In the figure shown, a hemispherical bowl of radius $R$ is shown. Electric field of intensity $E$ is present perpendicular to the circular cross section of the hemisphere. The electric flux through the hemisphere is

A ring of radius $R$ having a linear charge density $\lambda$ moves towards a solid imaginary sphere of radius $\frac{R}{2},$ so that the centre of ring passes through the centre of the sphere. The axis of the ring is perpendicular to the line joining the centres of the ring and the sphere. The maximum flux through the sphere in this process is

The electric field in a region is given by $\overrightarrow{E}=-4x\hat{\mathrm{i}}+6y\hat{\mathrm{j}}$. Then find the charge enclosed in the cube of side $1 \mathrm{m}$ oriented as shown in the diagram.

Eight point charges (can be assumed as small uniformly charged spheres and their centres at the corner of the cube) each having values $q$ are fixed at vertices of a cube. The electric flux through the square surface $ABCD$ of the cube is

A charge $Q$ is placed at a distance of $4R$ above the centre of a disc of radius $R.$ The magnitude of electric 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

The figure below shows a closed Gaussian surface in the shape of cube of edge length $3.0 \mathrm{m}$. There exists an electric field given by $\overrightarrow{\mathit{E}}=\left[\left(2.0x+4.0\right)\hat{\mathrm{i}}+8.0\hat{\mathrm{j}}+3.0\hat{\mathrm{k}}\right] \mathrm{N} {\mathrm{C}}^{-1}$ where $x$ is in metres in the region in which it lies. The net charge (in Coulombs) enclosed by the cube is equal to