$A$ charge of $4\,\mu C$ is to be divided into two parts. The distance between the two divided charges is constant. What should be the magnitude of the divided charges so that the electrostatic force between them is maximum?

Two point charges $q_2 = 3 \times 10^{-6} \ C$ and $q_1 = 5 \times 10^{-6} \ C$ are located at $B(3, 5, 1) \ m$ and $A(1, 3, 2) \ m$ respectively. Find the magnitude of the force on $q_1$ due to $q_2$.

$A$ total charge $Q$ is broken into two parts $Q_1$ and $Q_2$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur when:

$A$ 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?

Three charges $q$,$Q$,and $4q$ are placed in a straight line of length $l$ at points distant $0$,$\frac{l}{2}$,and $l$ respectively from one end. In order to make the net force on $q$ zero,the charge $Q$ must be equal to

Two point charges $+q_1$ and $+q_2$ repel each other with a force of $100 \ N$. $q_1$ is increased by $10 \%$ and $q_2$ is decreased by $10 \%$. If they are kept at their original positions, the change in the force of repulsion between them is: