Question

A circular wire loop of radius $a$ carries a total charge $Q$ distributed uniformly over its length. A small length $dL$ of the wire is cut - off. Find the electric field at the centre due to remaining wire.

• A. $\frac{QdL}{4{\pi }^2{ϵ}_0{a}^3}$
• B. $\frac{QdL}{8{\pi }^2{ϵ}_0{a}^3}$
• C. $\frac{QdL}{5{\pi }^2{ϵ}_0{a}^3}$
• D. $\frac{QdL}{10{\pi }^2{ϵ}_0{a}^3}$

Answer 1

The correct option is B $\frac{QdL}{8{\pi }^2{ϵ}_0{a}^3}$

We know that, the net electric field at the centre of the ring having uniformly distributed charge over its circumference will be zero.

$\therefore {\overrightarrow{E}}_{net}=0.......\left(1\right)$

Let,
${\overrightarrow{E}}_{dL}=$ Electric field at centre $O$ due to $dL$ part only
${\overrightarrow{E}}_L=$ Electric field at centre $O$ due to remaining part

From eq. (1),

${\overrightarrow{E}}_{dL}+{\overrightarrow{E}}_L=0$

$\Rightarrow {\overrightarrow{E}}_L=-{\overrightarrow{E}}_{dL}...........\left(2\right)$

Let's assume $dL$ part of the circle have $dq$ charge.

$\therefore dq=\frac{Q}{2\pi a}dL$

So, $dq$ is very small, and we can assume as a point charge.

${\overrightarrow{E}}_{dL}=\frac{dq}{4\pi {ϵ}_0{a}^2}=\frac{QdL}{8{\pi }^2{ϵ}_0{a}^3}$

putting the value in eq. (2)

$\Rightarrow {\overrightarrow{E}}_L=-\frac{QdL}{8{\pi }^2{ϵ}_0{a}^3}$

$\therefore |{\overrightarrow{E}}_L|=\frac{QdL}{8{\pi }^2{ϵ}_0{a}^3}$

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