$A$ negatively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge:

$A$ 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 $cm/s$.

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:

$A$ 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

$A$ charged particle $q$ is shot from a large distance with speed $v$ towards a fixed charged particle $Q$. 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:

$A$ 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: