$A$ 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

$A$ 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:

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 $q_3$?

Obtain the equation for the electric potential energy of a system of two electric charges in an external electric field.

$A$ 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: