$A$ charged particle $q$ is shot from a large distance with speed $v$ towards a fixed charged particle $Q$. 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:

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

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 of $2 \ \mu C$ moves from point $B$ to point $C$ along the path shown in the figure. Calculate the work done in $J$.

Explain electric potential energy. Show that the sum of kinetic energy and electric potential energy remains constant.

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