Q. If the electric flux entering and leaving an enclosed surface respectively are $\phi_{1}$ and $\phi_{2}$, the electric charge inside the surface will be
A) $\left( \phi_{2} - \phi_{1} \right)\varepsilon_{0}$
B) $\frac{\phi_{1} + \phi_{2}}{\varepsilon_{0}}$
C) $\frac{\phi_{1} - \phi_{2}}{\varepsilon_{0}}$
D) $\varepsilon_{0}\left( \phi_{1} - \phi_{2} \right)$
Explanation
(a) From Gauss’s law,
$\frac{charge\ enclosed\ }{\varepsilon_{0}}$
=Flux leaving the surface
$\frac{q}{\varepsilon_{0}} = \phi_{2} - \phi_{1}\ $
$\Longrightarrow q = \left( \phi_{2} - \phi_{1} \right)\varepsilon_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.