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 23a

Chapter 21, Problem 23a

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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First, identify the initial velocity $ v_i $ of the proton, which is given as $ 4.50 \times 10^6 $ m/s, and the final velocity $ v_f $ which is 0 m/s since the proton comes to rest.

Next, note the distance $ d $ over which the proton comes to rest, which is given as 3.20 cm. Convert this distance to meters by dividing by 100, resulting in $ 0.032 $ m.

Use the kinematic equation $ v_f^2 = v_i^2 + 2a d $ to solve for the acceleration $ a $. Rearrange the equation to $ a = \frac{v_f^2 - v_i^2}{2d} $. Substitute the known values to find $ a $.

Recall that the force $ F $ acting on the proton is related to the electric field $ E $ by the equation $ F = qE $, where $ q $ is the charge of the proton ($ 1.60 \times 10^{-19} $ C). Also, $ F = ma $, where $ m $ is the mass of the proton ($ 1.67 \times 10^{-27} $ kg).

Combine the equations $ F = ma $ and $ F = qE $ to solve for the electric field $ E $. Rearrange to $ E = \frac{ma}{q} $. Substitute the values for $ m $, $ a $, and $ q $ to find the magnitude of the electric field. The direction of the electric field will be opposite to the direction of the proton's initial velocity to bring it to rest.

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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 charged particles. It is represented by the vector E and measured in newtons per coulomb (N/C). The direction of the electric field is the direction of the force it would exert on a positive charge.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It involves equations that relate displacement, velocity, acceleration, and time. In this problem, kinematics helps determine how the proton's velocity changes as it travels through the electric field.

Work-Energy Principle

The work-energy principle states that the work done by forces on an object results in a change in its kinetic energy. For a proton brought to rest, the work done by the electric field equals the initial kinetic energy of the proton. This principle helps calculate the required electric field strength to stop the proton over a given distance.

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The Work-Energy Theorem

Related Practice

Textbook Question

Three point charges are arranged along the $x$ -axis. Charge $q sub 1 equals positive 3.00$ $μ$ C is at the origin, and charge $q_{2} = - 5.00$ $μ$ C is at $x equals 0.200$ m. Charge $q_{2} = - 8.00$ $μ$ C. Where is $q sub 3$ located if the net force on $q sub 1$ is $7.00$ N in the $- x$ -direction?

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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 magnitude of the electric field.

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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.

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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. 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.

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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.

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A proton is traveling horizontally to the right at 4.50×1064.50\(\times\)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 3.203.20 cm.

Verified video answer for a similar problem:

Key Concepts

Electric Field

Kinematics

Work-Energy Principle

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.