Two electrons are moving towards each other,each with a velocity of $10^6 \, m/s$. What will be the closest distance of approach between them in $m$?

Three equal charges $q_1, q_2$ and $q_3$ are placed on the three corners of a square of side $a$. If the force between $q_1$ and $q_2$ is $F_{12}$ and that between $q_1$ and $q_3$ is $F_{13}$,then the ratio of magnitudes $\left(\frac{F_{12}}{F_{13}}\right)$ is

$A$ paisa coin is made up of $Al-Mg$ alloy and weighs $0.75\,g$. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8\,kC$. Suppose that these equal charges were concentrated in two point charges separated by:
$(i)$ $1\,cm$ $(\sim \frac{1}{2} \times \text{diagonal of the one paisa coin})$
$(ii)$ $100\,m$ $(\sim \text{length of a long building})$
$(iii)$ $10^6\,m$ $(\text{radius of the earth})$.
Find the force on each such point charge in each of the three cases. What do you conclude from these results?

$A$ charge $Q$ is placed at the midpoint of the line joining two identical point charges $q$. For what value of $q$ will the system be in equilibrium?

Two point charges $+9e$ and $+e$ are placed at a distance of $16\, cm$ from each other. Where should a third charge $q$ be placed between them so that the system remains in equilibrium?

Two point charges $+10 \mu C$ and $+4 \mu C$ are placed $10 \ cm$ apart in air. The work required to be done to bring them $2 \ cm$ closer is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right)$. (in $J$)