Three point charges $+Q$,$+2Q$,and $q$ are placed at the vertices of an equilateral triangle. The value of charge $q$ in terms of $Q$,so that the electrical potential energy of the system is zero,is given by:

Charge $1.6 \times 10^{-7} \text{ C}$ is distributed uniformly over the surface of a spherical conductor of radius $R$. The ratio of the electric potential inside the spherical conductor to the electric field on the surface is . . . . . . .

Two spheres $A$ and $B$ of radius $a$ and $b$ respectively are at the same electric potential. The ratio of the surface charge densities of $A$ and $B$ is

If the electric potential of the inner shell of radius $a$ is $10\,V$ and that of the outer shell of radius $2a$ is $5\,V$,then the potential at the centre will be.....$V$

$A$ 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:

$A$ charge of $10 \,\mu C$ is placed at the origin of an $x-y$ coordinate system. The potential difference between two points $(0, a)$ and $(a, 0)$ in volts will be