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1. A test charge of +2µC is placed halfway between a charge of +6µC and another of +4µC separated by distance 2d = 7 cm. +4uC d d The free body diagram of q₂ from Part 1 should look like this: 91 = +4 μC +2μC F3-2 ✔N 92 = +2 μC F₁-2 +6µC 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 = +6 μC O 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. N (c) Calculate the net force on the test charge by both the charges.

1. A test charge of +2µC is placed halfway between a charge of +6µC and another of +4µC separated by distance 2d = 7 cm. +4uC d d The free body diagram of q₂ from Part 1 should look like this: 91 = +4 μC +2μC F3-2 ✔N 92 = +2 μC F₁-2 +6µC 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 = +6 μC O 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. N (c) Calculate the net force on the test charge by both the charges.

College Physics
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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1. A test charge of +2µC is placed halfway between a charge of +6µC and another of +4µC separated by distance 2d = 7 cm. +4uC d d The free body diagram of q₂ from Part 1 should look like this: 91 = +4 μC +2μC F3-2 ✔N 92 = +2 μC F₁-2 +6µC 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 = +6 μC O 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. N (c) Calculate the net force on the test charge by both the charges.
Transcribed Image Text:1. A test charge of +2µC is placed halfway between a charge of +6µC and another of +4µC separated by distance 2d = 7 cm. +4uC d d The free body diagram of q₂ from Part 1 should look like this: 91 = +4 μC +2μC F3-2 ✔N 92 = +2 μC F₁-2 +6µC 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 = +6 μC O 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. N (c) Calculate the net force on the test charge by both the charges.
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