Four closed surfaces and corresponding charge distributions are shown below Let the respective electric fluxes through the surfaces be \({\phi _1},{\phi _2},{\phi _3}\) and \({\phi _4}\) . Then

For surface \(S_{1}, \phi_{1}=\frac{1}{\varepsilon_{0}}(2 q)\)

For surface

\(\mathrm{S}_{2}, \phi_{2}=\frac{1}{\varepsilon_{0}}(\mathrm{q}+\mathrm{q}+\mathrm{q}-\mathrm{q})=\frac{1}{\varepsilon_{0}} 2 \mathrm{q}\)

For surface \(S_{3}, \phi_{3}=\frac{1}{\varepsilon_{0}}(q+q)=\frac{1}{\varepsilon_{0}}(2 q)\)

For surface

\(S_{4}, \phi_{4}=\frac{1}{\varepsilon_{0}}(8 q-2 q-4 q)=\frac{1}{\varepsilon_{0}}(2 q)\)

Hence, \(\phi_{1}=\phi_{2}=\phi_{3}=\phi_{4}\) i.e. netelectric flux is same for all surfaces.

Keep in mind, the electric field due to a charge outside \(\left(\mathrm{S}_{3} \text { and } \mathrm{S}_{4}\right),\) the Gaussian surface contributes zero net flux through the surface, because as many lines due to that charge enter the surface as leave it.