Ch 21: Electric Charge and Electric Field

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 21: Electric Charge and Electric Field

Problem 23c

Chapter 21, Problem 23c

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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

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.

Recommended video:

Guided course

08:25

Kinematics Equations

Related Practice

Textbook Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, $1.60$ cm distant from the first, in a time interval of $3.20$\times$10^{-6}$ s. Find the speed of the proton when it strikes the negatively charged plate.

views

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 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.

views

Textbook Question

A $negative 4.00$ -nC point charge is at the origin, and a second $negative 5.00$ -nC point charge is on the $x$ -axis at $x equals 0.800$ m. Find the electric field (magnitude and direction) at each of the following points on the $x$ -axis: (i) $x equals 0.200$ m; (ii) $x equals 1.20$ m; (iii) $x = - 0.200$ m.

views

Textbook Question

A $+8.75$ -mC point charge is glued down on a horizontal frictionless table. It is tied to a $-6.50$ -mC point charge by a light, nonconducting $2.50$ -cm wire. A uniform electric field of magnitude $1.85$\times$10^8$ $N/C$ is directed parallel to the wire, as shown in Fig. E $21.34$ . What would the tension be if both charges were negative?

views

Textbook Question

A $positive 8.75$ -mC point charge is glued down on a horizontal frictionless table. It is tied to a $negative 6.50$ -mC point charge by a light, nonconducting $2.50$ -cm wire. A uniform electric field of magnitude $1.85 \times 10^{8}$ $N / C$ is directed parallel to the wire, as shown in Fig. E $21.34$ . Find the tension in the wire.

views