Problem 9 - Master Resource Book in JEE Main Physics

Question

An oil drop carrying a charge \(q\) has a mass \(m \mathrm{~kg}\). It is falling freely in air with terminal speed \(v\). The electric field required to make, the drop move upwards with the same speed is (a) \(\frac{m g}{q}\) (b) \(\frac{2 m g}{q}\) (c) \(\frac{m g v}{q^{2}}\) (d) \(\frac{2 m g v}{q}\)

Step-by-Step Solution

Answer

(a) \( \frac{mg}{q} \)

Step 1: Understanding Terminal Velocity

When the oil drop reaches terminal velocity while falling, the upward force due to air resistance equals the downward gravitational force. The gravitational force is given by \( F_g = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity.

Step 2: Upward Movement Contrary to Gravity

When an electric field is introduced to move the oil drop upwards at the same terminal speed, the electric force must counteract gravity and the drag force. For the drop to move upwards with speed \( v \), the upwards electric force must equal the gravitational force while also overcoming the air resistance.

Step 3: Expression for Required Electric Field

The electric force \( F_e \) needed is \( F_e = Eq \), where \( E \) is the electric field strength and \( q \) is the charge of the drop. To move upwards, \( F_e \) must equal \( mg + F_{drag} \), but since the problem posits equivalent upward speed, it’s sufficient that \( F_e = mg \).

Step 4: Solving for Electric Field

Equating the electric force to the gravitational force for equilibrium, we have:\[ Eq = mg \]Solving for \( E \), we get:\[ E = \frac{mg}{q} \]

Step 5: Choosing the Correct Option

Reviewing the options given:(a) \( \frac{mg}{q} \)(b) \( \frac{2mg}{q} \)(c) \( \frac{mgv}{q^2} \)(d) \( \frac{2mgv}{q} \)The expression \( E = \frac{mg}{q} \) corresponds to option (a). Therefore, the correct answer is (a).

Key Concepts

Oil Drop Experiment

The Oil Drop Experiment is a classic physics experiment originally conducted by Robert Millikan. It was designed to measure the charge of a single electron, allowing physicists to gain insights into fundamental electric charges. In the experiment, small oil droplets are suspended within a chamber between two charged plates.
The goal is to balance gravitational forces with electric forces on the drop. When done correctly, the drop will remain stationary or move upward, showcasing the influence of the electric field. By analyzing the motion of the drops, Millikan was able to calculate the elementary charge — a crucial step in the development of atomic theory.

• This experiment exemplifies the delicate balance between electric and gravitational forces.
• It highlighted that electrons are discrete particles with a set charge.
• Led to the determination of the elementary electric charge: approximately 1.6 x 10^-19 Coulombs.

Terminal Velocity

Terminal velocity occurs when a falling object reaches a constant velocity due to the balance of forces acting upon it. For an oil drop falling through air, the forces involved are gravity pulling it downwards and air resistance pushing upwards.
At terminal velocity, these forces are equal, and thus the net force is zero, leading to steady motion.

• The gravitational force is calculated as: \( F_g = mg \) where \( m \) is mass and \( g \) is the acceleration due to gravity.
• Air resistance (or drag) increases with the velocity of the object until it matches the gravitational force.
• Once terminal velocity is achieved, no further acceleration of the drop occurs in the downward direction.

This concept is essential in understanding how objects interact with resistive forces in a fluid (like air).

Electric Force

Electric force is the force exerted by an electric field on a charged particle. In the context of the Oil Drop Experiment, understanding this force is key to manipulating the motion of the oil drops.
When a charged particle, such as an oil droplet with charge \( q \), is placed in an electric field \( E \), it experiences an electric force \( F_e \) that is given by \( F_e = Eq \). This force can be used to counteract gravitational forces on the particle when sustained.

• Electric force direction depends on the sign of the charge. Positive charges move in the direction of the field.
• For a particle with charge \( q \), changing the electric field \( E \) alters the force experienced by the particle.
• Balancing electric force with gravity and drag helps achieve desired motion for charged particles.

Motion in Electric Fields

Motion in electric fields describes the behavior of charged particles when subjected to such fields. The interaction between the charge and the electric field determines the particle's motion direction and velocity.
In our example, an oil drop with charge \( q \) in an electric field \( E \) will experience a force that influences its movement. If this force matches the combined gravitational and drag forces, the particle moves upward at constant speed.

• Ensuring the forces are equal allows precise control over the particle's velocity and direction.
• Electric fields can be adjusted to make particles move against gravity by providing the necessary upward force.
• This principle is foundational in various technological applications such as cathode ray tubes and particle accelerators.

Understanding these principles allows scientists and engineers to manipulate charged particles effectively, guiding technological and scientific advances.