$A$ charge $Q$ is enclosed by a Gaussian surface of radius $R$. If the radius is doubled,then the outward electric flux will

An electric field is given by $\vec{E} = (6 \hat{i} + 5 \hat{j} + 3 \hat{k}) \ N/C$. The electric flux through a surface area $\vec{A} = 30 \hat{i} \ m^2$ (in $SI$ units) is:

Eight dipoles of charges of magnitude $e$ are placed inside a cube. The total electric flux coming out of the cube will be

Consider a uniform electric field $\vec{E} = 3 \times 10^3 \hat{k} \text{ N C}^{-1}$. The electric flux of this field through a square of $20 \text{ cm}$ on a side whose plane is parallel to the $yz$-plane is $....... \text{ N m}^2 \text{ C}^{-1}$.

The dimensional formula of electric flux is . . . . . . .

$A$ 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 the flux through a closed Gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is the flux through a closed Gaussian surface $S_2$ whose radius is just greater than $R_1$,then: