,

${\int }_{l=\infty }^{l=0}-E.dl=V_{l=0}-V_{l=\infty }...\left(1\right)$

As per the usual convention, potential at infinity is zero.i.e.,
$V_{\infty }=0$

So, from eqution $\left(1\right)$, we have
${\int }_{l=\infty }^{l=0}-E.dl=V_{l=0}...\left(2\right)$

Now,
Potential at the centre of the ring , whether the charge distribution is uniform or non uniform,
$V_{l=0}=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R}$

From equation $\left(2\right)$,
${\int }_{l=\infty }^{l=0}-E.dl=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R}$

Substituting the values,
${\int }_{l=\infty }^{l=0}-E.dl=\frac{(9\times {10}^9)(1.11\times {10}^{-10})}{0.5}$

$\therefore {\int }_{l=\infty }^{l=0}-E.dl=2\text{V}$

Hence, option (a) is correct answer.

Why this question? This question will enable students to understand the physical significane of the concept of electric potential.