Two particles of equal mass m and charge q are placed at a distance of $16cm$. They do not experience any force. The value of $q/m$ is:
(A) 1
(B) $\sqrt {\pi {\varepsilon _0}/G} $
(C) $\sqrt {G/4\pi {\varepsilon _0}} $
(D) $\sqrt {4\pi {\varepsilon _0}G} $

Hint It is given that the charges are of equal mass and charge, and they do not experience any force. The only forces acting between the particles are the gravitational and electrostatic forces. Since they have similar charges, the electrostatic force between them would be repulsive. We know that the gravitational force is attractive in nature, both of these forces must be equal in magnitude.
Formula used:
${F_E} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
${F_G} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Here, ${F_E}$ is the electrostatic force between two particles.
${F_G}$ is the gravitational force between two particles.
$G$ is the universal gravitational constant.
${\varepsilon _0}$ is the permittivity of free space.
${q_1}$ and${q_2}$ are the charges of both particles.
${m_1}$ and ${m_2}$ are the masses of both particles.
$r$ is the distance between the center of mass of both particles.

Complete Step by step solution
The difference between the repulsive electrostatic force between the two particles and the attractive gravitational force between them should give the net force between the charges.
${F_G} - {F_E} = {F_{net}}$
It is given in the question that the mass and charge of both particles are same. And the net force is zero, the electrostatic force must be equal to the gravitational force.
${F_G} = {F_E}$
The force of gravity between the particles is given as-
${F_G} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
We know that,
${m_1} = {m_2}$
And $r = 0.16m$ ($1$ centimetre $ = 0.01$meter )
Therefore,
${F_G} = \dfrac{{G{m^2}}}{{{{(0.16)}^2}}}$
The electric force between the particles is given by,
${F_E} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
We know that,
${q_1} = {q_2}$
And $r = 0.16m$
Therefore,
${F_E} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q^2}}}{{{{(0.16)}^2}}}$
Since, ${F_G} = {F_E}$
$\dfrac{{G{m^2}}}{{{{(0.16)}^2}}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q^2}}}{{{{(0.16)}^2}}}$
Bringing $q$ and $m$on the same side,
${\left( {\dfrac{q}{m}} \right)^2} = G \times 4\pi {\varepsilon _0}$
$\dfrac{q}{m} = \sqrt {4\pi {\varepsilon _0}G} $

Therefore, option (D) is correct.

Note In this question, the distance between the particles is not important to calculate the charge to mass ratio. The electrostatic force and gravitational force are both inversely proportional to the square of the distance between the particles, this is why the term containing the distance between them is different.