Two metal spheres of radius $R$ and $3R$ have the same surface charge density $\sigma$. If they are brought in contact and then separated, the surface charge density on the smaller and bigger sphere becomes $\sigma_1$ and $\sigma_2$, respectively. The ratio $\frac{\sigma_1}{\sigma_2}$ is:

Inside a charged hollow sphere,at any point the electric field $(E)$ and potential $(V)$ are

Two spheres of radius $a$ and $b$ respectively are charged and joined by a wire. The ratio of the electric field at the surfaces of the spheres is

$A$ hollow metal sphere of radius $R$ is charged with a charge $Q$. The electric potential and intensity inside the sphere are respectively

Two isolated metallic solid spheres of radii $R$ and $2R$ are charged such that both have the same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$,then the ratio $\frac{\sigma^{\prime}}{\sigma}$ is:

$A$ point charge $Q$ is placed inside a cavity within a solid isolated conducting sphere. Consider points $A$, $B$ and $C$, where the magnitudes of the electric fields are $E_A$, $E_B$ and $E_C$, respectively. Point $A$ is inside the material of the conductor. Points $B$ and $C$ are at the same distance from the center of the solid sphere. The correct option is: