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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
Chapter 21, Problem 28a

The earth has a net electric charge that causes a field at points near its surface equal to 150150 N/CN/C and directed in toward the center of the earth. What magnitude and sign of charge would a 6060-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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1
Identify the forces acting on the human: the gravitational force (weight) and the electric force. The gravitational force can be calculated using the formula: Fg = 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: Fg = 60 \, \(\text{kg}\) \(\times\) 9.8 \, \(\text{m/s}\)^2.
Determine the electric force needed to balance the gravitational force: The electric force Fe must be equal in magnitude and opposite in direction to the gravitational force to overcome it. Therefore, Fe = Fg.
Use the formula for electric force: The electric force can be expressed as Fe = 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: qE = mg.
Solve for the charge q: Rearrange the equation to solve for q: q = mgE. 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.
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Intro to Electric Fields

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.
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Electric Field due to a Point Charge

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, 1.601.60 cm distant from the first, in a time interval of 3.20×1063.20\(\times\)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 4.50×1064.50\(\times\)10^6 m/s. How much time does it take the proton to stop after entering the field?

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Textbook Question

The earth has a net electric charge that causes a field at points near its surface equal to 150150 N/CN/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 100100 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 6060-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+x-axis) of the electric field in Example 21.621.6. Example 21.621.6: A point charge q=8.0q = -8.0 nC is located at the origin. Find the electric-field vector at the field point x=1.2x = 1.2 m, y=1.6y = -1.6 m.

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Textbook Question

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.

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Textbook Question

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