Q. The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11}$ volts. The electric field at the point is
A) $4\pi \in_{0}\ Q \times 10^{20}V/m$
B) $12\pi \in_{0}\ Q \times 10^{22}V/m$
C) $4\pi \in_{0}\ Q \times 10^{22}V/m$
D) $12\pi \in_{0}\ Q \times 10^{20}V/m$
Explanation
(c) Electric potential at a distance $r$ due to point charge $Q$ is $V = \frac{kQ}{r}$ and electric field at the same point $E = \frac{kQ}{r^{2}}$
$\Rightarrow E = \frac{kQ}{(kQ/V)^{2}} = \frac{V^{2}}{kQ} = \frac{\left( Q \times 10^{11} \right)^{2}}{kQ} = 4\pi\varepsilon_{0}\ Q \times 10^{22}V/m$
$\left\lbrack \because k = \frac{1}{4\pi\varepsilon_{0}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.