Two point charges of $12\ \mu C$ and $8\ \mu C$ are placed $10\ cm$ apart. The work done in bringing them $4\ cm$ closer is:

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

$A$ particle of mass $1 \text{ g}$ and charge $1 \mu\text{C}$ is held at rest on a frictionless horizontal surface at a distance of $1 \text{ m}$ from a fixed charge of $2 \text{ mC}$. If the particle is released, it will be repelled. What is the speed of the particle when it is at a distance of $10 \text{ m}$ from the fixed charge (in $\text{ m s}^{-1}$)?

Two point charges of $5 \ \mu C$ and $10 \ \mu C$ are separated by a distance of $1 \ m$. The work done to bring them to a distance of $0.5 \ m$ is:

The displacement of a charge $Q$ in the electric field $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$ is $\vec{r} = a \hat{i} + b \hat{j}$. The work done by the electric field is:

$A$ 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: