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Chapter 18: Problem 78

The maximum voltage at the center of a typical tandem electrostatic accelerator is \(6.0 \mathrm{MV}\). If the distance from one end of the acceleration tube to the midpoint is \(12 \mathrm{~m},\) what is the magnitude of the average electric field in the tube under these conditions? A. \(41,000 \mathrm{~V} / \mathrm{m}\) B. \(250,000 \mathrm{~V} / \mathrm{m}\) C. \(500,000 \mathrm{~V} / \mathrm{m}\) D. \(6,000,000 \mathrm{~V} / \mathrm{m}\)

Short Answer

Expert verified
The magnitude of the average electric field is 500,000 V/m, corresponding to option C.

Step by step solution

01

Understand the Formula for Electric Field

The average electric field \( E \) in a system where voltage \( V \) is applied across a distance \( d \) is given by the formula \( E = \frac{V}{d} \). This formula helps us find the electric field when voltage and distance are known.
02

Identify Given Values

In the problem, the voltage \( V \) at the center of the accelerator is given as \( 6.0 \text{ MV} \) which is \( 6.0 \times 10^6 \text{ V} \). The distance from one end of the tube to the midpoint where this voltage is measured is \( 12 \text{ m} \).
03

Substitute Values into the Formula

Using the formula for electric field, substitute the given values: \( E = \frac{V}{d} = \frac{6.0 \times 10^6 \text{ V}}{12 \text{ m}} \).
04

Perform the Calculation

Calculate the electric field using the formula: \( E = \frac{6,000,000 \text{ V}}{12 \text{ m}} = 500,000 \text{ V/m} \). So, the magnitude of the average electric field is \( 500,000 \text{ V/m} \).
05

Select the Correct Answer

Compare the calculated electric field with the given options. Option C matches our calculated value of \( 500,000 \text{ V/m} \).

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Key Concepts

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

Average Electric Field
The concept of an average electric field is crucial when exploring how voltage and distance interact in physics. In a tandem electrostatic accelerator, the voltage is applied across a specific distance, creating an electric field. The average electric field gives us an idea of the overall influence of the electric forces in that region.
By using the formula \( E = \frac{V}{d} \), where \( E \) is the electric field, \( V \) is the voltage, and \( d \) is the distance, you can easily determine how strong the electric field is in a given area. An average electric field is especially useful when considering complex systems where the field might not be uniform.
  • This concept helps in identifying how effectively the voltage influences particle movement over a distance.
  • In the case of the tandem electrostatic accelerator, knowing the average field aids in predicting the behavior of particles being accelerated.
This measure simplifies the overall effect of the electrostatic forces across the length of the tube.
Voltage Calculation
Voltage calculation is the basis for determining the electric force exerted over a particular distance. In the exercise we mined from, the maximum voltage was given as \(6.0 \text{ MV}\) or \(6.0 \times 10^6 \text{ V}\). Calculating voltage is straightforward when you understand the relationship between voltage and electric energy.

  • In physics, voltage represents the potential difference between two points, effectively measuring the potential energy per unit charge.
  • In tandem accelerators, this potential energy difference is what propels charged particles.
By mastering voltage calculations, you can proficiently determine the conditions required for specific electrostatic or energetic phenomena. Reliable calculations ensure experiments and theoretical predictions hold accurately within the designed frameworks.
Electric Field Formula
The electric field formula \( E = \frac{V}{d} \) is central to understanding how electric fields function in physics. This formula expresses the direct relationship between the voltage applied to a system and the distance across which it acts.
Let's examine the components:
  • \( E \) refers to the electric field strength, indicating how forcefully charges are pushed or pulled over a distance.
  • \( V \) symbolizes voltage, or electrical potential difference, between two points.
  • \( d \) stands for the distance, essentially how far the voltage extends its influence.
This relationship makes it possible to investigate how effective an electric field is in specific environments. Recognizing the interaction between these factors helps in setting up electronic devices, conducting experiments, or understanding natural phenomena that involve electrostatic forces.

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Most popular questions from this chapter

Neurons are the basic units of the ner- vous system. They contain long tubular structures called axons that propagate electrical signals away from the ends of the neurons. The axon contains a solution of potassium ions \(\mathrm{K}^{+}\) and large negative organic ions. The axon membrane prevents the large ions from leaking out, but the smaller \(\mathrm{K}^{+}\) ions are able to penetrate the membrane to some degree. (See Figure \(18.35 .)\) This leaves an excess of negative charge on the inner surface of the axon membrane and an excess of positive charge on the outer surface, resulting in a potential difference across the membrane that prevents further \(\mathrm{K}^{+}\) ions from leaking out. Measurements show that this potential difference is typically about \(70 \mathrm{mV}\). The thickness of the axon membrane itself varies from about 5 to \(10 \mathrm{nm}\), so we'll use an average of \(7.5 \mathrm{nm}\). We can model the membrane as a large sheet having equal and opposite charge densities on its faces. (a) Find the electric field inside the axon membrane, assuming (not too realistically) that it is filled with air. Which way does it point, into or out of the axon? (b) Which is at a higher potential, the inside surface or the outside surface of the axon membrane?

In the Bohr model of the hydrogen atom, a single electron revolves around a single proton in a circle of radius \(r .\) Assume that the proton remains at rest. (a) By equating the electric force to the electron mass times its acceleration, derive an expression for the electron's speed. (b) Obtain an expression for the electron's kinetic energy, and show that its magnitude is just half that of the electric potential energy. (c) Obtain an expression for the total energy, and evaluate it using \(r=5.29 \times 10^{-11} \mathrm{~m}\). Give your numerical result in joules and in electronvolts.

Two oppositely charged identical insulating spheres, each \(50.0 \mathrm{~cm}\) in diameter and carrying a uniform charge of magnitude \(175 \mu \mathrm{C},\) are placed \(1.00 \mathrm{~m}\) apart center to center (Figure 18.48 ). (a) If a voltmeter is connected between the nearest points \((a\) and \(b\) ) on their surfaces, what will it read? (b) Which point, \(a\) or \(b\), is at the higher potential? How can you know this without any calculations?

A parallel-plate air capacitor is made from two plates \(0.200 \mathrm{~m}\) square, spaced \(0.800 \mathrm{~cm}\) apart. It is connected to a \(120 \mathrm{~V}\) battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of \(1.60 \mathrm{~cm},\) what are the answers to parts (a), (b), (c), and (d)?

A \(5.80 \mu \mathrm{F}\) parallel-plate air capacitor has a plate separation of \(5.00 \mathrm{~mm}\) and is charged to a potential difference of \(400 \mathrm{~V}\). Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3}\).
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