Question

A non-conducting ring of radius $0.5\text{m}$ have charge $1.11\times {10}^{-10}\text{C}$ non-uniformly distributed over the circumference of the ring, which produces an electric field $E$ around itself. If $l=0$ is the center of the ring, then the value of $\underset{l=\infty }{\overset{l=0}{\int }}-\overrightarrow{E}\cdot \overrightarrow{dl}$ is

• A. $2\text{V}$
• B. $-2\text{V}$
• C. $-1\text{V}$
• D. Zero

Answer 1

The correct option is A $2\text{V}$


As we know that the relation between potential difference and electric field is given by$$dV=-E\cdot dr$$$\Rightarrow \underset{A}{\overset{B}{\int }}dV=\underset{r_A}{\overset{r_B}{\int }}-E\cdot dr$

$\Rightarrow V_B-V_A=\underset{r_A}{\overset{r_B}{\int }}-E.dr$

In the given case $r_A=\infty$ & $r_B=0\text{(centre)}$

So, $V_{center}-V_{\infty }=\underset{\infty }{\overset{0}{\int }}-E.dr$

we know that $V_{\infty }=0$

$\Rightarrow V_{center}=\underset{\infty }{\overset{0}{\int }}-E.dr$

Electric potential at the centre of a ring having charge $q$ distributed on its circumference is given by, $$V_{centre}=\frac{kq}{R}$$$\Rightarrow \underset{\infty }{\overset{0}{\int }}-E.dr=\frac{kq}{R}$

Substituting the values,

$\Rightarrow \underset{\infty }{\overset{0}{\int }}-E.dr=\frac{9\times {10}^9\times 1.11\times {10}^{-10}}{0.5}$

$\Rightarrow \underset{\infty }{\overset{0}{\int }}-E\cdot dr=1.998=2\text{Volt}$

Hence, option $\left(a\right)$ is the correct answer.

Why this problem: This problem enhances the understanding of definition of electric potential and its use for potential at the center of a charged ring.