$A$ bullet of mass $2\,g$ has a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$?

Charge is uniformly distributed on the surface of a hollow hemisphere. Let $O$ and $A$ be two points on the base of the hemisphere,where $O$ is the center of the base and $A$ is a point between the center and the rim. Let $V_O$ and $V_A$ be the electric potentials at $O$ and $A$ respectively. Then,

$A$ hollow metallic sphere of radius $10 \; cm$ is charged such that the potential at its surface is $80 \; V$. The potential at the centre of the sphere would be (in $; V$)

$64$ identical drops each charged up to a potential of $10 \, mV$ are combined to form a bigger drop. The potential of the bigger drop will be $.......... \, mV$.

$A$ particle of mass $m$ and charge $q$ is accelerated through a potential difference of $V$ volt. Its energy is

The variation of electrostatic potential $V$ with radial distance $r$ from the centre of a positively charged metallic thin shell of radius $R$ is given by the graph: