The work done in carrying a charge of $5\,\mu C$ from a point $A$ to a point $B$ in an electric field is $10\,mJ$. The potential difference $(V_B - V_A)$ is then

Two point charges $q_1 = 4 \times 10^{-8} \ C$ and $q_2 = -6 \times 10^{-8} \ C$ are placed at points $A$ and $B$ respectively,which are $50 \ cm$ apart. At what distance from point $A$ on the line $AB$ is the electric potential zero (in $cm$)?

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

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:

$A$ 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:

Two concentric hollow spherical conductors of radii $R$ and $2R$ have charges $Q$ and $-Q$ respectively. Find the electric potential at a distance of $\frac{3R}{2}$ from the centre. $\left[K=\frac{1}{4\pi\varepsilon_0}\right]$