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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 in the tube under these conditions is \(250,000 V/m\). Thus, the correct answer is (b) \(250.000 V/m\) .

Step by step solution

01

Understand and Convert Units

First, it's important to understand that MV stands for Mega Volts, which equals 1 million volts. Therefore 6.0 MV equals 6,000,000 V. Similarly, m refers to meters. The given distance is 12 m.
02

Use the Definition of Electric Field

Apply the formula for electric field (E), which is the change in voltage (dV) divided by the change in distance (d). The voltage from one end of the tube to the midpoint is half the maximum voltage. So, dV = V_max/2 = 6,000,000 V / 2 = 3,000,000 V. The given distance (d) is 12 m.
03

Calculate the Electric Field

Now, divide dV by d: E = dV / d = 3,000,000 V / 12 m = 250,000 V/m.

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Key Concepts

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

Voltage
In physics, voltage is an essential concept, often referred to as electric potential difference. It represents the potential energy difference between two points in an electric field. Voltage is a measure of how much work is needed to move a charge between these two points. Usually expressed in volts (V), this concept is pivotal in understanding electric fields and forces. For example, when working with a tandem electrostatic accelerator, the voltage indicates how much energy is imparted to charged particles as they travel through the accelerator. It's the driving force that accelerates these particles to high speeds. Understanding voltage is also crucial for solving physics textbook problems, where accurate unit conversion is often necessary, for instance, converting mega volts (MV) to volts (V).
Tandem Electrostatic Accelerator
A tandem electrostatic accelerator is a type of particle accelerator that uses high voltages to accelerate particles to high energies. These accelerators consist of an acceleration tube through which particles are directed, gaining kinetic energy due to the electric field generated by a high voltage at the center. They are primarily used in experimental and applied physics to create charged particles with specific properties, facilitating studies in nuclear physics, material science, and medical research. The tandem design allows for acceleration in two stages—a unique feature that permits higher energy levels. The change in voltage across the tube directly influences the energy particles gain. Understanding this process is instrumental in solving related physics problems and comprehending unit conversions, such as translating MV to V.
Physics Textbook Problem
Physics textbook problems are designed to challenge students' understanding of principles and their ability to apply them to real-world scenarios. These problems require a thorough understanding of concepts like voltage and electric fields, necessitating careful reading and analysis. Solving such problems typically involves a step-by-step approach: identifying known values, converting units where necessary, applying relevant formulas, and calculating the result. For instance, determining the electric field in a tandem electrostatic accelerator involves understanding the relationship between voltage, distance, and the resulting electric field. Problems often have multiple-choice answers, prompting students to examine and validate their solutions critically.
Unit Conversion
Unit conversion is a fundamental skill in physics, ensuring that different measurements can be accurately compared and calculated. This involves converting quantities from one unit to another, such as converting mega volts (MV) to volts (V), where 1 MV is equivalent to 1,000,000 V. Precision in conversion is vital for accuracy in calculations and understanding the magnitude of measurements. For example, in the context of an electric field calculation, knowing the proper units allows for the correct application of formulas, ensuring the resulting electric field value in volts per meter (V/m) accurately reflects the system's physical reality. Mastery of unit conversion simplifies solving textbook problems and helps in grasping complex concepts like those found in tandem electrostatic accelerators.

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Most popular questions from this chapter

(a) How much excess charge must be placed on a copper sphere \(25.0 \mathrm{~cm}\) in diameter so that the potential of its center is \(3.75 \mathrm{kV} ?\) Take the point where \(V=0\) to be infinitely far from the sphere, (b) What is the potential of the sphere's surface?

A very long insulating cylinder of charge of radius \(2.50 \mathrm{~cm}\) carries a uniform linear density of \(15.0 \mathrm{nC} / \mathrm{m}\). If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads \(175 \mathrm{~V} ?\)

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y^{-},\) and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

For a particular experiment, helium ions are to be given a kinetic energy of \(3.0 \mathrm{MeV}\). What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? (a) \(-3.0 \mathrm{MV}\) (b) \(+3.0 \mathrm{MV} ;(\mathrm{c})+1.5 \mathrm{MV} ;(\mathrm{d})+1.0 \mathrm{MV}\)

Identical charges \(q=+5.00 \mu \mathrm{C}\) are placed at opposite corners of a square that has sides of length \(8.00 \mathrm{~cm}\). Point \(A\) is at one of the empty corners, and point \(B\) is at the center of the square. A charge \(q_{0}=-3.00 \mu \mathrm{C}\) is placed at point \(A\) and moves along the diagonal of the square to point \(B\). (a) What is the magnitude of the net electric force on \(\bar{q}_{0}\) when it is at point \(A ?\) Sketch the placement of the charges and the direction of the net force. (b) What is the magnitude of the net electric force on \(q_{0}\) when it is at point \(B ?\) (c) How much work does the electric force do on \(q_{0}\) during its motion from \(A\) to \(B ?\) Is this work positive or negative? When it goes from \(A\) to \(B,\) does \(q_{0}\) move to higher potential or to lower potential?

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