A test charge of +2μC is placed halfway between a charge of +6µC and another of +4µC separated by distance 2d = 11 cm. +6μС X +4µC d The free body diagram of q2 from Part 1 should look like this: 9₁ = +4 µC +2μC F3-2 N q2 = +2 μC F1-2 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. q% = +6 μC F3+2 > 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. 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.

Physics for Scientists and Engineers: Foundations and Connections
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Chapter24: Electric Fields
Section: Chapter Questions
Problem 56PQ
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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 = 11 cm.
+4uC
9₁ = +4 μC
The free body diagram of q2 from Part 1 should look like this:
+2μC
F3-2
d
N
q2 = +2 μC
F1-2
F3→2 > 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.
(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
93 = +6 μC
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
Transcribed Image Text:A test charge of +2µC is placed halfway between a charge of +6µC and another of +4μC separated by distance 2d = 11 cm. +4uC 9₁ = +4 μC The free body diagram of q2 from Part 1 should look like this: +2μC F3-2 d N q2 = +2 μC F1-2 F3→2 > 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. (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 93 = +6 μC 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
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