Concept explainers
(a)
The surface charge density of the ground.
(a)
Answer to Problem 54AP
The surface charge density of the ground is
Explanation of Solution
Write the expression for the electric field due to charged earth immediately above its surface.
Here,
Write the expression for the surface charge density.
Here,
Substitute
Conclusion:
Substitute
Since the electric field is directed downward. Hence, the surface charge density will be negative.
Therefore, the surface charge density of the ground is
(b)
The charge of the whole surface of the earth.
(b)
Answer to Problem 54AP
The charge of the whole surface of the earth is
Explanation of Solution
Conclusion:
Substitute
Therefore, the charge of the whole surface of the earth is
(c)
The earth’s electric potential.
(c)
Answer to Problem 54AP
The earth’s electric potential is
Explanation of Solution
Write the expression for the electric potential.
Here,
Conclusion:
Substitute
Therefore, The earth’s electric potential is
(d)
The difference in potential between the head and foot of a person.
(d)
Answer to Problem 54AP
The difference in potential between the head and foot of a person is
Explanation of Solution
Write the relationship between the electric potential and the electric field.
Here,
Conclusion:
Substitute
Therefore, the difference in potential between the head and foot of a person is
(e)
The force that the earth would exert on the moon.
(e)
Answer to Problem 54AP
The force that the earth would exert on the moon is
Explanation of Solution
Write the expression for the magnitude of electrostatic force exerted by the earth on the moon.
Here,
Conclusion:
Substitute
Therefore, the force that the earth would exert on the moon is
(f)
The comparison between the gravitational force and the electrostatic force exerted by the earth on the moon.
(f)
Answer to Problem 54AP
The gravitational force is
Explanation of Solution
Write the expression for the gravitational force between the earth and moon.
Here,
Write the expression to calculate the ratio of gravitational force and the electrostatic force between the earth and the moon.
Conclusion:
Substitute
Substitute
Therefore, the gravitational force is
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Chapter 25 Solutions
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
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- Three particles with equal positive charges q are at the corners of an equilateral triangle of side a as shown in Figure P20.10. (a) At what point, if any, in the plane of the particles is the electric potential zero? (b) What is the electric potential at the position of one of the particles due to the other two particles in the triangle? Figure P20.10arrow_forwardA uniformly charged insulating rod of length 14.0 cm is bent into the shape of a semicircle as shown in Figure P20.29. The rod has a total charge of 7.50 C. Find the electric potential at O, the center of the semicircle. Figure P20.29arrow_forwardGiven two particles with 2.00-C charges as shown in Figure P25.19 and a particle with charge q = 1.28 10-18 C at the origin, (a) what is the net force exerted by the two 2.00-C; charges on the charge q? (b) What is the electric field at the origin due to the two 2.00-C particles? (c) What is the electric potential at the origin due to the two 2.00-C particles?arrow_forward
- A uniform electric field of magnitude 325 V/m is directed in the negative y direction in Figure P20.1. The coordinates of point are (0.200, 0.300) m, and those of point are (0.400, 0.500) m. Calculate the electric potential difference using the dashed-line path. Figure P20.1arrow_forwardA positive point charge q = +2.50 nC is located at x = 1.20 m and a negative charge of 2q = 5.00 nC is located at the origin as in Figure P16.18. (a) Sketch the electric potential versus x for points along the x-axis in the range 1.50 m x 1.50 m. (b) Find a symbolic expression for the potential on the x-axis at an arbitrary point P between the two charges. (c) Find the electric potential at x = 0.600 m. (d) Find the point along the x-axis between the two charges where the electric potential is zero.arrow_forwardA filament running along the x axis from the origin to x = 80.0 cm carries electric charge with uniform density. At the point P with coordinates (x = 80.0 cm, y = 80.0 cm), this filament creates electric potential 100 V. Now we add another filament along the y axis, running from the origin to y = 80.0 cm, carrying the same amount of charge with the same uniform density. At the same point P, is the electric potential created by the pair of filaments (a) greater than 200 V, (b) 200 V, (c) 100 V, (d) between 0 and 200 V, or (e) 0?arrow_forward
- Two particles each with charge +2.00 C are located on the x axis. One is at x = 1.00 m, and the other is at x = 1.00 m. (a) Determine the electric potential on the y axis at y = 0.500 m. (b) Calculate the change in electric potential energy of the system as a third charged particle of 3.00 C is brought from infinitely far away to a position on the y axis at y = 0.500 m.arrow_forwardFor the arrangement described in Problem 26, calculate the electric potential at point B, which lies on the perpendicular bisector of the rod a distance b above the x axis. Figure P20.26arrow_forwardThe three charged particles in Figure P20.11 are at the vertices of an isosceles triangle (where d = 2.00 cm). Taking q = 7.00 C, calculate the electric potential at point A, the midpoint of the base. Figure P20.11arrow_forward
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