A cubical box contains charge $+Q$ at its centre. The total electric flux emerging from the box is:

If a square coil is making an angle $60°$ with electric field $E$ according to figure, the electric flux passing through the square coil is (the side of square is $4 \mathrm{cm}$):

A square surface of side $L \mathrm{m}$ is in the plane of the paper. A uniform electric field $\overrightarrow{E }\left(\mathrm{V} {\mathrm{m}}^{-1}\right),$ also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is:

A charge $Q$ is placed at a distance of $4R$ above the centre of a disc of radius $R.$ The magnitude of flux through the disc is $\varphi .$ Now a hemispherical shell of radius $R$ is placed over the disc such that it forms a closed surface. The flux through the curved surface (taking the direction of area vector along outward normal as positive), is -

Gauss law is given by ${\in }_0\oint \overrightarrow{E}.\overrightarrow{ds}=q,$ if net charge enclosed by Gaussian surface is zero then-.