$A$ 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:

Charges $-q, q, q$ are placed at the vertices $A, B, C$ respectively of an equilateral triangle of side $a$ as shown in the figure. If charge $-q$ is released keeping the remaining two charges fixed,then the kinetic energy of charge $-q$ at the instant when it passes through the midpoint $M$ of side $BC$ 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

Two charges $7 \ \mu C$ and $-2 \ \mu C$ are placed at $(-9, 0, 0) \ cm$ and $(9, 0, 0) \ cm$ respectively in an external field $E = \frac{A}{r^2} \hat{r}$, where $A = 9 \times 10^5 \ N/C \cdot m^2$. Considering the potential at infinity is $0$, the electrostatic energy of the configuration is . . . . . . $J$.

$A$ point charge $q$ moves from point $P$ to point $S$ along the path $PQRS$ (as shown in the figure) in a uniform electric field $E$ pointing parallel to the positive direction of the $X$-axis. The coordinates of the points $P, Q, R,$ and $S$ are $(a, b, 0), (2a, 0, 0), (a, -b, 0),$ and $(0, 0, 0)$ respectively. The work done by the field in the above process is given by the expression:

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