Question

If the electric flux entering and leaving a closed surface are $6\times {10}^6$ and $9\times {10}^{6}SI$ units respectively, then the charge inside the surface of permittivity of free space ${\omega }_0$ is

• A. ${\omega }_0\times {10}^6$
• B. $-{\omega }_0\times {10}^6$
• C. $-2{\omega }_0\times {10}^6$
• D. $3{\omega }_0\times {10}^6$
• E. $2{\omega }_0\times {10}^6$

Answer 1

The correct option is D $3{\omega }_0\times {10}^6$

The net electric flux through the closed surface is $\varphi ={\varphi }_l-{\varphi }_e=(9\times {10}^6)-(6\times {10}^6)=3\times {10}^6$

By Gauss's law, electric flux through a closed surface is $\varphi =\frac{Q_{in}}{{\epsilon }_0}$ where $Q_{in}=$ charge inside the surface and ${\epsilon }_0=$ permittivity of free space.

Thus, $Q_{in}=\varphi {\omega }_0=3{\omega }_0\times {10}^6$