If the electric potential at a distance of $5 \, cm$ from the center of a charged conducting sphere of radius $10 \, cm$ is $V$,then what will be the potential at a point at a distance of $15 \, cm$ from the center?

Eight charges,each of magnitude $q$,are placed at the vertices of a cube placed in a vacuum. The electric potential at the centre of the cube due to this system of charges is . . . . . . . ($\varepsilon_0 = $ permittivity of vacuum,$a = $ length of each side of the cube.)

$A$ charge $Q$ is distributed over two concentric conducting thin spherical shells of radii $r$ and $R$ $(R > r)$. If the surface charge densities on the two shells are equal,the electric potential at the common centre is

Electric charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively,which are at a distance of $2L$ apart. If $C$ is the midpoint between $A$ and $B$,then the work done in moving a charge $+Q$ along the semi-circle $CRD$ is

$64$ drops of mercury,each charged to a potential of $10\,V$,are combined to form one bigger drop. The potential of this bigger drop will be.......$V$ (Assume all the drops to be spherical).

Two point charges $-q$ and $+q$ are located at points $(0,0,-a)$ and $(0,0, a)$ respectively. The electric potential at a point $(0,0, z)$,where $z>a$ is