For a closed surface $\oint {\overrightarrow {E \cdot } } \,\overrightarrow {ds} \,\, = \,\,0$, then
Electric field at every point on surface is zero
Electric field at every point on surface is uniform
Electric Field at every point on surface is parallel
The number of electric field lines entering the surface will be equal to number of electric field line exit the surface
An electric field, $\overrightarrow{\mathrm{E}}=\frac{2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{\sqrt{6}}$ passes through the surface of $4 \mathrm{~m}^2$ area having unit vector $\hat{\mathrm{n}}=\left(\frac{2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right)$. The electric flux for that surface is $\mathrm{Vm}$
A charge $Q$ is situated at the comer of a cube, the electric flux passed through all the six faces of the cube is
Electric flux through a surface of area $100$ $m^2$ lying in the $xy$ plane is (in $V-m$) if $\vec E = \hat i + \sqrt 2 \hat j + \sqrt 3 \hat k$
It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss’s theorem because
A cube is placed inside an electric field, $\overrightarrow{{E}}=150\, {y}^{2}\, \hat{{j}}$. The side of the cube is $0.5 \,{m}$ and is placed in the field as shown in the given figure. The charge inside the cube is $.....\times 10^{-11} {C}$