$A$ 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 $cm/s$.

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

Two point charges of $5 \ \mu C$ and $10 \ \mu C$ are separated by a distance of $1 \ m$. The work done to bring them to a distance of $0.5 \ m$ 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:

$A$ 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

Given $q_1 = +2 \times 10^{-8} \ C$ and $q_2 = -0.4 \times 10^{-8} \ C$. When a charge $q_3 = 0.2 \times 10^{-8} \ C$ is moved from $C$ to $D$,what is the change in the potential energy of $q_3$?