Question

A wire having a uniform linear charge density $\lambda$, is bent in the form of a ring of radius $R$. Point $A$ as shown in the figure, is in the plane of the ring but not at the centre. Two elements of the ring of lengths $a_1$ and $a_2$ subtend same (very small) angle at the point $A$. If they are at distances $r_1$ and $r_2$ from the point $A$ respectively, choose the correct statements.

• A. The ratio of charge of elements $a_1$ and $a_2\text{is}\frac{r_1}{r_2}.$
• B. The element $a_1$ produced greater magnitude of electric field at $A$ than element $a_2$.
• C. The elements $a_1$ and $a_2$ produce same potential at $A$.
• D. The direction of net electric field at $A$ is towards element $a_2$.

Answer 1

Answer: (A), (B), (C), (D)

The direction of net electric field at $A$ is towards element $a_2$.


Using length of arc formula
$a_1=r_1\theta ;a_2=r_2\theta$

Given charge density is $\lambda$. So,
$q_1=\lambda a_1;q_2=\lambda a_2$

$\Rightarrow \frac{q_1}{q_2}=\frac{a_1}{a_2}=\frac{r_1}{r_2}$
(Option A is correct)

The fields at $A$ due to $a_1$ is
$d{E}_1=\frac{k{q}_1}{r_1^2}=\frac{k\lambda r_1\theta }{r_1^2}$
The fields at $A$ due to $a_2$ is
$d{E}_2=\frac{k{q}_1}{r_2^2}=\frac{k\lambda r_2\theta }{r_2^2}$
$\frac{d{E}_1}{d{E}_2}=\frac{r_2}{r_1}>1\Rightarrow d{E}_1>d{E}_2$
(Option B is correct)

Net field at point $A$ is
$d{E}_{net}=d{E}_1-d{E}_2$
It is towards $a_2$, as $d{E}_1>d{E}_2$
(Option D is correct)

The potential at $A$ due to $a_1$ is
$d{V}_1=\frac{k{q}_1}{r_1}=\frac{k\lambda r_1\theta }{r_1}=k\lambda \theta$
The potential at $A$ due to $a_2$ is
$d{V}_2=\frac{k{q}_2}{r_2}=\frac{k\lambda r_2\theta }{r_2}=k\lambda \theta$
$\Rightarrow d{V}_1=d{V}_2$
(Option C is correct)