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

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 $q_3$?

$A$ particle $A$ has charge $+q$ and a particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference,the ratio of their speeds $\frac{v_A}{v_B}$ will be:

$A$ 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 $J$)

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:

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?