Answered: E=0 b+ b+ Q Search 1 Y FS F6 F7 L D

Science

Physics

E=0 b+ b+ Q Search 1 Y FS F6 F7 L D

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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The picture below shows two concentric rings. The outer ring has a charge of -300nC and a radius of 18mm. The inner ring has a linear charge density of 12uC/m and a radius of 9mm. What is the net electric field at point P, which is 30mm above both rings.
We have calculated the electric field due to a uniformly charged disk of radius R, along its axis. Note that the final result does not contain the integration variable r: R. Q/A 2€0 Edisk (x² +R*)* Edisk perpendicular to the center of the disk Uniform Q over area A (A=RR²) Show that at a perpendicular distance R from the center of a uniformly negatively charged disk of CA and is directed toward the disk: Q/A radius R, the electric field is 0.3- 2€0 4.4.1b
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Problems 4-5 refer to the following diagram and situation. A point charge (with charge Q) at the end of an insulating cord is observed to be in equilibrium in a uniform horizontal electric field of when the pendulum's angle with the vertical is 0 = 38.6°, is as shown in the Figure. The electric field is to the right and the strength of the electric field is 9500 N/C. The mass of the point charge is m = 0.0200 kg and the acceleration due to gravity is 9.8 m/sec². Problem 4: What is the tension on the cord? a. 0.251 N Problem 5: What is the charge Q? a. (1.65 x 10-5)C E b. 0.451 N b. (3.65 x 10-5)C 0 Q m c. 0.651 N C. (5.65 x 10-5) C d. 0.851 N d. (7.65 x 10-5)C
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Problem An electric charge Q is distributed uniformly along a thick and enormously long conducting wire with radius R and length L. Using Gauss's law, what is the electric field at distance r perpendicular to the wire? (Consider the cases inside and outside the wire) Solution To find the electric field inside at r distance from the wire we will use the Gauss's law which is expressed as PĒ - dÀ= We will choose a symmetric Gaussian surface, which is the surface a cylinder excluding its ends, then evaluate the dot product to obtain A = (Equation 1) Case 1: Inside the wire Since, r falls inside the wire, then all the enclosed charge must be: denc = On the other hand, the Gaussian surface inside the wire is given by A = Using Equation 1, the electric field in simplified form is E = Case 2: Outside the wire Since, r falls outside the wire, then, all the charge must be enclosed, thus denc = On the other hand, the Gaussian surface outside the wire is given by A = Using Equation 1, the electric…
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