$A$ charge of $2 \ \mu C$ moves from point $B$ to point $C$ along the path shown in the figure. Calculate the work done in $J$.

$A$ particle $A$ has charge $+q$ and a particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference,the ratio of their speeds $\frac{v_A}{v_B}$ will be:

$A$ region contains a uniform electric field $E = (10 \hat{i} + 30 \hat{j}) \ Vm^{-1}$. $A$ and $B$ are two points in the field at $(1, 2, 0) \ m$ and $(2, 1, 3) \ m$,respectively. The work done when a charge of $0.8 \ C$ moves from $A$ to $B$ in a parabolic path is (in $J$)

Explain electric potential energy. Show that the sum of kinetic energy and electric potential energy remains constant.

Two point charges $A = +3 \text{ nC}$ and $B = +1 \text{ nC}$ are placed $5 \text{ cm}$ apart in air. The work done to move charge $B$ towards $A$ by $1 \text{ cm}$ is

$A$ negatively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge: