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

Answer

D

Explanation

Let `-q_(1)` be the charge, due to which flux, `phi_(1)` is entering the surace.
`phi_(1)=(-q_(1))/(epsilon_(0)) or q_(1)=-phi_(1)epsilon_(0)`
Let `+q_(2)` be the charge, due to which flux `phi_(2)`, is leaving the surface
`phi_(2)=(q_(2))/(epsilon_(0)) or q_(2)=epsilon_(0)phi_(2)`
Electric charge inside the surface
`=q_(2)-q_(1)=epsilon_(0)phi_(2)+epsilon_(0)phi_(1)=epsilon_(0)(phi_(2)+phi_(2))`