Problem 41 - University Physics with Modern Physics

Question

(a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A\) ? (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s},\) east. What is the speed of the proton at point \(B\) ?

Step-by-Step Solution

Answer

(a) The electron's speed at point B is \(6.33 \times 10^5\) m/s. (b) The proton's speed at point B is \(2.17 \times 10^4\) m/s.

Step 1: Understand the forces acting on the electron

The electron experiences a force due to the electric field, calculated as \( F = qE \), where \( q \) is the charge of the electron (\(-1.6 \times 10^{-19} \) C) and \( E \) is the electric field strength (\(1.50 \) N/C). The electron moves against the direction of the electric field, experiencing a force to the east with magnitude \( F = 1.6 \times 10^{-19} \times 1.50 = 2.4 \times 10^{-19} \) N.

Step 2: Find the acceleration of the electron

Using Newton's second law, \( F = ma \), where \( m \) is the mass of the electron \( (9.11 \times 10^{-31} \) kg). Solve for acceleration \( a \):\[ a = \frac{F}{m} = \frac{2.4 \times 10^{-19}}{9.11 \times 10^{-31}} = 2.64 \times 10^{11} \text{ m/s}^2 \].

Step 3: Calculate the final velocity at point B using kinematics

The initial velocity at point A is \( v_0 = 4.50 \times 10^5 \) m/s. The electron travels a distance \( d = 0.375 \) m. Using the equation \( v^2 = v_0^2 + 2ad \), solve for \( v \):\[ v^2 = (4.50 \times 10^5)^2 + 2 \times 2.64 \times 10^{11} \times 0.375 \]Calculate \( v \):\[ v = \sqrt{2.025 \times 10^{11} + 1.98 \times 10^{11}} = \sqrt{4.005 \times 10^{11}} \approx 6.33 \times 10^5 \text{ m/s} \].

Step 4: Understand the forces acting on the proton

The proton is in the same electric field and experiences a force with the same magnitude but opposite direction due to its positive charge. Calculate the force on the proton as \( F = qE = 1.6 \times 10^{-19} \times 1.50 = 2.4 \times 10^{-19} \) N, acting west.

Step 5: Find the acceleration of the proton

Using the mass of a proton \( (1.67 \times 10^{-27} \) kg), compute the acceleration:\[ a = \frac{F}{m} = \frac{2.4 \times 10^{-19}}{1.67 \times 10^{-27}} = 1.44 \times 10^{8} \text{ m/s}^2 \].

Step 6: Calculate the final velocity of the proton at point B

The initial velocity of the proton at point A is \( v_0 = 1.90 \times 10^4 \) m/s. Using the kinematic equation \( v^2 = v_0^2 + 2ad \) with a distance of \( 0.375 \) m, solve for \( v \):\[ v^2 = (1.90 \times 10^4)^2 + 2 \times 1.44 \times 10^{8} \times 0.375 \]Calculate \( v \):\[ v = \sqrt{3.61 \times 10^{8} + 1.08 \times 10^{8}} = \sqrt{4.69 \times 10^{8}} \approx 2.17 \times 10^4 \text{ m/s} \].

Key Concepts

Electron Motion

Electrons are negatively charged particles that face forces when placed in electric fields. In this exercise, an electron moves east against an electric field directed west. Since the electron is moving opposite to the field direction, it experiences a force to the east.
This force is calculated with the formula:

• \( F = qE \)

Where:

• \( q \) is the charge of the electron \((-1.6 \times 10^{-19} \) C)
• \( E \) is the electric field strength (1.50 N/C)

The magnitude of the force is \( 2.4 \times 10^{-19} \) N. This force alters the motion of the electron. Knowing this helps relate the concept of motion affected by forces in an electric field.

Proton Motion

Protons, unlike electrons, are positively charged and move in the same direction as the electric field within which they are placed. Here, the electric field still points west, and the proton experiences a force in the west direction due to its positive charge.
Similarly, we use the equation for force:

• \( F = qE \)

Where:

• \( q \) is now a positive charge \(1.6 \times 10^{-19} \) C
• \( E \) remains 1.50 N/C

The result is the same as with the electron: \( 2.4 \times 10^{-19} \) N, but directed west. This fundamental understanding helps us comprehend how different charges behave oppositely in the same electric field.

Kinematics

Kinematics is the branch of physics that describes motion without considering forces. In this exercise, the motion of the electron and proton over a distance of 0.375m can be calculated using kinematic equations. The specific formula used is:

• \( v^2 = v_0^2 + 2ad \)

Where:

• \( v_0 \) is the initial velocity
• \( a \) is the acceleration
• \( d \) is the displacement

These factors allow us to find the final velocity \( v \) of both particles at point B. Calculating motion in such detail requires understanding changes due to forces (acceleration) and how these changes relate in terms of distance traveled.

Newton's Second Law

Newton's Second Law is pivotal in understanding how forces affect motion. It states that when a net force \( F \) acts on an object, it produces an acceleration \( a \) proportional to the force and inversely proportional to the object's mass \( m \):

• \( F = ma \)

For the electron:

• The calculated force \( F \) is \( 2.4 \times 10^{-19} \) N
• \( m \) (mass of electron) is \( 9.11 \times 10^{-31} \) kg

Acceleration \( a \) is found using:

• \( a = \frac{F}{m} \approx 2.64 \times 10^{11} \text{ m/s}^2 \)

For the proton, its mass \( m \) (\(1.67 \times 10^{-27} \) kg) results in a different acceleration. This law provides the acceleration values essential for solving the kinematics equations outlined earlier, directly connecting how forces influence motion.