When a $10 \mu C$ charge is enclosed by a closed surface,the flux passing through the surface is $\phi$. Now,another $10 \mu C$ charge is placed inside the closed surface,then the flux passing through the surface is . . . . . . .

$A$ sphere of radius $R$ and charge $Q$ is placed inside a concentric imaginary sphere of radius $2R$. The flux associated with the imaginary sphere is

$A$ charge $q$ is placed at the center of one of the faces of a cube. What is the electric flux linked with the cube?

If the electric field is given by $\vec{E} = (5 \hat{i} + 4 \hat{j} + 9 \hat{k})$. The electric flux through a surface of area $20$ units lying in the $Y-Z$ plane will be (in units):

$A$ square of side $20 \ cm$ is enclosed by a spherical surface of radius $80 \ cm$. The centers of the square and the sphere are the same. Four charges $2 \times 10^{-6} \ C, -5 \times 10^{-6} \ C, -3 \times 10^{-6} \ C$,and $6 \times 10^{-6} \ C$ are placed at the four corners of the square. The total flux coming out of the spherical surface in $N \cdot m^2/C$ is:

Draw the electric field lines for a negative point charge.