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Chapter 5: Problem 27

An electron with a speed of \(1.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\) moves horizontally into a region where a constant vertical force of \(4.5 \times\) \(10^{-16} \mathrm{~N}\) acts on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg}\). Determine the vertical distance the electron is deflected during the time it has moved 30 mm horizontally.

Short Answer

Expert verified
The electron is deflected approximately 1.54 mm vertically.

Step by step solution

01

Calculate the Time Taken

First, calculate the time it takes for the electron to move 30 mm horizontally. Use the formula for time, which is:\[ t = \frac{d}{v} \]where \( d = 30 \text{ mm} = 0.03 \text{ m} \) and \( v = 1.2 \times 10^{7} \text{ m/s} \).\[ t = \frac{0.03}{1.2 \times 10^{7}} = 2.5 \times 10^{-9} \text{ s} \]
02

Determine the Acceleration

Calculate the acceleration of the electron due to the force acting on it. Use Newton's Second Law:\[ F = ma \]or\[ a = \frac{F}{m} \]Substitute \( F = 4.5 \times 10^{-16} \text{ N} \) and \( m = 9.11 \times 10^{-31} \text{ kg} \):\[ a = \frac{4.5 \times 10^{-16}}{9.11 \times 10^{-31}} = 4.94 \times 10^{14} \text{ m/s}^{2} \]
03

Calculate the Vertical Displacement

Use the formula for the vertical displacement, starting from rest, with constant acceleration:\[ s = \frac{1}{2} a t^2 \]Substitute \( a = 4.94 \times 10^{14} \text{ m/s}^2 \) and \( t = 2.5 \times 10^{-9} \text{ s} \):\[ s = \frac{1}{2} \times 4.94 \times 10^{14} \times (2.5 \times 10^{-9})^2 \]\[ s = \frac{1}{2} \times 4.94 \times 10^{14} \times 6.25 \times 10^{-18} \]\[ s = 1.544 \times 10^{-3} \text{ m} \]Convert \( s \) to mm by multiplying by 1000:\[ s \approx 1.544 \text{ mm} \]

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Key Concepts

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

Constant Force
A constant force keeps acting on an electron once it enters a specific region. The term **constant force** means that the force remains the same in magnitude and direction throughout the electron's journey. In physics, when we talk about force, we often refer to Newton's Second Law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, or simply, \( F = ma \). Consequently, constant force results in a constant acceleration.
  • In this exercise, a vertical force of \(4.5 \times 10^{-16}\) N continuously acts on the electron.
  • This force impacts the electron's motion, causing it to deviate vertically as it travels horizontally.
Remember, the direction of this force is crucial as it tells us which way the electron will accelerate, altering its otherwise straight path.
Vertical Displacement
When an electron is subjected to a constant vertical force, it experiences what we call vertical displacement. This simply refers to the change in position of the electron in the vertical direction due to that force.To compute the vertical displacement, we need to consider the time for which the force acts and the acceleration caused by this force. The procedure is to use the equation: \[ s = \frac{1}{2} a t^2 \]
  • "\(s\)" is the displacement we're trying to find.
  • "\(a\)" is the acceleration from the force.
  • "\(t\)" is the time the force acts.
For the given problem, we find that the displacement from this vertical force is \(1.544\) mm. This means that during the electron's horizontal motion of 30 mm, it is also pushed \(1.544\) mm upwards.
Electron Acceleration
Electron acceleration is the change in velocity that an electron experiences when acted upon by a force. The acceleration is directly related to the force applied and inversely to the electron's mass, as stated by Newton鈥檚 Second Law \( a = \frac{F}{m} \).
  • In this particular problem, the force is \(4.5 \times 10^{-16}\) N, and the mass of the electron is \(9.11 \times 10^{-31}\) kg.
  • The result is an electron acceleration of \(4.94 \times 10^{14}\) \(\mathrm{m/s}^2\).
This massive acceleration allows the electron to gain significant velocity in the vertical direction in a very short period. If not for other forces or fields, the electron would follow a straight path, but due to this acceleration, its path curves instead!

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Most popular questions from this chapter

A customer sits in an amusement park ride in which the compartment is to be pulled downward in the negative direction of a \(y\) axis with an acceleration magnitude of \(1.24 g\), with \(g=\) \(9.80 \mathrm{~m} / \mathrm{s}^{2}\), A \(0.567 \mathrm{~g}\) coin rests on the customer's knee. Once the motion begins and in unit-vector notation, what is the coin's acceleration relative to (a) the ground and (b) the customer? (c) How long does the coin take to reach the compartment ceiling, \(2.20 \mathrm{~m}\) above the knee? In unit-vector notation, what are (d) the actual force on the coin and (c) the apparent force according to the customer's measure of the coin's acceleration?

Compute the weight of a \(75 \mathrm{~kg}\) space ranger (a) on Earth, (b) on Mars, where \(g=3.7 \mathrm{~m} / \mathrm{s}^{2}\), and (c) in interplanetary space, where \(g=0 .(\mathrm{d})\) What is the ranger's mass at each location?

If the \(1 \mathrm{~kg}\) standard body has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(20.0^{\circ}\) to the positive direction of an \(x\) axis, what are (a) the \(x\) component and (b) the \(y\) component of the net force acting on the body, and (c) what is the net force in unit-vector notation?

A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. What are the magnitudes of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

Figure \(5-47\) shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time \(t=0\), container 1 has mass \(1.30 \mathrm{~kg}\) and container 2 has mass \(2.80 \mathrm{~kg}\), but container 1 is losing mass (through a leak) at the constant rate of \(0.200 \mathrm{~kg} / \mathrm{s}\). At what rate is the acceleration magnitude of the containers changing at (a) \(t=0\) and (b) \(t=3.00 \mathrm{~s}\) ? (c) When does the acceleration reach its maximum value?
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