Physics for Scientists and Engineers (AP Edition)
Physics for Scientists and Engineers (AP Edition)
9th Edition
ISBN: 9781133953951
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 25, Problem 25.54AP

Review. In fair weather, the electric field in the air at a particular location immediately above the Earth's surface is 120 N/C directed downward, (a) What is the surface charge density on the ground? Is it positive or negative? (b) Imagine the surface charge density is uniform over the planet. What then is the charge of the whole surface of the Earth? (c) What is the Earth’s electric potential due to this charge? (d) What is the difference in potential between the head and the feet of a person 1.75 m tall? (Ignore any charges in the atmosphere.) (e) Imagine the Moon, with 27.3% of the radius of the Earth, had a charge 27.3% as large, with the same sign. Find the electric force the Earth would then exert on the Moon, (f) State how the answer to part (e) compares with the gravitational force the Earth exerts on the Moon.

(a)

Expert Solution
Check Mark
To determine

The surface charge density on the ground.

Answer to Problem 25.54AP

The surface charge density on the ground is 1.06nC/m2 .

Explanation of Solution

Given info: The electric field above the earth’s surface is 120N/C directed towards downward.

Write the expression for the surface charge density.

σ=Eε0

Here,

σ is the surface charge density.

E is the electric field.

ε0 is the permittivity of free space.

The value of ε0 is 8.85×1012C2/Nm2 .

Substitute 120N/C for E and 8.85×1012C2/Nm2 for ε0 in above equation.

σ=120N/C×8.85×1012C2/Nm2=1.06×109C/m2(109nC1C)=1.06nC/m2

The direction of the electric field is downwards, so the direction of the surface charge density is negative.

Conclusion:

Therefore, the surface charge density on the ground is 1.06nC/m2 .

(b)

Expert Solution
Check Mark
To determine

The charge of whole earth.

Answer to Problem 25.54AP

The charge of whole earth is 541kC .

Explanation of Solution

Given info: The electric field above the earth’s surface is 120N/C directed towards downward.

From part (a) the surface charge density on the ground is 1.06nC/m2 .

The radius of the earth is 6.37×106m .

The area of the earth is expressed as,

A=4πr2

Here,

r is the radius of the earth.

Substitute 6.37×106m for r in above equation.

A=4π(6.37×106m)2=5.099×1014m2

Thus, the area of the earth is 5.099×1014m2 .

Write the expression for the charge of whole the earth.

Q=σA

Here,

Q is the charge.

A is the area that encloses the charge.

Substitute 5.099×1014m2 for E and 1.06nC/m2 for σ in above equation.

Q=1.06nC/m2(109nC1C)×5.099×1014m2=540498.62C(1kC1000C)=540.498kC541kC

Conclusion:

Therefore, the charge of whole earth is 541kC .

(c)

Expert Solution
Check Mark
To determine

The electric potential of earth.

Answer to Problem 25.54AP

The electric potential of earth is 7.64×102MV .

Explanation of Solution

Given info: The electric field above the earth’s surface is 120N/C directed towards downward.

From part (b) the charge of the earth is 541kC .

Write the expression for the electric potential of earth.

V=keQr

Here,

V is the electric potential of earth.

ke is the coulomb’s constant.

The value of k is 8.99×109Nm2/C2 .

Substitute 541kC for Q , 8.99×109Nm2/C2 for ke and 6.37×106m for r in above equation.

V=8.99×109Nm2/C2(541kC(1000C1kC))6.37×106m=7.635×108V(1MV106V)7.64×102MV

Conclusion:

Therefore, the electric potential of earth is 7.64×102MV .

(d)

Expert Solution
Check Mark
To determine

The electric potential difference between the head and feet of a 1.75m tall person.

Answer to Problem 25.54AP

The electric potential difference between the head and feet of a 1.75m tall person is 210V .

Explanation of Solution

Given info: The electric field above the earth’s surface is 120N/C directed towards downward.

Write the expression for the electric potential difference between the head and feet of a person.

ΔV=Ed

Here,

ΔV is the electric potential difference between the head and feet of a person.

d is the height of the person.

Substitute 120N/C for E and 1.75m for d in above equation.

ΔV=120N/C×1.75m=210V

Conclusion:

Therefore, the electric potential difference between the head and feet of a 1.75m tall person is 210V .

(e)

Expert Solution
Check Mark
To determine

The electric force between the earth and moon.

Answer to Problem 25.54AP

The electric force between the earth and moon is 4.87×103N .

Explanation of Solution

Given info: The electric field above the earth’s surface is 120N/C directed towards downward, the radius of the moon is 27.3% of radius of earth and the charge of the moon is 27.3% of charge of earth.

The distance between moon and earth is 3.84×108m .

From part (b) the charge of earth is 541kC .

The charge of moon is given by,

Qm=27.3%(541kC)=147.693kC

Thus, the value of Qm is 147.693kC .

Write the expression for the electric force between moon and earth.

F=keQQmr2

Here,

F is the electric force between moon and earth.

Qm is the charge of moon.

r is the distance between moon and earth.

Substitute 8.99×109Nm2/C2 for ke , 541kC for Q , 147.693kC for Qm and 3.84×108m for r in above equation.

F=8.99×109Nm2/C2×(541kC)×(147.693kC)(3.84×108m)2(1000C1kC)2=4.87×103N

Conclusion:

Therefore, the electric force between the earth and moon is 4.87×103N .

(f)

Expert Solution
Check Mark
To determine

The comparison of the gravitation force and the electric force

Answer to Problem 25.54AP

The gravitation force is 4.08×1016 times larger than the electric force and electric force is negligible in comparison to gravitation force.

Explanation of Solution

Given info: The electric field above the earth’s surface is 120N/C directed towards downward, the radius of the moon is 27.3% of radius of earth and the charge of the moon is 27.3% of charge of earth.

The mass of earth is 5.98×1024kg and the mass of the moon is 7.36×1022kg .

From part (e) the electric force between the earth and moon is 4.87×103N .

Write the expression for the gravitational force between moon and earth.

Fg=Gmemmr2

Here,

Fg is the gravitational force between moon and earth.

mm is the mass of moon.

me is the mass of earth.

G is the gravitational constant.

The value of G is 6.67×1011Nm2/kg2 .

Substitute 6.67×1011Nm2/kg2 for G , 7.36×1022kg for mm , 5.98×1024kg for me and 3.84×108m for r in above equation.

Fg=6.67×1011Nm2/kg2×(7.36×1022kg)×(5.98×1024kg)(3.84×108m)2=1.99×1020N

Thus, the value of Fg is 1.99×1020N .

Divide Fg with F to compare.

FgF

Substitute 1.99×1020N for Fg and 4.87×103N for F in above equation.

1.99×1020N4.87×103N=4.08×1016

The gravitation force is 4.08×1016 times larger than the electric force and opposite in direction.

Conclusion:

Therefore, the gravitation force is 4.08×1016 times larger than the electric force and electric force is negligible in comparison to gravitation force.

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Chapter 25 Solutions

Physics for Scientists and Engineers (AP Edition)

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