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 very small and equal angles at point $A$. They are at distances, $r_1$ and $r_2$ from point $A$ respectively. Then -

• A. The ratio of charge on elements $a_1$ and $a_2$ is $\frac{r_1}{r_2}$.
  
• B. The element $a_1$ produces a 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 elements $a_1$ produces greater potential at $A$ than element $a_2$.

Answer 1

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

The elements $a_1$ and $a_2$ produce same potential at $A$.



Charge on $a_1$ is $r_1\theta \lambda$
$[$ Assuming small angle subtended by elements at $A$ to be $\theta ]$

Similarly, charge on $a_2$ is $r_2\theta \lambda$

So, ratio of charges is $\frac{r_1}{r_2}$.

Now, electric field produced by $a_1$ at $A$,
$E_1=\frac{k{r}_1\theta \lambda }{r_1^2}=\frac{k\theta \lambda }{r_1}$

Similarly, by $a_2$,
$E_2=\frac{k\theta \lambda }{r_2}$

As $r_2>r_1\Rightarrow E_1>E_2$

Further, electric potential by $a_1$ at $A$,
$V_1=\frac{k{r}_1\theta \lambda }{r_1}=k\theta \lambda$

Similarly, by $a_2$,
$V_2=\frac{k{r}_2\theta \lambda }{r_2}=k\theta \lambda$

$\therefore V_1=V_2$