Question

A thin non-conducting rod of length $50cm$ has a positive charge of uniform linear density $10^{-12}C/m$. Find the electric potential due to the rod at a point, which is at a perpendicular distance of $1.0cm$ from one-end of the rod $\left({\epsilon }_0=8.8\times 10^{-12}F/m\right)$.

• A. 0.02 V
• B. 0.04 V
• C. 0.06 V
• D. 1.02 V

Answer 1

Answer: (B)

The given, $\lambda =10^{-12}C/m$,
$l=50cm$,
$r=1\times 10^{-2}m$
and ${\epsilon }_0=8.8\times 10^{-12}F/m$
We know that, $\lambda =\frac{Q}{l}$
$10^{-12}=\frac{Q}{50cm}$
or $10^{-12}=\frac{2Q}{2\times 50cm}$
$10^{-12}=\frac{2Q}{100cm},$
$10^{-12}=\frac{2Q}{1m}$,
$Q=\frac{1}{2}\times 10^{-12}C$,
Electric potential $V=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q}{r}$
$=\frac{1}{4\times 3.14\times 8.8\times 10^{-12}}\cdot \frac{\frac{1}{2}\times 10^{-12}}{1\times 10^{-2}}$
$=\frac{100}{221.05}=0.4V$