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-.

The fig. shows lines of constant potential in a region in which an electric field is present. The value of the potential are written in brackets of the points $A, B$ and $C,$ the magnitude of the electric field is greatest at the point -