Four equal charges $q$ are placed at centre of a conducting hollow sphere. If they are displaced $1.5 \mathrm{cm}$ from the centre, the change in flux will be (Radius of sphere $>1.5 \mathrm{cm}$) :

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-.