Question

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) \((\mathrm{Q} / 2)\)

Short answer

The charge on the second oil drop is \(\frac{Q}{4}\), so the correct answer is (B).

Step-by-step solution

Recall the electric force and gravitational force formulas

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: \(F_e = F_g\).

Relate the electric force and gravitational force for the first oil drop

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, \(E_1 \cdot Q = m_1 \cdot g\)

Relate the electric force and gravitational force for the second oil drop

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, \(E_2 \cdot Q_2 = m_2 \cdot g\)

Recall the relationship between mass, volume, and density for spherical objects

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: \(V = \frac{4}{3}\pi r^3\)

Set up two equations and perform substitution

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: \(\frac{2400}{600} = \frac{\rho \cdot \frac{4}{3}\pi (2r_2)^3 \cdot g}{\rho \cdot \frac{4}{3}\pi r_2^3 \cdot g}\)

Solve for charge Q_2

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).

Key concepts

Electric Force

Electric force plays a crucial role in Millikan's Oil Drop Experiment. This force is exerted by the electric field on a charged particle. In the experiment, electric force acts upward on the oil drop against the downward gravitational force. To keep the oil drop stationary, the electric force must balance the gravitational force. This is expressed by Coulomb’s law as:

• \(F_e = E \cdot Q\)

Here, \(F_e\) is the electric force, \(E\) is the electric field strength, and \(Q\) is the charge of the oil drop. The electric field \(E\) is derived from the potential difference \(V\) across the plates divided by their separation \(d\), i.e., \(E = \frac{V}{d}\). Therefore, changes in the electric force can be made by adjusting the potential difference between the plates.

Gravitational Force

Gravitational force is the second fundamental force acting on the oil drop in Millikan's experiment. It pulls the oil drop downward towards the Earth. Mathematically, this force is given by:

• \(F_g = m \cdot g\)

Where \(F_g\) denotes the gravitational force, \(m\) is the mass of the oil drop, and \(g\) is the acceleration due to gravity (approximately 9.8 m/s²). In the equilibrium condition, when the drop is motionless, the gravitational force is exactly balanced by the electric force. This relationship sets up an equality: \(F_e = F_g\). The gravitational force is dependent on the mass of the drop, which is determined by its volume and density.

Charge Quantization

Charge quantization is a significant concept that underlies the results of Millikan's Oil Drop Experiment. It refers to the idea that charge comes in discrete packets, or multiples of the elementary charge \(e\). Robert Millikan's experiment provided visual evidence of this fact by determining the charge of individual oil droplets, showing them to be multiples of \(e\). During the experiment, the specific charge \(Q\) on the oil drop is calculated by observing the balance of electric and gravitational forces and observing changes in potential differences and droplet sizes. This helped build the foundation for understanding that charge is always quantized.

Potential Difference

In Millikan's Oil Drop Experiment, potential difference is key in controlling the electric field that influences the oil drop. The electric field is directly proportional to the potential difference between two plates and acts on the charged oil drop. It is expressed by the formula:

• \(E = \frac{V}{d}\)

Where \(E\) is the electric field, \(V\) is the potential difference, and \(d\) is the distance between the plates. When the potential difference is modified, it alters the electric field, affecting the balancing act between gravitational and electric forces on the oil drop. For example, if the potential difference is halved (from 2400 V to 600 V in the given problem), the electric field strength is also reduced, impacting the required charge on the drop to maintain equilibrium.