Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is

Three charges each of magnitude $q$ are placed at the corners of an equilateral triangle. The electrostatic force on the charge $Q$ placed at the center is (each side of the triangle is $L$):

$(a)$ In a quark model of elementary particles,a neutron is made of one up quark [ charge $\frac{2}{3}e$ ] and two down quarks [ charges $-\frac{1}{3}e$ ]. Assume that they have a triangle configuration with side length of the order of ${10^{ - 15}} \ m$. Calculate the electrostatic potential energy of the neutron and compare it with its mass $939 \ MeV$. $(b)$ Repeat the above exercise for a proton which is made of two up and one down quark.

Two spheres of radii $3 \ cm$ and $1 \ cm$ are placed at a distance of $10 \ cm$ from each other in a vacuum. If each sphere is charged to a potential of $10 \ V$,the force of repulsion between them is:

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 identical tennis balls each having mass $m$ and charge $q$ are suspended from a fixed point by threads of length $l$. What is the equilibrium separation when each thread makes a small angle $\theta$ with the vertical?