Two point charges of $12\ \mu C$ and $8\ \mu C$ are placed $10\ cm$ apart. The work done in bringing them $4\ cm$ closer 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:

Two positive point charges of $12\,\mu C$ and $8\,\mu C$ are $10\,cm$ apart. The work done in bringing them $4\,cm$ closer is

$A$ 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?

$A$ particle of mass $1 \text{ g}$ and charge $1 \mu\text{C}$ is held at rest on a frictionless horizontal surface at a distance of $1 \text{ m}$ from a fixed charge of $2 \text{ mC}$. If the particle is released, it will be repelled. What is the speed of the particle when it is at a distance of $10 \text{ m}$ from the fixed charge (in $\text{ m s}^{-1}$)?

$A$ 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: