Two point charges $A$ and $B$,having charges $+Q$ and $-Q$ respectively,are placed at a certain distance apart and the force acting between them is $F$. If $25 \%$ of the charge of $A$ is transferred to $B$,then the force between the charges becomes

$A$ particle of mass $1 \, mg$ and charge $q$ is placed at the midpoint of two stationary particles,each carrying the same charge $q$,kept at a distance of $2 \, m$. If the free charged particle is displaced from its equilibrium position by a small distance $x$ $(x \ll 1 \, m)$,it executes $SHM$. Its angular frequency of oscillation will be $.... \times 10^{8} \, rad/s$ if $q^{2} = 10 \, C^{2}$.

$A$ charge $q_1$ exerts some force on a second charge $q_2$. If a third charge $q_3$ is brought near,the force of $q_1$ exerted on $q_2$:

$A$ 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 each of the opposite corners of a square. $A$ charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero,then $Q/q$ equals:

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 $m$?