The electric flux for Gaussian surface $A$ that encloses the charged particles in free space is (given $q_1 = -14\, nC$,$q_2 = 78.85\, nC$,$q_3 = -56\, nC$):

$A$ 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:

You are given a dipole of charge $+q$ and $-q$ separated by a distance $2R$. $A$ sphere '$A$' of radius ' $R$ ' passes through the centre of the dipole as shown below and another sphere '$B$' of radius ' $2R$ ' passes through the charge $+q$. Then the electric flux through the sphere '$A$' is

If the electric flux entering and leaving an enclosed surface respectively is $\phi_1$ and $\phi_2$, the electric charge inside the surface will be:

$A$ charge $q$ is placed inside a hollow cylinder. The electric flux associated with the curved surface $B$ is $\phi$. The flux associated with a plane surface (either $A$ or $C$) will be:

Two concentric spherical surfaces $P_1$ and $P_2$ enclose charges $\frac{Q}{2}$ and $4Q$ as shown in the figure. If $\phi_1$ and $\phi_2$ are the electric fluxes linked with the surfaces $P_1$ and $P_2$ respectively,then: