Question

A rod of uniform charge density $\lambda$ and length L is placed vertically as shown. A point charge $Q_0$ is placed at point O at a distance L from its one end. Potential energy of charge $Q_0$ is equal to

• A. $\frac{\lambda Q_{0}ln2}{4\pi {ϵ}_0}$
• B. $\frac{\lambda Q_0}{4\pi {ϵ}_0}$
• C. $\frac{\lambda Q_0}{2\pi {ϵ}_0}$
• D. $\frac{\lambda Q_{0}ln2}{2\pi {ϵ}_0}$

Answer 1

The correct option is A $\frac{\lambda Q_{0}ln2}{4\pi {ϵ}_0}$

$V_0=\frac{\left(\lambda L\right)ln2}{4\pi {ϵ}_{0}L}=\left(\frac{\lambda ln2}{4\pi {ϵ}_0}\right)$
$U=\frac{\lambda Q_{0}ln2}{4\pi {ϵ}_0}$