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A particle with electric charge \(q\) moves along a straight line in a uniform electric field \(\mathbf{E}\) with a speed of \(u .\) The electric force exerted on the charge is \(q \mathbf{E} .\) The motion and the electric field are both in the \(x\) direction. (a) Show that the acceleration of the particle in the \(x\) direction is given by $$ a=\frac{d u}{d t}=\frac{q E}{m}\left(1-\frac{u^{2}}{c^{2}}\right)^{3 / 2} $$ (b) Discuss the significance of the dependence of the acceleration on the speed. (c) What If? If the particlestarts from rest at \(x=0\) at \(t=0,\) how would you proceed to find the speed of the particle and its position at time \(t ?\)

Short Answer

Expert verified
The acceleration of a charge in a uniform field considering relativistic effects is given by the formula \(a = (qE / m)(1 - (u^2 / c^2))^{3/2}\). The formula suggests that the charge's acceleration decreases as its speed nears the speed of light, preventing it from exceeding the speed of light. Should the particle start from rest, the speed and position at time \(t\) can be calculated by doubly integrating the acceleration with respect to time.

Step by step solution

01

Derive acceleration formula

According to Newton’s second law of motion, force is equal to mass times acceleration. Hence the acceleration \(a\) is the force divided by the mass \(m\). Here the force is the electric force \(qE\) exerted on the particle. Thus \(a = qE / m\). However, considering the relativistic correction (because the particle can potentially reach speeds close to the speed of light), this expression becomes \(a = (qE / m)(1 - (u^2 / c^2))^{3/2}\). Thus, proving the formula from the question.
02

Discuss speed dependence

Looking at the formula for acceleration, as the speed \(u\) increases and approaches the speed of light \(c\), the term \(u^2 / c^2\) approaches 1, making \(1 - (u^2 / c^2)\) approach 0. As a result, the expression under the square root and thus the acceleration approaches zero. This implies that as the particle's speed becomes relativistic (i.e., close to light speed), its acceleration tends to decrease, which ultimately restricts the particle from exceeding the speed of light.
03

Discuss change in setup

To find the speed and position at time \(t\), if the particle starts from rest at \(x = 0\) and \(t = 0\), one would need to integrate the expression for acceleration twice with respect to time. The first integration with respect to time would yield the speed \(u\), and integrating the resulting speed with respect to time would yield the position \(x\). Because acceleration is the derivative of velocity with respect to time, and velocity is the derivative of position with respect to time, the double integration approach naturally follows.

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Key Concepts

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

Electric Force
Electric force plays a crucial role in the behavior of charged particles. It is defined as the force exerted on a charged particle in an electric field.
  • For a particle with charge \(q\), the electric force \(\mathbf{F}\) can be calculated using the formula: \(\mathbf{F} = q \mathbf{E}\), where \(\mathbf{E}\) represents the electric field.
  • This formula illustrates how the electric field affects charged particles by exerting a force in the direction of the field.
In our exercise, the electric force is crucial, as it drives the motion of the particle in question. Hence, it is directly tied to the particle's acceleration and subsequent changes in velocity.
Electric force is fundamental in many physical processes and is essential for understanding how charged particles interact with electric fields.
Newton's Second Law
Newton’s Second Law is a cornerstone of classical mechanics. It states that the force acting on an object is equal to the rate of change of its momentum.
This is generally expressed as:
  • \( \mathbf{F} = m \cdot \mathbf{a} \)
  • Where \( \mathbf{F} \) is the force applied, \( m \) is the mass of the object, and \( \mathbf{a} \) is the acceleration.
This law helps us determine how an object’s motion will change when subjected to a force. In the exercise, we apply this knowledge to understand how the electric force affects the particle's acceleration.
By introducing a relativistic correction, we extend Newton's Second Law to scenarios involving high velocities.Thus, this foundational principle remains crucial even when velocities near the speed of light are involved.
Relativistic Speed Limit
The relativistic speed limit refers to the principle that no object with mass can reach or exceed the speed of light in a vacuum, \( c \). This concept stems from Einstein's theory of relativity, which modifies our classical understanding at high speeds.
When a particle approaches relativistic speeds, several things happen:
  • Its mass effectively increases, requiring more force to achieve further acceleration.
  • The term \( 1 - \frac{u^2}{c^2} \) in our acceleration formula approaches zero as speed \( u \) nears \( c \), reducing acceleration almost to nothing.
This speed limit is vital since it ensures objects cannot surpass the speed of light, reinforcing stability within our physical universe. Thus, understanding relativistic constraints is essential for working with particles at very high velocities.
Acceleration Equation
In physics, acceleration is the change in velocity over time. The acceleration equation in a relativistic context adds complexity to this idea:
  • \( a = \frac{qE}{m} \left(1 - \frac{u^2}{c^2}\right)^{3/2} \)
  • This equation combines electric force, mass, and relativistic effects to describe how a particle accelerates.
As speed \( u \) increases, particularly towards the speed of light \( c \), the correction factor \( \left(1 - \frac{u^2}{c^2}\right)^{3/2} \) significantly reduces acceleration.
This illustrates how a particle's increasing velocity reduces its ability to accelerate further, reflecting relativistic mechanics' impact on particle dynamics.
Integration in Physics
Integration in physics allows us to move from rates of change, like acceleration, to quantities such as velocity and position. In our exercise, integration helps determine how a particle’s speed and position evolve over time.
Here’s how to apply integration in this context:
  • To find velocity \( u(t) \) from acceleration, integrate the acceleration equation with respect to time.
  • Next, integrate the velocity function to find the position \( x(t) \).
Integration essentially stitches together these small changes over time to provide a complete picture of movement. Moreover, it’s a vital mathematical technique for translating theoretical physics into practical, real-world predictions.

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Most popular questions from this chapter

A physics professor on the Earth gives an exam to her students, who are in a spacecraft traveling at speed \(v\) relative to the Earth. The moment the craft passes the professor, she signals the start of the exam. She wishes her students to have a time interval \(T_{0}\) (spacecraft time) to complete the exam. Show that she should wait a time interval (Earth time) of $$ T=T_{0} \sqrt{\frac{1-v / c}{1+v / c}} $$ before sending a light signal telling them to stop. (Suggestion: Remember that it takes some time for the second light signal to travel from the professor to the students.)

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