Question

A thin rod placed long x-axis from x=-a to x=+a, the rod carries a change uniformly distributed along its length with linear charge density $\lambda$. The potential at the point P (2a, 0, 0) will be

• A. $\frac{\lambda }{\pi {ϵ}_0}In3$
• B. $\frac{\lambda }{4\pi {ϵ}_0}In2$
• C. $\frac{\lambda }{4\pi {ϵ}_0}In3$
• D. $\frac{\lambda }{\pi {ϵ}_0}In2$

Answer 1

The correct option is C $\frac{\lambda }{4\pi {ϵ}_0}In3$

$x=-a$ to $x=+a$

charge density $=\lambda$

Potential at the point $P(2a,0,0)$

linear charge density $\lambda$

$\lambda =\frac{Q}{2a}$

To find the $dE$, we will need

$dQ=\lambda dy$

$=\frac{Q}{4\pi {ϵ}_0}dy=\frac{Q_a}{4\pi }=0$

$dE=\frac{\lambda }{4\pi {ϵ}_0}ln3$

$=\frac{\lambda }{4\pi {ϵ}_0}ln3=2a+a=3a$