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Chapter 21: Problem 41

(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 \(B\) ?

Short Answer

Expert verified
(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 by step solution

01

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.
02

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 \( a \):\[ a = \frac{F}{m} = \frac{2.4 \times 10^{-19}}{9.11 \times 10^{-31}} = 2.64 \times 10^{11} \text{ m/s}^2 \].
03

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 \( v \):\[ v = \sqrt{2.025 \times 10^{11} + 1.98 \times 10^{11}} = \sqrt{4.005 \times 10^{11}} \approx 6.33 \times 10^5 \text{ m/s} \].
04

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.
05

Find the acceleration of the proton

Using the mass of a proton \( (1.67 \times 10^{-27} \) kg), compute the acceleration:\[ a = \frac{F}{m} = \frac{2.4 \times 10^{-19}}{1.67 \times 10^{-27}} = 1.44 \times 10^{8} \text{ m/s}^2 \].
06

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 \( v \):\[ v = \sqrt{3.61 \times 10^{8} + 1.08 \times 10^{8}} = \sqrt{4.69 \times 10^{8}} \approx 2.17 \times 10^4 \text{ m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

CP A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 \(\mathrm{cm}\) distant from the first, in a time interval of \(1.50 \times 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

If Atoms Were Not Neutral... Because the charges on the electron and proton have the same absolute value, atoms are electrically neutral. Suppose this were not precisely true, and the absolute value of the charge of the electron were less than the charge of the proton by 0.00100\(\% .\) (a) Estimate what the net charge of this textbook would be under these circumstances. Make any assumptions you feel are justified, but state clearly what they are. (Hint: Most of the atoms in this textbook have equal numbers of electrons, protons, and neutrons.) (b) What would be the magnitude of the electric force between two textbooks placed 5.0 \(\mathrm{m}\) apart? Would this force be attractive or repulsive? Estimate what the acceleration of each book would be if the books were 5.0 \(\mathrm{m}\) apart and there were no non-electric forces on them. (c) Discuss how the fact that ordinary matter is stable shows that the absolute values of the charges on the electron and proton must be identical to a very high level of accuracy.

CP An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 \(\mathrm{m}\) in the first 3.00\(\mu\) s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively,

Two very small \(8.55-\) g spheres, 15.0 \(\mathrm{cm}\) apart from center to center, are charged by adding equal numbers of electrons to each of them. Disregarding all other forces, how many electrons would you have to add to each sphere so that the two spheres will accelerate at 25.0 \(\mathrm{g}\) when released? Which way will they accelerate?

Two small plastic spheres are given positive electrical charges. When they are 15.0 \(\mathrm{cm}\) apart, the repulsive force between them has magnitude 0.220 \(\mathrm{N} .\) What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?
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