Charges $5 \mu C$ and $10 \mu C$ are placed $1 \ m$ apart. Work done to bring these charges at a distance $0.5 \ m$ from each other is . . . . . . .
$(k = 9 \times 10^9 \ SI)$

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

$A$ charge $(-q)$ and another charge $(+Q)$ are kept at two points $A$ and $B$ respectively. Keeping the charge $(+Q)$ fixed at $B$,the charge $(-q)$ at $A$ is moved to another point $C$ such that $ABC$ forms an equilateral triangle of side $\ell$. The net work done in moving the charge $(-q)$ is

Obtain the equation for the electric potential energy of a system of two electric charges in an external electric field.

Charges $-q, q, q$ are placed at the vertices $A, B, C$ respectively of an equilateral triangle of side $a$ as shown in the figure. If charge $-q$ is released keeping the remaining two charges fixed,then the kinetic energy of charge $-q$ at the instant when it passes through the midpoint $M$ of side $BC$ is:

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