Q. A point charge of 1.8μC is at the centre of cubical Gaussian surface 55cm on edge. What is the net electric flux through the surface?
A) $1.0 \times 10^{5}N - m^{2}C^{- 1}$
B) $3.0 \times 10^{5}N - m^{2}C^{- 1}$
C) $2.0 \times 10^{5}N - m^{2}C^{- 1}$
D) $4.0 \times 10^{5}N - m^{2}C^{- 1}$
Explanation
(c) By Gauss’s law
$\varepsilon_{0}\phi = q$
$\phi = \frac{q}{\varepsilon_{0}}$
$= \frac{1.8 \times 10^{- 6}}{8.85 \times 10^{- 12}}$
$= 2.0 \times 10^{5}N - m^{2}C^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.