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

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

If a charge is moved against the Coulomb force of an electric field, then:

$A$ 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

An electron is released from a distance of $4 \ m$ from a stationary point charge $20 \ nC$. What will be the speed of the electron when it is $2 \ m$ away from the point charge?
[Charge of electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \ m^2 \ C^{-2}$]

$A$ 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: