A test charge of +2μC is placed halfway between a charge of +6μC and another of +4µC separated by distance 2d = 10 cm. +4μC d The free body diagram of q2 from Part 1 should look like this: -35.66281 9₁ = +4 μC 23.77521 d -11.8876 +2μC F3-2 N ! No, that's not the correct answer. q2 = +2 μC 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. F1-2 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. X N (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. +6μС F32 > F1→2 X q3 = +6 μC

A test charge of +2μC is placed halfway between a charge of +6μC and another of +4µC separated by distance 2d = 10 cm. +4μC d The free body diagram of q2 from Part 1 should look like this: -35.66281 9₁ = +4 μC 23.77521 d -11.8876 +2μC F3-2 N ! No, that's not the correct answer. q2 = +2 μC 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. F1-2 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. X N (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. +6μС F32 > F1→2 X q3 = +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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A test charge of +2μC is placed halfway between a charge of +6μC and another of +4µC separated
by distance 2d = 10 cm.
+4uC
d
The free body diagram of q₂ from Part 1 should look like this:
9₁ = +4 µC
-35.66281
23.77521
-11.8876
d
+2μC
F3-2
N
! No, that's not the correct answer.
q2 = +2 μC
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.
F1-2
N
(b) Calculate the force on the test charge by the +4μC charge.
X N
(c) Calculate the net force on the test charge by both the charges.
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.
+6μС
F3→2 > F1→2
X
93 = +6 μC

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