In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a p.d. \(2400 \mathrm{v}\) between the plates. To keep a drop of half the radius stationary the potential difference had to be made $600 \mathrm{v}$. What is the charge on the second drop? (A) \([(3 Q) / 2]\) (B) \((\mathrm{Q} / 4)\) (C) \(Q\) (D) \((\mathrm{Q} / 2)\)

The electric force acting on an oil drop can be found using Coulomb's law: \(F_e = E \cdot Q\), where \(E\) is the electric field between the plates, and \(Q\) is the charge on the oil drop. For parallel plates, the electric field's magnitude is equal to the potential difference (\(V\)) divided by the distance between the plates (\(d\)): \(E=V/d\). The gravitational force acting on the oil drop can be found using the formula: \(F_g = m \cdot g\), where \(m\) is the mass of the oil drop, and \(g\) is the acceleration due to gravity (approximately \(9.8 \frac{\mathrm{m}}{\mathrm{s^2}}\)). For each oil drop, we know that the electric force and the gravitational force are balanced. That is, the forces are equal in magnitude but in opposite directions: \(F_e = F_g\).

Using the formula for the electric force acting on the first oil drop carrying a charge of \(Q\) and held stationary by a potential difference of \(2400 \mathrm{v}\), we have: \(F_{e1} = E_1 \cdot Q\) Additionally, for the gravitational force acting on the first oil drop, we can write \(F_g = m_1 \cdot g\), where \(m_1\) is the mass of the first drop. Then, since the first oil drop is held stationary, \(E_1 \cdot Q = m_1 \cdot g\)

Now, we need to find the relationship between the electric force and gravitational force for the second oil drop, with half the radius and held stationary by a potential difference of \(600 \mathrm{v}\): \(F_{e2} = E_2 \cdot Q_2\) The gravitational force acting on the smaller oil drop can be written as \(F_{g2} = m_2 \cdot g\), where \(m_2\) is the mass of the second drop. Since the drop is held stationary, \(E_2 \cdot Q_2 = m_2 \cdot g\)

For a spherical object, the mass (\(m\)) is related to its density (\(\rho\)) and volume (\(V\)) as follows: \(m = \rho \cdot V\) The volume of a sphere is given by: \(V = \frac{4}{3}\pi r^3\)

Now, let's substitute these relationships into the equations for the first and second oil drops: \(E_1 \cdot Q = \rho \cdot \frac{4}{3}\pi r_1^3 \cdot g\) \(E_2 \cdot Q_2 = \rho \cdot \frac{4}{3}\pi r_2^3 \cdot g\) We know that the radius of the first oil drop is twice the radius of the second oil drop: \(r_1 = 2r_2\). Let's substitute the electric fields, which are related to the potential differences, into the equations: \(\frac{2400}{d} \cdot Q = \rho \cdot \frac{4}{3}\pi (2r_2)^3 \cdot g\) \(\frac{600}{d} \cdot Q_2 = \rho \cdot \frac{4}{3}\pi r_2^3 \cdot g\) The distance between the plates is the same for both drops, so the denominator (\(d\)) cancels out. We can now set up a proportion: \(\frac{2400}{600} = \frac{\rho \cdot \frac{4}{3}\pi (2r_2)^3 \cdot g}{\rho \cdot \frac{4}{3}\pi r_2^3 \cdot g}\)

Simplify the proportion: \(\frac{2400}{600} = \frac{(2r_2)^3}{r_2^3}\) \(4 = 8\) Now solve for \(Q_2\), the charge on the second oil drop: \(Q_2 = \frac{Q}{4}\) The charge on the second oil drop is \(\frac{Q}{4}\), so the correct answer is (B).