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:

$A$ 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:

$A$ test charge $q$ is moved in the electric field of a point charge $Q$ along two different closed paths as shown in the figure. The first path has sections along and perpendicular to the lines of the electric field. The second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?

Two fixed charges $A$ and $B$ of $5 \mu C$ each are separated by a distance of $6 \text{ m}$. $C$ is the midpoint of the line joining $A$ and $B$. $A$ charge $Q$ of $-5 \mu C$ is shot perpendicular to the line joining $A$ and $B$ through $C$ with a kinetic energy of $0.06 \text{ J}$. The charge $Q$ comes to rest at a point $D$. The distance $CD$ is

$A$ 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 $(-q)$ 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