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 32a

Chapter 21, Problem 32a

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.

Verified step by step guidance

Start by understanding the motion of the proton. Since it is released from rest, its initial velocity is zero. The proton is accelerated by the electric field between the plates.

Use the kinematic equation for uniformly accelerated motion:

d = v

₀t + 12at², where

d

is the distance,

v

₀ is the initial velocity,

a

is the acceleration, and

t

is the time.

Since the initial velocity

v

₀ is zero, the equation simplifies to

d = \frac{1}{2} a t

². Plug in the given values:

= 1.60 imes 10^{-2} \(\text{ m}\)

and

= 3.20 imes 10^{-6} \(\text{ s}\)

Solve for acceleration

a

using the equation:

a = rac{2 d}{t

²}. Substitute the values to find

a

Relate the acceleration to the electric field using the formula

a = rac{q E}{m}

, where

q

is the charge of the proton and

m

is the mass of the proton. Rearrange to solve for the electric field

E

E = rac{m a}{q}

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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 object where other charged objects experience a force. It is represented by the vector quantity E, which indicates the force per unit charge. In this scenario, the uniform electric field between the plates exerts a constant force on the proton, influencing its motion.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. To solve this problem, we use kinematic equations to relate the proton's initial velocity, acceleration due to the electric field, distance traveled, and time taken to reach the opposite plate.

Force on a Charged Particle

The force on a charged particle in an electric field is given by F = qE, where q is the charge of the particle and E is the electric field strength. For a proton, this force causes acceleration, which can be calculated using Newton's second law, F = ma, allowing us to determine the electric field's magnitude.

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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.

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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 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

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?

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?

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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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