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

Twenty-seven drops of water of the same size are equally and similarly charged. They are then united to form a bigger drop. By what factor will the electrical potential change?

$A$ charge of $10 \, \text{esu}$ is placed at a distance of $2 \, \text{cm}$ from a charge of $40 \, \text{esu}$ and at a distance of $4 \, \text{cm}$ from another charge of $-20 \, \text{esu}$. The potential energy of the $10 \, \text{esu}$ charge in ergs is:

If eight identical drops are joined to form a bigger drop,the potential on the bigger drop as compared to that on the smaller drop will be:

$A$ hollow metal sphere is charged with a charge of $3.2 \times 10^{-19} \ C$. If the radius of the sphere is $10 \ cm$,then the electric potential at a distance of $4 \ cm$ from its center will be:

$A$ ball of mass $1\, g$ and charge $10^{-8}\, C$ moves from a point $A$,where the potential is $600\, V$,to a point $B$,where the potential is $0\, V$. The velocity of the ball at point $B$ is $20\, cm/s$. The velocity of the ball at point $A$ will be: