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 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)

${10}^{-9}C$

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
$q^{\prime }=4\pi {\epsilon }_0{V}_r=\frac{1}{9\times {10}^9}\times 12\times 0.5=0.667\times {10}^{-9}C={10}^{-9}C$