Question

Infinite line charge of charge per unit length $\lambda$ is passing through one of the edges of cube of side a as shown in figure. A point charge q is placed at vertex A then.

• A. Work done to take q from A to B is zero
• B. Work done to take q from A to C is $\frac{q\lambda }{2\pi {ϵ}_o}\ln \left(\sqrt{2}\right)$
• C. Work done to take q from A to D is zero
• D. Work done to take q from A to E is $\frac{q\lambda }{4\pi {ϵ}_o}$

Answer 1

Answer: (B), (C)

The correct options are
B Work done to take q from A to C is $\frac{q\lambda }{2\pi {ϵ}_o}\ln \left(\sqrt{2}\right)$
C Work done to take q from A to D is zero

$△V=-{\int }_{r_i}^{r_f}\overrightarrow{E}.\overrightarrow{dr}=-\left[\frac{\lambda }{2\pi {ϵ}_o}ln\right(r){]}_{r_i}^{r_f}$
Work done from A to B ($r_f=a,r_i=a\sqrt{2}$)
$=q△V=-\frac{\lambda q}{2\pi {ϵ}_o}\ln (r{)}_{a\sqrt{2}}^a=\frac{\lambda q}{2\pi {ϵ}_o}\ln \left(\sqrt{2}\right)$
Work done from A to C ($r_f=a,r_i=a\sqrt{2}$)
$=q△V=-\frac{\lambda q}{2\pi {ϵ}_o}\ln (r{)}_{a\sqrt{2}}^a=\frac{\lambda q}{2\pi {ϵ}_o}\ln \left(\sqrt{2}\right)$
Work done from A to D,
$r_i=r_f$, so work done = 0
Work done from A to E ($r_f=a,r_i=a\sqrt{2}$)
$=q△V=-\frac{\lambda q}{2\pi {ϵ}_o}\ln (r{)}_{a\sqrt{2}}^a=\frac{\lambda q}{2\pi {ϵ}_o}\ln \left(\sqrt{2}\right)$