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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 = 13 cm.   The free body diagram of q2 from Part 1 should look like this:   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. ?3→2>?1→2?3→2>?1→2 because q3>q1.    (a) Calculate the force on the test charge by the +6μC charge.

A test charge of +2μC is placed halfway between a charge of +6μC and another of +4μC separated by distance 2d = 13 cm.   The free body diagram of q2 from Part 1 should look like this:   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. ?3→2>?1→2?3→2>?1→2 because q3>q1.    (a) Calculate the force on the test charge by the +6μC charge.

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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 = 13 cm.

 

The free body diagram of q2 from Part 1 should look like this:

 

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.

?3→2>?1→2?3→2>?1→2
because q3>q1. 

 

(a) Calculate the force on the test charge by the +6μC charge.

 

A test charge of +2µC is placed halfway between a charge of +6μC and another of +4µC separated by distance 2d = 13 cm. +4uC 9₁ = +4 µC d The free body diagram of q₂ from Part 1 should look like this: +2μC F3-2 ✔N d q2 = +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 > F12 F3→2 > F1→2 q3 = +6 μC 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.
Transcribed Image Text:A test charge of +2µC is placed halfway between a charge of +6μC and another of +4µC separated by distance 2d = 13 cm. +4uC 9₁ = +4 µC d The free body diagram of q₂ from Part 1 should look like this: +2μC F3-2 ✔N d q2 = +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 > F12 F3→2 > F1→2 q3 = +6 μC 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.
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