Five conducting parallel plates having area $A$ and separation between them $d$, are placed as shown in the figure . Plate number $2$ and $4$ are connected wire and between point $A$ and $B$, a cell of emf $E$ is connected . The charge flown through the cell is :-
$\frac{3}{4}\frac{{{\varepsilon _0}AE}}{d}$
$\frac{2}{3}\frac{{{\varepsilon _0}AE}}{d}$
$\frac{{4{\varepsilon _0}AE}}{d}$
$\frac{{{\varepsilon _0}AE}}{2d}$
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
The figure shows two parallel equipotential surfaces $A$ and $B$ kept a small distance $r$ apart from each other. $A$ point charge of $q$ coulomb is taken from the surface $A$ to $B$. The amount of net work done will be
The equivalent capacitance between $A$ and $B$ is (in $\mu\, F$)
Two spherical conductors $A$ and $B$ of radii $1\, mm$ and $2\, mm$ are separated by a distance of $5\, cm$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres $A$ and $B$ is-
A square plate of side $'a'$ is placed in $xy$ plane having centre at origin if charge density of square plate is $\sigma = xy$ then. Total charge on the plate will be.