$A$ 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:

Two point charges $100\,\mu C$ and $5\,\mu C$ are placed at points $A$ and $B$ respectively,with $AB = 40\,cm$. Calculate the work done by an external force in displacing the charge $5\,\mu C$ from $B$ to $C$,where $BC = 30\,cm$ and $\angle ABC = \frac{\pi}{2}$. Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,N m^2/C^2$.

$A$ particle of mass $100\, g$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force. (Result in $m/s$) (in $.73$)

$A$ mass of $1 \ kg$ carrying a charge of $2 \ C$ is accelerated through a potential difference of $1 \ V$. The velocity acquired by it 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: