Two uniformly charged spherical conductors $A$ and $B$ of radii $5 \ mm$ and $10 \ mm$ are separated by a distance of $2 \ cm$. If the spheres are connected by a conducting wire,then in equilibrium condition,the ratio of the magnitudes of the electric fields at the surface of the sphere $A$ and $B$ will be.

Consider three concentric metallic spheres $A, B$ and $C$ of radii $a, b, c$ respectively,where $a < b < c$. $A$ and $B$ are connected,whereas $C$ is grounded. The potential of the middle sphere $B$ is raised to $V$. Then the charge on the sphere $C$ is

Two conducting spheres $A$ and $B$ have radii $1 \, mm$ and $2 \, mm$ respectively and are charged. They are placed at a distance of $5 \, cm$. When they are connected by a conducting wire,the ratio of the electric field intensities on their surfaces at equilibrium is:

$A$ conducting sphere of radius $10\, cm$ is charged with $10\,\mu C$. Another uncharged sphere of radius $20\, cm$ is allowed to touch it for some time. After the spheres are separated,the ratio of the surface charge densities on the spheres will be:

Five identical conducting plates,each of face area $A$,are placed parallel to each other with a separation of $d$ between two adjacent plates as shown in the figure. The plates $M$ and $N$ are given charges $Q_1$ and $Q_2$ respectively and the remaining plates are neutral. If the outermost plates are grounded,the potential difference between the plates $M$ and $N$ is . . . . . .

Assertion : In a cavity within a conductor,the electric field is zero.
Reason : Charges in a conductor reside only at its surface.