Question

An infinite line of charge $\lambda$ per unit length is placed along the y-axis. The work done in moving a charge q from A(a, 0) to B(2a,0) is :

• A. $\frac{q\lambda }{2\pi {\epsilon }_0}ln2$
• B. $\frac{q\lambda }{2\pi {\epsilon }_0}ln\left(\frac{1}{2}\right)$
• C. $\frac{q\lambda }{4\pi {\epsilon }_0}ln\sqrt{2}$
• D. $\frac{q\lambda }{4\pi {\epsilon }_0}ln2$

Answer 1

The correct option is B $\frac{q\lambda }{2\pi {\epsilon }_0}ln\left(\frac{1}{2}\right)$

Potential due to a line charge at a distance r is given by:

$V-V_o=\frac{\lambda }{2\pi {\epsilon }_o}ln\left(\frac{r}{r_o}\right)$

Hence, potential difference between A and B is:

$V_A-V_B=\frac{\lambda }{2\pi {\epsilon }_o}ln\left(\frac{1}{2}\right)$

Change in potential energy of the charge on moving from A to B is:

$U_B-U_A=q(V_B-V_A)$

Work done is negative of change in potential energy.

$W=U_A-U_B=q(V_A-V_B)$

$W=\frac{q\lambda }{2\pi {\epsilon }_o}ln\left(\frac{1}{2}\right)$