Question

A particle having a charge of \(1.6 \times 10^{-19} \mathrm{C}\) enters between the plates of a parallel plate capacitor. The initial velocity of the particle is parallel to the plates. A potential difference of \(300 \mathrm{v}\) is applied to the capacitor plates. If the length of the capacitor plates is \(10 \mathrm{~cm}\) and they are separated by \(2 \mathrm{~cm}\), Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is \(12 \times 10^{-24} \mathrm{~kg}\). (A) \(10^{4}(\mathrm{~m} / \mathrm{s})\) (B) \(10^{2}(\mathrm{~m} / \mathrm{s})\) (C) \(10^{-1}(\mathrm{~m} / \mathrm{s})\) (D) \(10^{3}(\mathrm{~m} / \mathrm{s})\)

Short answer

The greatest initial velocity for which the particle will not be able to come out of the plates is (A) \(10^{4}(\mathrm{m}/\mathrm{s})\).

Step-by-step solution

Calculate the electric field inside the capacitor

We can use the formula for the electric field inside a parallel plate capacitor: \[ E = \frac{V}{d} \] Where: - \( E \) is the electric field (in \( N/C \)) - \( V \) is the potential difference between the plates (in V) - \( d \) is the distance between the plates (in m) Given that \( V = 300 \, V \) and \( d = 2 \, cm = 0.02 \, m \), we can calculate the electric field \( E \): \[ E = \frac{300 \, V}{0.02 \, m} = 15\,000 \, N/C \]

Calculate the electric force acting on the particle

Using the formula for electric force: \[ F_e = qE \] Where: - \( F_e \) is the electric force (in N) - \( q \) is the charge of the particle (in C) - \( E \) is the electric field (in \( N/C \)) Given that \( q = 1.6 \times 10^{-19} \, C \) and \( E = 15\,000 \, N/C \), we can calculate the electric force \( F_e \): \[ F_e = (1.6 \times 10^{-19} \, C)(15\,000 \, N/C) = 2.4 \times 10^{-15} \, N \]

Calculate the acceleration of the particle

Using the formula for acceleration: \[ a = \frac{F_e}{m} \] Where: - \( a \) is the acceleration (in \( m/s^2 \)) - \( F_e \) is the electric force (in N) - \( m \) is the mass of the particle (in kg) Given that \( F_e = 2.4 \times 10^{-15} \, N \) and \( m = 12 \times 10^{-24} \, kg \), we can calculate the acceleration \( a \): \[ a = \frac{2.4 \times 10^{-15} \, N}{12 \times 10^{-24} \, kg} = 2 \times 10^{8} \, m/s^2 \]

Calculate the time to reach the top plate and compare it with the time to travel the plate length.

We can use the formula for the motion of the particle to calculate the time it takes to reach the top plate: \[ d = \frac{1}{2}at^2 \] Where: - \( d \) is the distance between the plates (in m) - \( a \) is the acceleration (in \( m/s^2 \)) - \( t \) is the time (in s) Given that \( d = 0.02 \, m \) and \( a = 2 \times 10^{8} \, m/s^2 \), we can solve for \( t \): \[ 0.02 \, m = \frac{1}{2}(2 \times 10^{8} \, m/s^2)t^2 \] Solving for \( t \): \[ t = \sqrt{\frac{0.02 \, m}{10^8 \, m/s^2}} = 10^{-4} \, s \] Now we need to calculate the time required to travel the plate length. We can use the formula: \[ x = vt \] Where: - \( x \) is the length of the plate (in m) - \( v \) is the initial velocity of the particle (in \( m/s \)) - \( t \) is the time (in s) Given that \( x = 10 \, cm = 0.1 \, m \) and \( t = 10^{-4} \, s \), we can now solve for \( v \) \[ 0.1 \, m = v(10^{-4} \, s) \] Solving for \( v \): \[ v = 10^4 \, m/s \] The greatest initial velocity for which the particle will not be able to come out of the plates is \( 10^4 \, m/s \). The correct answer is (A) \(10^{4}(\mathrm{~m} / \mathrm{s})\).

Key concepts

Electric Force

When a charged particle enters an electric field, it experiences a force called the electric force. This force is crucial for understanding how charged particles behave in fields created by devices like capacitors.
The electric force can be calculated using the equation:\[ F_e = qE \]

• \( F_e \) is the force acting on the particle, measured in newtons (N).
• \( q \) represents the charge of the particle, measured in coulombs (C).
• \( E \) is the electric field strength, measured in newtons per coulomb (N/C).

The particle in the exercise has a charge of \(1.6 \times 10^{-19} \, C\), and it is subject to an electric field of \(15,000 \, N/C\). Using these values, the electric force can be calculated to determine how the particle will behave under the influence of this field.

Acceleration

Once the electric force acting on a charged particle is known, its acceleration can be determined. Acceleration indicates how fast the velocity of the particle changes over time. It is calculated using Newton's second law of motion.The formula for acceleration is:\[ a = \frac{F_e}{m} \]

• \( a \) is the acceleration, measured in meters per second squared (m/s²).
• \( F_e \) is the electric force, which we have already calculated.
• \( m \) is the mass of the particle, given as \(12 \times 10^{-24} \, kg\).

For the charged particle in the problem, the calculated acceleration is \(2 \times 10^8 \, m/s²\). This large value helps us understand how quickly the particle can change its speed due to the electric force in the capacitor's field.

Parallel Plate Capacitor

A parallel plate capacitor is a common device used to create a uniform electric field. This field is useful for controlling the movement of charged particles, such as in the exercise provided. In a parallel plate capacitor:

• Two metal plates are placed parallel to each other, creating an area between them where the electric field is nearly uniform.
• The field's strength depends on the potential difference (voltage) and the distance between the plates.

The equation that relates these characteristics is:\[ E = \frac{V}{d} \]Here, the potential difference \( V = 300 \) volts and the distance between the plates \( d = 0.02 \) meters, which gives an electric field strength \( E = 15,000 \, N/C\).
This setup enables precise control of the charged particles, influencing their motion within the field.

Kinematics of Charged Particles

Understanding the kinematics—or motion—of charged particles in an electric field involves examining how they move under various forces.In case of a parallel plate capacitor, the particle's path is influenced by:

• Electric forces that alter its velocity perpendicular to the field.
• Its initial velocity, which could be parallel or at an angle to the plates.

In this specific exercise, we determine whether the particle will traverse the entire length of the plate or curve and hit one of the plates. This is done by comparing the time the particle takes to reach the top plate and the time to travel the plate length using the formula:\[ d = \frac{1}{2}at^2 \]If it takes less time to travel the plate length than reaching the top plate, it indicates that the particle will exit. Solving for the initial velocity gives us the critical speed for confinement within the capacitor, crucial for applications requiring precise control over particle paths.