If the electric flux entering and leaving an enclosed surface respectively are $ϕ_{1} and ϕ_{2}$ , the electric charge inside the surface will be

$\frac{ϕ_{2} - ϕ_{1}}{ε_{o}}$

$\frac{ϕ_{2} + ϕ_{1}}{ε_{o}}$

$\frac{ϕ_{1} - ϕ_{2}}{ε_{0}}$

$ε_{0} (ϕ_{1} + ϕ_{2})$

Solution

$ε_{0} (ϕ_{1} + ϕ_{2})$

According to Gauss theorem "the net electric flux through any closed surface is equal to the net charge inside the surface divided by ε_o''.

Therefore

$ϕ \propto \frac{q}{ε_{o}}$

Let-q₁ be the charge, due to which flux, is entering the surface,

$ϕ_{1} = \frac{- q_{1}}{ε_{o}}$

$- q_{1} = ε_{o} ϕ_{1}$

Let +q₂ be the charge, due to which flux $ϕ_{2}$ is entering the surface.

$ϕ_{2} = \frac{q_{2}}{ε_{0}}$

⇒ q₂ = $ε_{o} ϕ_{2}$

So, electric charge inside the surface

= $q_{2} - q_{1}$

= ${ε_{0}}_{} ϕ_{2} + ε_{0} ϕ_{1}$

= $ε_{0} (ϕ_{2} - ϕ_{1})$

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