Q. The inward and outward electric flux for a closed surface in units of $N - m^{2}/C$ are respectively $8 \times 10^{3}$ and $4 \times 10^{3}.$ Then the total charge inside the surface is [where $\varepsilon_{0} =$ permittivity constant]
A) $4 \times 10^{3}\ C$
B) $- 4 \times 10^{3}\ C$
C) $\frac{( - 4 \times 10^{3})}{\varepsilon}C$
D) $- 4 \times 10^{3}\varepsilon_{0}\ C$
Explanation
(d) By Gauss’s law $\phi = \frac{1}{\varepsilon_{0}}\lbrack Q_{enclosed}\rbrack$
$\Rightarrow Q_{enclosed} = \phi\varepsilon_{0} = \left\lbrack - 8 \times 10^{3} + 4 \times 10^{3} \right\rbrack\varepsilon_{0}$
$= - 4 \times 10^{3}\varepsilon_{0}\ Coulomb$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.