Question

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

• A. $\frac{Qdl}{8{\pi }^2{ϵ}_0{r}^3}$
• B. $\frac{Qdl}{8\pi {ϵ}_0{r}^3}$
• C. $\frac{Qdl}{4{\pi }^2{ϵ}_0{r}^3}$
• D. $\frac{Qdl}{2{\pi }^2{ϵ}_0{r}^3}$

Answer 1

The correct option is A $\frac{Qdl}{8{\pi }^2{ϵ}_0{r}^3}$

Linear charge density $\left(\lambda \right)$ will be

$\lambda =\frac{Q}{2\pi r}$

For a complete loop with uniform charge $\text{Q}$, electric field at the centre will be zero.

${\overrightarrow{E}}_c=0$

Now $dl$ is being cut from circumference so that the system will have some electric field at centre $\text{C}$.

Let,
${\overrightarrow{E}}_{dl}=$ electric field due to $dl$

${\overrightarrow{E}}_{rw}=$ electric field due to remaining wire

$\therefore {\overrightarrow{E}}_{dl}+{\overrightarrow{E}}_{rw}=0$

$\Rightarrow {\overrightarrow{E}}_{rw}=-{\overrightarrow{E}}_{dl}$

Considering only magnitude, we get

$\Rightarrow |{\overrightarrow{E}}_{rw}|=|{\overrightarrow{E}}_{dl}|...........\left(1\right)$

Let's say $dq$ charge is removed

So, $dq=\lambda dl$

From Coulomb's law, considering $dq$ is a point charge

$E_{dl}=\frac{1}{4\pi {ϵ}_0}\frac{dq}{r^2}=\frac{1}{4\pi {ϵ}_0}\frac{\lambda dl}{r^2}$
$\Rightarrow E_{dl}=\frac{1}{4\pi {ϵ}_0}\frac{Qdl}{2\pi r}\frac{1}{r^2}=\frac{Qdl}{8{\pi }^2{ϵ}_0{r}^3}$

putting the value in eq.$\left(1\right)$, we get

$\Rightarrow |{\overrightarrow{E}}_{rw}|=\frac{Qdl}{8{\pi }^2{ϵ}_0{r}^3}$

$\begin{array}{c}Ans\left(a\right)\end{array}$