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 28a

Chapter 21, Problem 28a

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?

Verified step by step guidance

Identify the forces acting on the human: the gravitational force (weight) and the electric force. The gravitational force can be calculated using the formula:

F

_g = mg, where

m

is the mass of the human (60 kg) and

g

is the acceleration due to gravity (approximately 9.8 m/s²).

Calculate the gravitational force: Substitute the given values into the formula for gravitational force:

F

_g = 60 \, $\text{kg}$ $\times$ 9.8 \, $\text{m/s}$^2.

Determine the electric force needed to balance the gravitational force: The electric force

F

_e must be equal in magnitude and opposite in direction to the gravitational force to overcome it. Therefore,

F

_e = F_g.

Use the formula for electric force: The electric force can be expressed as

F

_e = qE, where

q

is the charge and

E

is the electric field (150 N/C). Set this equal to the gravitational force to find the required charge:

q E = m g

Solve for the charge

q

: Rearrange the equation to solve for

q

q = m g E

. Substitute the known values to find the magnitude and sign of the charge needed.

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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 charges experience a force. It is represented by the vector quantity E, measured in newtons per coulomb (N/C). The direction of the field is the direction of the force it would exert on a positive test charge placed in the field.

Force Due to Electric Field

The force exerted by an electric field on a charge is given by F = qE, where F is the force, q is the charge, and E is the electric field strength. This relationship shows that the force is directly proportional to both the magnitude of the charge and the strength of the electric field.

Gravitational Force

The gravitational force acting on an object near the Earth's surface is given by F = mg, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s²). This force is directed towards the center of the Earth and is responsible for the object's weight.

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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 magnitude of the electric field.

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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

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

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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