$A$ 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:

$A$ charge $(-q)$ and another charge $(+Q)$ are kept at two points $A$ and $B$ respectively. Keeping the charge $(+Q)$ fixed at $B$,the charge $(-q)$ at $A$ is moved to another point $C$ such that $ABC$ forms an equilateral triangle of side $\ell$. The net work done in moving the charge $(-q)$ is

$A$ region contains a uniform electric field $E = (10 \hat{i} + 30 \hat{j}) \ Vm^{-1}$. $A$ and $B$ are two points in the field at $(1, 2, 0) \ m$ and $(2, 1, 3) \ m$,respectively. The work done when a charge of $0.8 \ C$ moves from $A$ to $B$ in a parabolic path is (in $J$)

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:

$A$ ball of mass $1 \, g$ and charge $10^{-8} \, C$ moves from point $A$ $(V_A = 600 \, V)$ to point $B$ where the potential is zero. The velocity of the ball at point $B$ is $20 \, cm/s$. Find the velocity of the ball at point $A$ in $cm/s$.