Problem 90 - Master Resource Book in JEE Main Physics

Question

In Millikan's oil drop experiment on oil drop of mass \(16 \times 10^{-6} \mathrm{~kg}\) is balanced by a electric field of \(10^{6} \mathrm{Vm}^{-1}\). The charge in coulomb on the drop will be \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(16 \times 10^{-13} \mathrm{C}\) (b) \(16 \times 10^{-11} \mathrm{C}\) (c) \(6.2 \times 10^{-11} \mathrm{C}\) (d) \(16 \times 10^{-9} \mathrm{C}\)

Step-by-Step Solution

Answer

The charge on the oil drop is \(16 \times 10^{-11} \mathrm{C}\), which is option (b).

Step 1: Understanding the problem

In Millikan's experiment, a charged oil drop is balanced using an electric field. The force due to the electric field balances the gravitational force on the drop. We are given the mass of the oil drop, the strength of the electric field, and need to find the charge on the drop.

Step 2: Identify given values

The mass of the oil drop is given as \( m = 16 \times 10^{-6} \mathrm{~kg} \). The electric field strength is \( E = 10^6 \mathrm{Vm}^{-1} \), and the acceleration due to gravity is \( g = 10 \mathrm{~ms}^{-2} \).

Step 3: Set up the equilibrium equation

At equilibrium, the gravitational force \( F_g = mg \) is equal to the electric force \( F_e = qE \). Therefore, we can write the equation: \[ qE = mg \]

Step 4: Solve for the charge \( q \)

Rearrange the equation \( qE = mg \) to find the charge on the oil drop: \[ q = \frac{mg}{E} \]Substitute the given values: \[ q = \frac{16 \times 10^{-6} \mathrm{~kg} \times 10 \mathrm{~ms}^{-2}}{10^{6} \mathrm{Vm}^{-1}} \]

Step 5: Calculate the charge

Perform the calculation: \[ q = \frac{16 \times 10^{-6} \times 10}{10^{6}} = 16 \times 10^{-11} \mathrm{C} \]

Step 6: Select the correct option

The calculated charge is \(16 \times 10^{-11} \mathrm{C}\). This matches option (b).

Key Concepts

Electric Field

The concept of an electric field is crucial in understanding Millikan's oil drop experiment. An electric field is essentially a region around a charged particle where other charged particles experience a force. In this experiment, an electric field is applied to balance the gravitational force acting on an oil drop. This balance allows for the calculation of the charge on the drop.

The electric field strength is measured in volts per meter (V/m) and tells us how much force would be exerted on a charge in that field. For instance, an electric field of one million volts per meter ( 10^6 V/m) indicates a very strong field, capable of exerting significant force. This force can counteract the force of gravity pulling the oil drop downward. Understanding how the electric field works gives us a way to measure tiny electric charges in particles such as oil drops.

Gravitational Force

Gravitational force is a fundamental force of nature that attracts two bodies towards each other. In Millikan's oil drop experiment, the gravitational force acting on the drop is given by the product of its mass and the acceleration due to gravity. It is calculated using the formula:

\[ F_g = mg \]

where \( F_g \) is the gravitational force, \( m \) is the mass of the oil drop, and \( g \) is the gravitational acceleration, typically measured as 10 m/s² in this problem.

This force attempts to pull the drop downwards. The purpose of the experiment is to use an upward electric force to balance this downward gravitational force, enabling us to determine the charge on the drop through the equilibrium of both forces.

Charge Calculation

Charge calculation is an essential aspect of understanding Millikan's oil drop experiment. The aim is to calculate the charge \( q \) on a single oil drop when it's in equilibrium. This means that the gravitational force acting downward is exactly balanced by the electric force acting upward.

The formula used to calculate the charge is:

\[ q = \frac{mg}{E} \]

Here, \( mg \) represents the gravitational force, \( E \) is the electric field strength, and \( q \) is the charge of the oil drop.

By substituting the given values: mass \( m = 16 \times 10^{-6} \) kg, gravity \( g = 10 \) m/s², and electric field \( E = 10^6 \) V/m, we can solve for \( q \), yielding \( 16 \times 10^{-11} \) C. This method provides a means to calculate the basic electric charge using known forces and field strengths.

Equilibrium of Forces

The equilibrium of forces is a key principle in Millikan's oil drop experiment. It involves balancing the gravitational force acting on the oil drop with the electric force exerted by the electric field. This state of equilibrium allows us to determine the charge on the oil drop.

At equilibrium, these forces are equal in magnitude but opposite in direction. This can be mathematically expressed as:

\[ qE = mg \]

The left side of the equation, \( qE \), represents the upward electric force. The right side, \( mg \), denotes the downward gravitational force. When these two forces are equal, the oil drop remains suspended in mid-air without moving, indicating equilibrium.

By achieving this balance, scientists can accurately calculate the charge \( q \) on the drop by rearranging the equation as \( q = \frac{mg}{E} \). Understanding this equilibrium concept is fundamental to grasping how the oil drop can be studied without external interference.