The radius of a soap bubble whose potential is $16\,V$ is doubled. The new potential of the bubble will be.....$V$

In the following diagram,the work done in moving a point charge from point $P$ to points $A, B$ and $C$ are $W_A, W_B$ and $W_C$ respectively. Then ($A, B, C$ are points on a semicircle and point charge $q$ is at the centre of the semicircle):

$A$ charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ $(R > r)$ such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

$A$ spherical charged conductor has surface charge density $\sigma$. The electric field on its surface is $E$ and the electric potential of the conductor is $V$. Now,the radius of the sphere is halved while keeping the charge constant. The new values of the electric field and potential would be:

The electric potential at the centre of two concentric half rings of radii $R_1$ and $R_2$,having same linear charge density $\lambda$ is

Two hollow conducting spheres of radii $R_{1}$ and $R_{2}$ $(R_{1} >> R_{2})$ have equal charges. The potential would be: