Two point charges $+Q$ and $+q$ repel each other with a force of $100 \,N$. Keeping the distance between them unchanged, if $Q$ is increased by $10 \%$ and $q$ is decreased by $10 \%$, the force of repulsion between them will:

$A$ 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 point charges are placed at the corners of an equilateral triangle with sides of $10 \ cm$ as shown in the figure. The resultant force on the charge at $B$ will be:

The charges on two spheres are $+7\,\mu C$ and $-5\,\mu C$ respectively. They experience a force $F$. If each of them is given an additional charge of $-2\,\mu C$,the new force of attraction will be

Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and the distance between them is doubled,then the new force acting on each charge is

$A$ certain charge $Q$ is divided into two parts $q$ and $(Q-q)$. How should the charges $Q$ and $q$ be divided so that $q$ and $(Q-q)$ placed at a certain distance apart experience maximum electrostatic repulsion?