1. A test charge of +2µC is placed halfway between a charge of +6µC and another of +4µCseparated by distance 2d = 9 cm.+4uCThe free body diagram of q2 from Part 1 should look like this:9₁ = +4 µCd35.52+2μCF3-2✔N92 = +2 μCF₁-2+6μС XThe force on q2 due to q₁ is repulsive and acts to the right. The force on q2 due to q3 isrepulsive and acts to the left.F3-2 > F1-293 =because 93 91.(a) Calculate the force on the test charge by the +6μC charge.Hint: The force in the positive x-direction is positive and the negative x direction is negative.✔N(b) Calculate the force on the test charge by the +4µC charge.(c) Calculate the net force on the test charge by both the charges.+6 μC

Given data - Charge q1 = +4μC = Columb Charge q2 = +2μC =…

1. A test charge of +2μC is placed halfway between a charge of +6μC and another of +4µC separated by distance 2d = 9 cm. +4uC The free body diagram of q2 from Part 1 should look like this: 9₁ = +4µC +2μC F3-2 35.52 q2 = +2 μC N F1-2 +6μС The force on q2 due to q₁ is repulsive and acts to the right. The force on q2 due to q3 is repulsive and acts to the left. F32 F1-2 (b) Calculate the force on the test charge by the +4μC charge. N (c) Calculate the net force on the test charge by both the charges. X because 9391. (a) Calculate the force on the test charge by the +6μC charge. Hint: The force in the positive x-direction is positive and the negative x direction is negative. N 93 = +6 μC

1. A test charge of +2μC is placed halfway between a charge of +6μC and another of +4µC separated by distance 2d = 9 cm. +4uC The free body diagram of q2 from Part 1 should look like this: 9₁ = +4µC +2μC F3-2 35.52 q2 = +2 μC N F1-2 +6μС The force on q2 due to q₁ is repulsive and acts to the right. The force on q2 due to q3 is repulsive and acts to the left. F32 F1-2 (b) Calculate the force on the test charge by the +4μC charge. N (c) Calculate the net force on the test charge by both the charges. X because 9391. (a) Calculate the force on the test charge by the +6μC charge. Hint: The force in the positive x-direction is positive and the negative x direction is negative. N 93 = +6 μC

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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Question

1. A test charge of +2µC is placed halfway between a charge of +6µC and another of +4µC
separated by distance 2d = 9 cm.
+4uC
The free body diagram of q2 from Part 1 should look like this:
9₁ = +4 µC
d
35.52
+2μC
F3-2
✔N
92 = +2 μC
F₁-2
+6μС X
The force on q2 due to q₁ is repulsive and acts to the right. The force on q2 due to q3 is
repulsive and acts to the left.
F3-2 > F1-2
93 =
because 93 91.
(a) Calculate the force on the test charge by the +6μC charge.
Hint: The force in the positive x-direction is positive and the negative x direction is negative.
✔N
(b) Calculate the force on the test charge by the +4µC charge.
(c) Calculate the net force on the test charge by both the charges.
+6 μC

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