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An electron is acted upon by a force of 5.00 \(\times\) 10\(^{-15}\) N due to an electric field. Find the acceleration this force produces in each case: (a) The electron's speed is 1.00 km/s. (b) The electron's speed is 2.50 \(\times\) 10\(^8\) m/s and the force is parallel to the velocity.

Short Answer

Expert verified
(a) 5.49 × 10¹⁵ m/s², (b) 3.27 × 10¹⁵ m/s² (relativistic).

Step by step solution

01

Understand the problem

We need to find the acceleration of the electron when acted upon by a force of \( 5.00 \times 10^{-15} \) N. We will use Newton's second law of motion \( F = ma \) where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. The mass of an electron is approximately \( 9.11 \times 10^{-31} \) kg.
02

Calculate acceleration for case (a)

For case (a), the electron's speed is \( 1.00 \) km/s. However, since acceleration is independent of velocity in classical mechanics (unless relativistic effects come into play), we directly use Newton's second law: \[ a = \frac{F}{m} \]So, the acceleration \( a = \frac{5.00 \times 10^{-15}}{9.11 \times 10^{-31}} \) m/s\(^2\).
03

Solve for acceleration in case (a)

Compute the acceleration:\[a = \frac{5.00 \times 10^{-15}}{9.11 \times 10^{-31}} = 5.49 \times 10^{15} \text{ m/s}^2\]
04

Understand relativity effects for case (b)

For case (b), the electron's speed is \( 2.50 \times 10^8 \) m/s, which is close to the speed of light (\( c = 3.00 \times 10^8 \) m/s). At this speed, relativistic effects become significant, meaning we need to introduce the Lorentz factor \( \gamma \).
05

Calculate Lorentz factor \( \gamma \)

The Lorentz factor \( \gamma \) is used to account for relativistic effects and is given by \[ \gamma = \frac{1}{\sqrt{1 - (v/c)^2}} \]Substituting the electron's speed, \( v = 2.50 \times 10^8 \) m/s:\[ \gamma = \frac{1}{\sqrt{1 - (2.50 \times 10^8/3.00 \times 10^8)^2}} = 1.29 \]
06

Calculate acceleration for case (b) considering relativity

In relativistic mechanics, the force is related to acceleration via:\[ F = \gamma^3 ma \]So the acceleration \( a = \frac{F}{\gamma^3 m} \). Substituting the values:\[ a = \frac{5.00 \times 10^{-15}}{(1.29)^3 \, \times 9.11 \times 10^{-31}} = 3.27 \times 10^{15} \, \text{m/s}^2 \]

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Key Concepts

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

Newton's Second Law
Newton's second law of motion is a fundamental principle that describes the relationship between the force applied to an object, its mass, and the resulting acceleration. The law is mathematically expressed as \( F = ma \), where \( F \) stands for force, \( m \) for mass, and \( a \) for acceleration. This formula implies that the acceleration of an object is directly proportional to the net force acting upon it, and inversely proportional to its mass.

When applying this law to calculate acceleration, it's crucial to know both the force applied and the mass of the object. For instance, in the given problem, the force acting on the electron is \( 5.00 \times 10^{-15} \) N and the mass is approximately \( 9.11 \times 10^{-31} \) kg. By substituting these values into Newton's second law, we calculate the acceleration \( a = \frac{F}{m} \). This is a straightforward application in classical mechanics when velocities are much lower than the speed of light. However, as speeds increase closer to the speed of light, relativistic effects, which are not accounted for in Newtonian mechanics, become significant.

Newton's second law thus serves as the cornerstone for understanding motion, providing insight into how forces affect the motion of objects.
Relativistic Mechanics
Relativistic mechanics comes into play when objects move at speeds close to that of light. Classical mechanics, like Newton's laws, assume that time and space are absolute. However, this assumption fails at very high velocities, where relativistic effects are significant.

According to Einstein's theory of relativity, as the velocity of an object approaches the speed of light, time dilation and length contraction occur. These effects must be incorporated into new equations of motion, often making the simple relations of Newtonian physics inadequate.

In the exercise, when the electron is moving at a velocity of \( 2.50 \times 10^8 \) m/s, relativistic mechanics is essential. The equations involve an additional factor known as the Lorentz factor, \( \gamma \), which adjusts the calculations to reflect these relativistic effects. The inclusion of the Lorentz factor alters the simple relationship between force and acceleration, demanding adjustments to accurately calculate the resultant acceleration in such scenarios.
Lorentz Factor
The Lorentz factor, denoted by \( \gamma \), is a crucial element in relativistic physics. It accounts for the modifications required in physical calculations at velocities nearing the speed of light. The Lorentz factor is defined by the equation:
\[\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\]
where \( v \) is the velocity of the object, and \( c \) is the speed of light in a vacuum, approximately \( 3.00 \times 10^8 \) m/s.

In the given problem, the Lorentz factor is necessary to find the effective acceleration of an electron moving at a high speed of \( 2.50 \times 10^8 \) m/s. By substituting the given velocity, the Lorentz factor can be calculated as \( \gamma = 1.29 \).

The Lorentz factor effectively shows how much the behavior of objects deviates from Newtonian predictions as their speed increases close to \( c \). It particularly influences calculations concerning time intervals, energy, and momentum, making it an indispensable concept in the realm of relativistic mechanics.

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Most popular questions from this chapter

Inside a spaceship flying past the earth at three-fourths the speed of light, a pendulum is swinging. (a) If each swing takes 1.80 s as measured by an astronaut performing an experiment inside the spaceship, how long will the swing take as measured by a person at mission control (on earth) who is watching the experiment? (b) If each swing takes 1.80 s as measured by a person at mission control, how long will it take as measured by the astronaut in the spaceship?

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot flies past you in her spaceracer at a constant speed of 0.800c relative to you. At the instant the spaceracer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled 1.20 \(\times\) 10\(^8\) m past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of 0.890c. Both particles travel at the same speed as measured in the laboratory. What is the speed of each particle, as measured in the laboratory?

(a) Consider the Galilean transformation along the \(x\)-direction : \(x' = x - vt\) and \(t' = t\). In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $${\partial^2E(x, t)\over \partial x^2} - {1 \over c^2} {\partial^2E(x, t)\over \partial t^2} = 0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S'\) is found to be $$(1 - {v^2 \over c^2} ) {\partial^2E(x', t') \over \partial{x'^2}} + {2v \over c^2} {\partial^2E(x', t') \over \partial{x'} \partial{t'}} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'^2}} = 0$$ This has a different form than the wave equation in \(S\). Hence the Galilean transformation \(violates\) the first relativity postulate that all physical laws have the same form in all inertial reference frames. (\(Hint\): Express the derivatives \(\partial/\partial{x}\) and \(\partial/\partial{t}\) in terms of \(\partial/\partial{x'}\) and \(\partial/\partial{t'}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S'\) the wave equation has the same form as in frame \(S\): $${\partial^2E(x', t') \over \partial{x'}^2} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'}^2} = 0$$ Explain why this shows that the speed of light in vacuum is c in both frames \(S\) and \(S'\).

The positive muon (\(\mu^+\)), an unstable particle, lives on average 2.20 \(\times\) 10\(^{-6}\) s (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of 0.900c, what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

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