Two point charges $q_1$ and $q_2$ are '$l$' distance apart. If one of the charges is doubled and the distance between them is halved, the magnitude of the force becomes $n$ times, where $n$ is:

Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$-axis from the origin,at which the net electric field due to these two point charges is zero is

Two identical pendulums $A$ and $B$ are suspended from the same point. The bobs are given positive charges,with $A$ having more charge than $B$. They diverge and reach equilibrium,with $A$ and $B$ making angles $\theta_1$ and $\theta_2$ with the vertical,respectively. Then:

The ratio of electrostatic and gravitational forces acting between an electron and a proton separated by a distance $5 \times 10^{-11} \, m$ will be (Charge on electron $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$,mass of proton $= 1.6 \times 10^{-27} \, kg$,$G = 6.7 \times 10^{-11} \, N m^2/kg^2$).

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:

Two identical spheres have charges $+q$ and $-q$ and are placed at a certain distance. The force between them is $F$. If a third identical sphere with charge $+q$ is placed exactly between the two spheres, what is the magnitude and direction of the force experienced by it?