Problem 75 - Arihant AIEEE Physics

Question

An oil drop of charge of 2 electrons fall freely with a terminal speed. The mass of oil drop so, it can move upward with same terminal speed, if electric field of \(2 \times 10^{3} \mathrm{~V} / \mathrm{m}\) is applied, is : (a) \(3.0 \times 10^{-17} \mathrm{~kg}\) (b) \(3.2 \times 10^{-17} \mathrm{~kg}\) (c) \(2.5 \times 10^{-17} \mathrm{~kg}\) (d) \(3.3 \times 10^{-17} \mathrm{~kg}\)

Step-by-Step Solution

Answer

None of the provided options are correct. Calculated mass is approximately \(6.53 \times 10^{-17} \mathrm{~kg}\).

Step 1: Understand the Problem

We need to find the mass of the oil drop that can move upwards with the same terminal speed when an electric field is applied. The oil drop has a charge of 2 electrons and the electric field is given as \(2 \times 10^{3} \ \mathrm{V/m}\).

Step 2: Determine the Charge

Each electron has a charge of \(1.6 \times 10^{-19} \ \mathrm{C}\). Since the oil drop has a charge of 2 electrons, the total charge \(q\) on the oil drop is: \[ q = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \ \mathrm{C} \]

Step 3: Apply Newton's Law of Motion

In equilibrium at terminal speed, gravitational force \((mg)\) and the electric force \((qE)\) must be equal in magnitude for the drop to move with the same speed upwards. Therefore, we have: \[ mg = qE \].

Step 4: Substitute Known Values

Substitute the values of charge \(q\) and the electric field \(E = 2 \times 10^{3} \mathrm{~V/m}\) into the equation: \( mg = 3.2 \times 10^{-19} \times 2 \times 10^{3} \).

Step 5: Calculate the Mass

Using \( g = 9.8 \ \mathrm{m/s^2} \) which is the acceleration due to gravity, solve for \( m \):\[ m = \frac{3.2 \times 10^{-19} \times 2 \times 10^{3} }{9.8} = \frac{6.4 \times 10^{-16}}{9.8} \approx 6.53 \times 10^{-17} \ \mathrm{kg} \].

Step 6: Compare with Options

As the calculated mass \(6.53 \times 10^{-17} \mathrm{~kg}\) is not perfectly matching with any given options, rechecking the provided options could be suggested. As an alternative, a slightly adjusted correct parameter or data should be identified.

Key Concepts

Charge of Electron

The charge of an electron is a fundamental constant and plays an essential role in understanding electrical interactions. Electrons are negatively charged subatomic particles found in atoms. Each electron carries an elemental charge of approximately

• \(1.6 \times 10^{-19} \ \mathrm{Coulombs} \) (C)

When dealing with multiple electrons, as in the case of this oil drop experiment, the total charge is calculated by multiplying the charge of one electron by the number of electrons. For instance, if an oil drop has the charge of 2 electrons, the total charge \(q\) is computed as:\[ q = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \ \mathrm{C} \] Understanding the charge is crucial for calculating electric forces and determining how particles will interact with electric fields.

Electric Field

An electric field represents the force exerted per unit charge in a specific region of space. It's a vector field, meaning it has both magnitude and direction. When a charged particle is placed in an electric field, it experiences a force that's directly proportional to its charge and the strength of the electric field.

• The formula for this force \(F\) is:

\[ F = qE \] where \(q\) is the charge of the particle and \(E\) is the electric field strength, measured in volts per meter \( \mathrm{V/m} \). For the oil drop, with an electric field applied, the force will act to counteract the force of gravity, facilitating the upward movement with original terminal speed when balanced perfectly with gravitational forces. This is critical when applying Newton's laws to find equilibrium conditions.

Terminal Velocity

Terminal velocity is the constant speed that a falling object eventually reaches when the resistance of the medium (such as air) prevents further acceleration. At terminal velocity, the force of gravity \(mg\) is balanced by drag force or any other opposing forces.

• For a particle in an electric field, terminal velocity might involve balancing the electric force \(qE\) against gravity \(mg\) if the particle is charged.

In our case, while moving upwards, the balance occurs according to:\[ mg = qE \] The object will stay at this velocity when forces are balanced according to this equation. It's essential for understanding how the oil drop can maintain the same speed whether moving upward or downward, given the correct balance of forces.

Newton's Laws of Motion

Newton's Laws of Motion play a fundamental role in analyzing an oil drop's motion within an electric field. The first law, or the law of inertia, suggests that an object will maintain its state of motion unless acted upon by an external force.

• The second law expresses the relationship between force, mass, and acceleration: \( F = ma \).

For the oil drop experiment, we applied this concept by setting the net force to zero at terminal velocity, such that:\[ mg = qE \] By manipulating this equation, you can solve for the mass \(m\) required for the oil drop to maintain an upward terminal velocity when an electric field is applied. Finally, understanding these principles is key to grasping how charged particles react within energy fields, embodying concepts that are crucial to physics and engineering fields.