Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium,assuming that the angle made by each thread is small,the separation $x$ between the balls is proportional to

Two positively charged spheres of masses $m_1$ and $m_2$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$. Charges on the two spheres are $q_1$ and $q_2$,respectively. At equilibrium,both strings make the same angle $\theta$ with the vertical. Then

The electrostatic force on a small sphere of charge $0.4 \; \mu C$ due to another small sphere of charge $-0.8 \; \mu C$ in air is $0.2 \; N$.
$(a)$ What is the distance between the two spheres?
$(b)$ What is the force on the second sphere due to the first?

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:

The law governing the force between electric charges is known as

Two identical spheres,each carrying a charge $Q = 10 \ \mu C$,are suspended from a single rigid support by strings of length $L = 1 \ m$. In the equilibrium position,if the angle between the two strings is $60^\circ$,what is the tension $T$ in the strings (in $N$)? (Given: $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$)