$A$ 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 . . . . . . $cm/s$.

$A$ 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 . . . . . . $\text{J}$.

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?

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:

$A$ 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:

$A$ charge $q$ is fired towards another charged particle $Q$,which is fixed,with a speed $v$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be