Question

The arc AB with the centre C and the infinitely long wire having linear charge density $\lambda$ are lying in the same plane. The minimum amount of work to be done to move a point charge $q_0$ from point A to B through a circular path AB of radius $a$ is equal to:

• A. $\frac{q_0^2}{2\pi {\epsilon }_0}\log \left(\frac{2}{3}\right)$
• B. $\frac{q_0\lambda }{2\pi {\epsilon }_0}\log \left(\frac{3}{2}\right)$
• C. $\frac{q_0\lambda }{2\pi {\epsilon }_0}\log \left(\frac{2}{3}\right)$
• D. $\frac{q_0\lambda }{\sqrt{2}\pi {\epsilon }_0}$

Answer 1

The correct option is B $\frac{q_0\lambda }{2\pi {\epsilon }_0}\log \left(\frac{3}{2}\right)$

Given :- $CA=a$ distance potential at any point $x$ with respect to any other point $y$ is given by,

$U_x-U_y=\frac{\lambda }{2\pi {\epsilon }_0}ln\left(\frac{x}{y}\right)$

In this case $x\to A\&y\to B$

$\Rightarrow U_A-U_B=\frac{\lambda }{2\pi {\epsilon }_0}ln\left[\frac{3a}{2a}\right]$

$=\frac{\lambda }{2\pi {\epsilon }_0}ln\left[\frac{3}{2}\right]$

Now, work done in moving charge from $A$ to $B$ will be given by,

$w=q_0(U_A-U_B)$

$=\frac{q_0\lambda }{2\pi {\epsilon }_0}ln\left[\frac{3}{2}\right]$

$\Rightarrow$ Option (B) is the correct option.