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 at $\text{A}$, then the direction of electric field at the centre due to the remaining wire is along.

• A. $\text{OA}$
• B. $\text{OB}$
• C. $\text{OC}$
• D. $\text{OD}$

Answer 1

The correct option is C $\text{OC}$

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.

Since, $dq$ is negative, so the direction of the electric field at centre $O$ due to $dq$ will be toward $\text{OA}$.

So, from eq. (2), the direction of the electric field at centre $O$ due to remaining part will be toward $\text{OC}$.

Hence, $\left(c\right)$ is the correct choice.