$A$ ball of mass $1\, g$ and charge $10^{-8}\ C$ moves from point $A$ at a potential of $600\, V$ to point $B$ at a potential of $0\, V$. If the velocity of the ball at point $B$ is $20\, cm/s$,then the velocity of the ball at point $A$ is . . . . . . $cm/s$.

$A$ charge of $2 \ \mu C$ moves from point $B$ to point $C$ along the path shown in the figure. Calculate the work done in $J$.

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:

$A$ 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:

$A$ point charge of $10^{-8} \text{ C}$ is placed at the origin. The work done in moving a point charge of $2 \mu\text{C}$ from point $A(4, 4, 2) \text{ m}$ to $B(2, 2, 1) \text{ m}$ is . . . . . . $\text{J}$.

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