If the electric flux entering and leaving an enclosed surface respectively is `phi_1` and `phi_2`, the electric charge inside the surface will be

Options

A. `(phi_(2)-phi_(1))/(epsilon_(0))`
B. `(phi_(2)+phi_(1))/(epsilon_(0))`
C. `(phi_(1)-phi_(2))/(epsilon_(0))`
D. `epsilon_(0)(phi_(2)-phi_(1))`

Answer

D

Explanation

According to gauss theorem, the net electric flux through any closed surface is equal to the net charge inside the surface divided by `epsilon_(0)`
Therefore `phi=q/ (epsilon_(0))`
Let `q_(1)` be the charge due to which flux `f_(1)` id entering the surface
`phi_(1)=(-q_(1))/(epsilon_(0))` or `-q_(1)=epsilon_(0)phi_(1)`
Let `+q_(2)` be the charge due to which flux `f_(2)` is leaving the surface
`phi_(2)=(q_(2))/(epsilon_(0))` or `q_(2)=epsilon_(0) phi_(2)`
So electric charge inside the surface
`=q_(2)-q_(10)=epsilon_(0)phi_(2)+epsilon_(0)phi_(1)=epsilon_(0)(phi_(2)-phi_(1))`