Two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is $r$. Now,the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:

Two point charges of equal magnitude but opposite sign are placed at a certain distance. The force between them is $F$. If $75\%$ of the charge from one is transferred to the other,what will be the new force between them?

Write the general equation for the Coulombian force on a charge ${q_1}$ due to a system of charges ${q_1}, {q_2}, \dots, {q_n}$.

Two point charges $A$ and $B$ with magnitudes $+4q$ and $-4q$ are placed along a line separated by a distance $r$. The force acting between them is $F$. If $25\%$ of the charge from point $A$ is transferred to point $B$,the force between the charges now becomes:

The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is $2.4 \times 10^{39}$. The ratio of the proportionality constant $K = \frac{1}{4 \pi \varepsilon_0}$ to the gravitational constant $G$ is nearly (Given that the charge of the proton and electron each $= 1.6 \times 10^{-19} \; C$,the mass of the electron $= 9.11 \times 10^{-31} \; kg$,the mass of the proton $= 1.67 \times 10^{-27} \; kg$):

Two beads,each with charge $q$ and mass $m$,are on a horizontal,frictionless,non-conducting,circular hoop of radius $R$. One of the beads is glued to the hoop at some point,while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [ $\varepsilon_0$ is the permittivity of free space.]