Question

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

Answer 1

Answer: (D)

$- 4 \times 10^{3}\varepsilon_{0}$C

Answer 2

Sol. By Gauss’s law$\varphi = \frac{1}{\varepsilon_{0}}$ (Q_enclosed)

⇒ $Q_{enclosed} = \varphi\varepsilon_{0} = ( - 8 \times 10^{3} + 4 \times 10^{3})\varepsilon_{0}$

$= - 4\mspace{6mu} \times \mspace{6mu} 10^{3}\varepsilon_{0}$ Coulomb.