A positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is flux through closed gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is flux through closed gaussian surface $S_2$ whose radius is just greater than $R_1$ then:-
$\phi_1 > \phi_2$
$\phi_2 > \phi_1$
$\phi_1 = \phi_2 = \frac {q}{\varepsilon_0}$
$\phi_1 = \phi_2 = \frac {kq}{\varepsilon_0}$
A square surface of side $L$ metres is in the plane of the paper. A uniform electric field $\vec E(V/m) $, 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
In a cuboid of dimension $2 L \times 2 L \times L$, a charge $q$ is placed at the centre of the surface ' $S$ ' having area of $4 L ^2$. The flux through the opposite surface to ' $S$ ' is given by
A charge $Q$ is enclosed by a Gaussian spherical surface of radius $R$. If the radius is doubled, then the outward electric flux will
The spatial distribution of the electric field due to charges $(A, B)$ is shown in figure. Which one of the following statements is correct
How much electric flux will come out through a surface $S = 10\hat j$ kept in an electrostatic field $\vec E = 2\hat i + 4\hat j + 7\hat k$.........$units$