4. Two charged objects with charges q; = - 500 µC and q; = - 400 µC are placed 0.50 m with each other. a. What is the electric force on q, exerted by q;? b. What is the electric force on q; exerted by q,? 0.50 m q1 = - 500 µC q2 = - 400 µC

4. Two charged objects with charges q; = - 500 µC and q; = - 400 µC are placed 0.50 m with each other. a. What is the electric force on q, exerted by q;? b. What is the electric force on q; exerted by q,? 0.50 m q1 = - 500 µC q2 = - 400 µC

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter22: Electric Fields

Section: Chapter Questions

Problem 19P: Three point charges are located on a circular arc as shown in Figure P22.19. (a) What is the total...

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Particle A of charge 3.00 104 C is at the origin, particle B of charge 6.00 101 C is at (4.00 m, 0), and particle C of charge 1.00 104 C is at (0, 3.00 in). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C? (b) What is the y component of the force exerted by A on C? (c) Kind the magnitude of the force exerted by B on C. (d) Calculate the x component of the force exerted by B on C. (e) Calculate the y component of the force exerted by B on C. (f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. (g) Similarly, find the y component of the resultant force vector acting on C. (h) Kind the magnitude and direction of the resultant electric force acting on C.
(a) Two point charges q1 and q23.00 m apart, and their total charge is 20 C. (a) If the force of repulsion between them is 0.075N, what are magnitudes of the two charges? (b) If one charge attracts the other with a force of 0.150 N, what are the magnitudes of the two charges? Note that you may need to solve a quadratic equation to reach your answer.
Earth has a net charge that produces an electric field of approximately 150 N/C downward at its surface. (a) What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? (b) What acceleration will the field produce on a free electron near Earth’s surface? (c) What mass object with a single extra electron will have its weight supported by this field?
Particle A of charge 3.00 104 C is at the origin, particle B of charge 6.00 104 C is at (4.00 m, 0), and particle C of charge 1.00 104 C is at (0, 3.00 m). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C? (b) What is the y component of the force exerted by A on C? (c) Find the magnitude of the force exerted by B on C. (d) Calculate the x component of the force exerted by B on C. (e) Calculate the y component of the force exerted by B on C. (f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. (g) Similarly, find the y component of the resultant force vector acting on C. (h) Find the magnitude and direction of the resultant electric force acting on C.
(a) Two point charges totaling 8.00 C exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the charge on each? (b) What is the charge on each if the force is attractive?
Assume the charged objects in Figure OQ23.10 are fixed. Notice that there is no sight line from the location of q2 to the location of q1. If you were at q1, you would be unable to see q2 because it is behind q3. How would you calculate the electric force exerted on the object with charge q1? (a) Find only the force exerted by q2 on charge q1. (b) Find only the force exerted by q3 an charge q1. (c) Add the force that q2 would exert by itself on charge q1 to the force that q3 would exert by itself on charge q1. (d) Add the force that q3 would exert by itself to a certain fraction of the force that q2 would exert by itself. (e) There is no definite way to find the force on charge q1.
A parallel-plate capacitor has a charge Q and plates of area A. What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is E = Q/A0, you might think the force is F = QE = Q2/A0. This conclusion is wrong because the field E includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F = Q2/2A0. Suggestion: Let C = 0A/x for an arbitrary plate separation x and note that the work done in separating the two charged plates is W=Fdx.
(a) What is the electric field 5.00 m from die center of the terminal of a Van de Graaff with a 3.00-mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal? (b) At this distance, what force does the field exert on a 2.00C charge on the Van de Graaff’s belt?
(a) What is the electric field 5.00 m from the center of the terminal of a Van de Graaff with a 3.00 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal? (b) At this distance, what force does the field exert on a 2.00 C charge on the Van de Graaff’s belt?
Panicle A of charge 3.00 104 C is at the origin, particle B of charge 6.00 104 C is at (4.00 m, 0), and panicle C of charge 1.00 104 C is at (0, 3.00 m). (a) What is the x-component of the electric force exerted by A on C? (b) What is the y-component of the force exerted by A on C? (c) Find the magnitude of the force exerted by B on C. (d) Calculate the x-component of the force exerted by B on C. (e) Calculate the y-component of the force exerted by B on C. (f) Sum the two x-components to obtain the resultant x-component of the electric force acting on C. (g) Repeat part (f) for the y-component. (h) Find the magnitude and direction of the resultant electric force acting on C.
Two point charges attract each other with an electric force of magnitude F. If the charge on one of the particles is reduced to one-third its original value and the distance between the particles is doubled, what is the resulting magnitude of the electric force between them? (a) 112F (b) 13F (c) 16F (d) 34F (e) 32F
Assume the charged objects in Figure OQ19.15 are fixed. Notice that there is no sight line from the location of q2 to the location of q1. If you were at q1, you would be unable to see q2 because it is behind q3. How would you calculate the electric force exerted on the object with charge q1? (a) Find only the force exerted by q2 on charge q1. (b) Find only the force exerted by q3 on charge q1. (c) Add the force that q2 would exert by itself on charge q1 to the force that q3 would exert by itself on charge q1. (d) Add the force that q3 would exert by itself to a certain fraction of the force that q2 would exert by itself. (e) There is no definite way to find the force on charge q1. Figure OQ19.15