Ch 21: Electric Charge and Electric Field

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 21: Electric Charge and Electric Field

Problem 25c

Chapter 21, Problem 25c

A proton is traveling horizontally to the right at $4.50$\times$10^6$ m/s. What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)? Note: Part (a) asks for how much time does it take the proton to stop after entering the field.

Verified step by step guidance

First, understand that the problem involves stopping an electron using an electric field. The electron has an initial velocity and we need to find the electric field that will decelerate it to a stop.

Recall the formula for the force exerted by an electric field on a charged particle:

F = q E

, where

q

is the charge of the electron and

E

is the electric field.

Use Newton's second law,

F = m a

, where

m

is the mass of the electron and

a

is the acceleration. Set the force from the electric field equal to the force needed to stop the electron:

q E = m a

Determine the acceleration needed to stop the electron using the kinematic equation:

v_{f} = v_{i} + a t

. Since the final velocity

v_{f}

is 0, solve for

a

Finally, solve for the electric field

E

using the equation

E = \frac{m}{a}

divided by

q

. Consider the direction of the field, which should be opposite to the electron's initial velocity to decelerate it.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

An electric field is a region around a charged particle where a force would be exerted on other charges. It is defined by the force per unit charge and is measured in volts per meter (V/m). The direction of the field is the direction of the force it would exert on a positive charge.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through an electric and magnetic field. It is given by F = q(E + v × B), where q is the charge, E is the electric field, v is the velocity, and B is the magnetic field. This concept is crucial for understanding how fields can influence particle motion.

Kinematics of Charged Particles

Kinematics involves the motion of particles without considering the forces that cause the motion. For charged particles, understanding their velocity and acceleration in response to electric fields is essential. The change in velocity can be calculated using the equation v = u + at, where u is initial velocity, a is acceleration, and t is time.

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Related Practice

Textbook Question

A proton is traveling horizontally to the right at $4.50$\times$10^6$ m/s. How much time does it take the proton to stop after entering the field?

Textbook Question

A positive charge $q$ is fixed at the point $x equals 0$ , $y equals 0$ , and a negative charge $negative 2 q$ is fixed at the point $x = a$ , $y equals 0$ . Show the positions of the charges in a diagram.

Textbook Question

The earth has a net electric charge that causes a field at points near its surface equal to $150$ $N/C$ and directed in toward the center of the earth. What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of $100$ m? Is use of the earth's electric field a feasible means of flight? Why or why not? Note: Part (a) asked for what magnitude and sign of charge would a $60$ -kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field.

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Textbook Question

Calculate the magnitude and direction (relative to the $+ x$ -axis) of the electric field in Example $21.6$ . Example $21.6$ : A point charge $q = - 8.0$ nC is located at the origin. Find the electric-field vector at the field point $x equals 1.2$ m, $y = - 1.6$ m.

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Textbook Question

A proton is traveling horizontally to the right at $4.50 \times 10^{6}$ m/s. Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of $3.20$ cm.

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Textbook Question

The earth has a net electric charge that causes a field at points near its surface equal to $150$ $N / C$ and directed in toward the center of the earth. What magnitude and sign of charge would a $60$ -kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field?