$A$ 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:

The distance between the plates of a capacitor is $1 \ mm$. An electric field of $5 \times 10^5 \ V/m$ is produced between the two plates. What will be the change in the potential energy $(PE)$ of an electron if it is moved from one plate to the other?

Obtain the equation for the electric potential energy of a single charge in an external electric field.

$A$ point charge $2 \times 10^{-2} \, C$ is moved from $P$ to $S$ in a uniform electric field of $30 \, N C^{-1}$ directed along the positive $x$-axis. If the coordinates of $P$ and $S$ are $(1, 2, 0) \, m$ and $(0, 0, 0) \, m$ respectively,the work done by the electric field will be $......... \, mJ$.

$A$ 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:

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 $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,N m^2/C^2$.