Question

The electric field intensity at a point $P$ due to point charge $q$ kept at point $Q$ is $24N{C}^{-1}$ and the electric potential at point $P$ due to same charge is $12J{C}^{-1}$. The order of magnitude of charge $q$ is

• A. $10^{-6}C$
• B. $10^{-7}C$
• C. $10^{-10}C$
• D. $10^{-9}C$

Answer 1

Answer: (D)

Electric field of a point charge,
$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}=24N{C}^{-1}$
Electric potential of a point charge,
$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}=12J{C}^{-1}$
The distance $PQ$ is
$r=\frac{V}{E}=\frac{12}{24}=0.5m$
$\therefore$ Magnitude of charge
$q^{\prime }=4\pi {\epsilon }_{0}Vr$
$=\frac{1}{9\times 10^9}\times 12\times 0.5$
$=0.667\times 10^{-9}C$
$\approx 10^{-9}C$