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College Physics (Urone)
Found in: Page 666
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

A simple and common technique for accelerating electrons is shown in Figure 18.55, where there is a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. (a) Calculate the acceleration of the electron if the field strength is \(2.50 \times {10^4}{\rm{ N/C}}\). (b) Explain why the electron will not be pulled back to the positive plate once it moves through the hole.

Figure 18.55 Parallel conducting plates with opposite charges on them create a relatively uniform electric field used to accelerate electrons to the right. Those that go through the hole can be used to make a TV or computer screen glow or to produce X-rays.

(a) The acceleration of the electron is \(4.39 \times {10^{15}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\).

(b) The electron will not be pulled back to the positive plate once it moves through the hole because there is no electric field outside the plate.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Coulomb force

The force experienced by the charge when it is placed is placed in an electric field created by some other charge is known as Coulomb force or electrostatic force. The expression for the Coulomb force is,

\(F = qE\)

Here, \(q\) is the magnitude of charge on electron, and \(E\) is the electric field.

Step 2: Acceleration of the electron

The force acting on the particle is,

\(F = ma\)

Here, \(m\) is the mass of the electron, and \(a\) is the acceleration of the electron.

Comparing equations (1.1) and (1.2) we get,

\(ma = qE\)

Therefore, the expression for the acceleration of the electron is,

\(a = \frac{{qE}}{m}\)

Substitute \(1.6 \times {10^{ - 19}}{\rm{ }}C\) for \(q\), \(2.50 \times {10^4}{\rm{ N/C}}\) for \(E\), and \(9.1 \times {10^{ - 31}}{\rm{ kg}}\) for \(m\).

\(\begin{array}{c}a = \frac{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ }}C} \right) \times \left( {2.50 \times {{10}^4}{\rm{ N/C}}} \right)}}{{\left( {9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)}}\\ = 4.39 \times {10^{15}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

Hence, the acceleration of the electron is \(4.39 \times {10^{15}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\).

Step 3: Reason the electron will not be pulled back to the positive place

Once the electron passes the hole, there will be no electric field exist outside the plates.

Hence, the electron will not be pulled back to the positive plate once it moves through the hole because there is no electric field outside the plate.

Most popular questions for Physics Textbooks

A \(5.00{\rm{ g}}\) charged insulating ball hangs on a \(30.0{\rm{ cm}}\) long string in a uniform horizontal electric field as shown in Figure 18.56. Given the charge on the ball is \(1.00{\rm{ }}\mu {\rm{C}}\), find the strength of the field.

Figure 18.56 A horizontal electric field causes the charged ball to hang at an angle of \(8.00^\circ \).

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one ball and where does the field become negligibly small? Among the things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule.

(a) Find the electric field at the center of the triangular configuration of charges in Figure 18.54, given that \({q_a} = + {\rm{2}}{\rm{.50 nC}}\), \({q_b} = - {\rm{8}}{\rm{.00 nC}}\), and \({q_c} = + {\rm{1}}{\rm{.50 nC}}\). (b) Is there any combination of charges, other than \({q_a} = {q_b} = {q_c}\), that will produce a zero-strength electric field at the center of the triangular configuration?

Figure 18.54 Point charges located at the corners of an equilateral triangle \(25.0{\rm{ cm}}\) on a side.

Suppose a woman carries an excess charge. To maintain her charged status can she be standing on ground wearing just any pair of shoes? How would you discharge her? What are the consequences if she simply walks away?

Sketch the electric field lines in the vicinity of the conductor in Figure 18.49 given the field was originally uniform and parallel to the object’s long axis. Is the resulting field small near the long side of the object?

Figure 18.49
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