Problem 33 - College Physics

Question

\(\bullet\) A uniform electric field exists in the region between two oppositely charged plane parallel plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.20 \(\mathrm{cm}\) distant from the first, in a time interval of \(1.5 \times 10^{-8} \mathrm{s}\) (a) Find the magnitude of this electric field. (b) Find the speed of the elec- tron when it strikes the second plate.

Step-by-Step Solution

Answer

The magnitude of the electric field is approximately \(4.56 \times 10^3 \text{ N/C}\) and the speed of the electron is approximately \(2.74 \times 10^7 \text{ m/s}\).

Step 1: Understanding the Problem

An electron moves from a negatively charged plate to a positively charged plate in a uniform electric field, covering a distance of 3.20 cm in a time interval of \(1.5 \times 10^{-8} \text{s}\). Our task is to find the magnitude of the electric field and the speed of the electron when it reaches the other plate.

Step 2: Understanding the Electron Motion

The electron is initially at rest and experiences uniform acceleration due to the electric field. The electric field exerts an electrostatic force on the electron, given by \(F = eE\), where \(e\) is the charge of the electron and \(E\) is the magnitude of the electric field.

Step 3: Determine the Acceleration

Since \(F = ma\) and \(F = eE\), equate these to find \(a = \frac{eE}{m}\), where \(m\) is the mass of the electron. Substituting \(a\) into the kinematic equation \(s = ut + \frac{1}{2}at^2\), where initial velocity \(u = 0\), distance \(s = 3.20 \times 10^{-2} \text{m}\), we get:\[ s = \frac{1}{2} \left( \frac{eE}{m} \right) t^2 \].

Step 4: Calculate Electric Field Magnitude

Rearrange the equation from Step 3 to solve for \(E\):\[ E = \frac{2s m}{e t^2} \].Substituting \(s = 3.20 \times 10^{-2} \text{m}, \ e = 1.60 \times 10^{-19} \text{C}, \ m = 9.11 \times 10^{-31} \text{kg}, \ t = 1.5 \times 10^{-8} \text{s} \), calculate \(E\).

Step 5: Calculate Final Speed

Use the kinematic equation \(v = u + at\) to find the speed \(v\) of the electron when it strikes the second plate. Since \(u = 0\), \[ v = \frac{eE}{m}t \]. Substitute the previously calculated \(E\) and other known values to find \(v\).

Step 6: Summary of Results

Conclude by stating the calculated values for the magnitude of the electric field and the speed of the electron when it reaches the second plate.

Key Concepts

Electron Motion

In an electric field, the movement of electrons can be intriguing. When an electron is released from rest near a negatively charged plate, it begins to move towards a positively charged plate due to the influence of the electric field.
This movement is characterized by the electron gaining speed as it is accelerated by the electrostatic force.
Key points about electron motion:

• The electron starts from rest, implying its initial velocity is zero.
• It experiences a constant acceleration because the electric field is uniform.
• The electrostatic force acting on the electron is proportional to the electric field and the charge of the electron. This force causes the electron to accelerate towards the positive plate.

Understanding electron motion in electric fields helps in grasping how charged particles behave under electrostatic forces.

Kinematic Equations

Kinematic equations describe the motion of objects under uniform acceleration. They are crucial in solving problems involving moving objects, such as our electron moving in the electric field.
For this scenario:

• Initial velocity (\(u\)) of the electron is 0 since it starts from rest.
• The distance (\(s\)) traveled by the electron is 3.20 cm, or \(3.20 \times 10^{-2} \text{ m}\).
• The time (\(t\)) taken to cover this distance is \(1.5 \times 10^{-8} \text{s}\).

Using the kinematic equation:\[s = ut + \frac{1}{2}at^2\] We relate the distance traveled to the time taken and the acceleration (\(a\)) experienced by the electron. With \(u = 0\), this simplifies to:\[s = \frac{1}{2}at^2\]This equation plays a vital role in determining acceleration, which we need to find the electric field magnitude and the electron's speed.

Electrostatic Force

The concept of electrostatic force is key in understanding the electron's motion in an electric field. An electrostatic force arises when the electron is placed in the electric field between charged plates.
This force is given by the equation:\[ F = eE \]Where:

• \(F\) is the force experienced by the electron.
• \(e\) is the charge of the electron, approximately \(1.60 \times 10^{-19} \text{C}\).
• \(E\) is the electric field's magnitude.

This force is what accelerates the electron across the gap between the plates. By knowing this force and the electron's mass, we can find the acceleration using Newton's second law, \(a = \frac{F}{m}\). With this acceleration, the electron's subsequent motion can be analyzed further using kinematic equations, allowing us to determine various aspects such as the electron's final speed. Electrostatic force, thus, is the driving force of the electron's journey.