Question

The arc $AB$ with the centre $C$ and the infinitely long wire having linear charge density $\lambda$ are lying in the same plane. A point charge $q_o$ moves from point $A$ to $B$ through a circular path $AB$ of radius $a$, then

• A. Electric potential at $A<$ Electric potential at $B$
• B. Minimum amount of work done will be $\frac{q_o\lambda }{2\pi {ϵ}_o}\ln \left(3/2\right)$
• C. Electric potential at $A>$ Electric potential at $B$
• D. Minimum amount of work done will be $\frac{q_o\lambda }{2\pi {ϵ}_o}\ln \left(2/3\right)$

Answer 1

Answer: (A), (B)

Minimum amount of work done will be $\frac{q_o\lambda }{2\pi {ϵ}_o}\ln \left(3/2\right)$


Work done,

$W=q_o(V_B-V_A)=q_o(V_C-V_A)...\left(i\right)$

Also,

$dV=-E.dr$

So,

${\int }_C^{A}dV=-{\int }_{2a}^{3a}\frac{\lambda }{2\pi {ϵ}_{o}r}dr$

$\Rightarrow V_A-V_C=-\frac{\lambda }{2\pi {ϵ}_o}\ln \left(3/2\right)$

$\Rightarrow V_C-V_A=\frac{\lambda }{2\pi {ϵ}_o}\ln \left(3/2\right)...\left(ii\right)$

$\Rightarrow W=\frac{q_o\lambda }{2\pi {ϵ}_o}\ln \left(3/2\right)\text{[From}\left(i\right)]$

Hence, option $\left(A\right)$ is correct.

From figure, $V_C=V_B$

From $\left(ii\right)$,

$V_C-V_A=V_B-V_A>0\Rightarrow V_B>V_A$

Therefore, option $\left(D\right)$ is also correct.