Question

A thin non-conducting rod of length $50 \,cm$ 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 $1.0 \,cm$ from one-end of the rod $\left(\varepsilon_{0}=8.8 \times 10^{-12} F / m \right)$.

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

Answer 1

Answer: (B)

The given, $\lambda=10^{-12}\, C / m$,
$l=50\, cm$,
$r=1 \times 10^{-2} \,m$
and $\varepsilon_{0}=8.8 \times 10^{-12} \,F / m$
We know that, $\lambda=\frac{Q}{l}$
$10^{-12}=\frac{Q}{50\, cm }$
or $10^{-12}=\frac{2 Q}{2 \times 50\,cm }$
$10^{-12}=\frac{2 Q}{100\, cm }, $
$10^{-12}=\frac{2 Q}{1 \,m}$,
$Q=\frac{1}{2} \times 10^{-12} \,C$,
Electric potential $V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r}$
$=\frac{1}{4 \times 3.14 \times 8.8 \times 10^{-12}} \cdot \frac{\frac{1}{2} \times 10^{-12}}{1 \times 10^{-2}}$
$=\frac{100}{221.05}=0.4 \,V$