If a charge is moved against the Coulomb force of an electric field, then:

$A$ 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

$A$ charged particle $q$ is shot from a large distance with speed $v$ towards a fixed charged particle $Q$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be:

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

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

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 . . . . . . .
$(k = 9 \times 10^9 \ SI)$