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)`
B. `(( phi_1 + phi_2))/(varepsilon_0)`
C. `((phi_2-phi_1))/(varepsilon_0)`
D. `(phi_1+phi)varepsilon_0`

Answer

A

Explanation

`Delta oint= q/4`
`rArr phi_2 -phi_1 = q/(varepsilon_0)`
So, the eelctric charge inside the surface will be
`rArr q = (phi_2 - phi_1) varepsilon_0`.