$A$ 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:

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:

Three particles of each charge $q$ are placed at the vertices of an equilateral triangle of side $L$. The work to be done to decrease the side of the triangle to $\frac{L}{2}$ is

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

Two fixed charges $A$ and $B$ of $5 \mu C$ each are separated by a distance of $6 \text{ m}$. $C$ is the midpoint of the line joining $A$ and $B$. $A$ charge $Q$ of $-5 \mu C$ is shot perpendicular to the line joining $A$ and $B$ through $C$ with a kinetic energy of $0.06 \text{ J}$. The charge $Q$ comes to rest at a point $D$. The distance $CD$ is

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: