$A$ 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

Two point charges $100\,\mu C$ and $5\,\mu C$ are placed at points $A$ and $B$ respectively,with $AB = 40\,cm$. Calculate the work done by an external force in displacing the charge $5\,\mu C$ from $B$ to $C$,where $BC = 30\,cm$ and $\angle ABC = \frac{\pi}{2}$. Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,N m^2/C^2$.

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

The distance between the plates of a capacitor is $1 \ mm$. An electric field of $5 \times 10^5 \ V/m$ is produced between the two plates. What will be the change in the potential energy $(PE)$ of an electron if it is moved from one plate to the other?

$A$ region contains a uniform electric field $E = (10 \hat{i} + 30 \hat{j}) \ Vm^{-1}$. $A$ and $B$ are two points in the field at $(1, 2, 0) \ m$ and $(2, 1, 3) \ m$,respectively. The work done when a charge of $0.8 \ C$ moves from $A$ to $B$ in a parabolic path is (in $J$)

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: