Look at example 2 on p. 535 in the 9th edition or 487 in the 10th edition, and c

Question

Look at example 2 on p. 535 in the 9th edition or 487 in the 10th edition, and consider what would happen if you did not know the distance between the charges, however you did know that the distance was twice as far as a distance you had previously computed the force for. What would be true about the force on the charges when they are twice as far apart compared with the force you'd already computed? When they are at the distance twice as far away, the force of attraction would be twice as strong. When they are at the distance twice as far away, the force of attraction would be four times stronger. When they are at the distance twice as far away, the force of attraction would be half as strong. When they are at the distance twice as far away, the force of attraction would be one-fourth as strong. UESTION 2 Consider Example 5 (on page 538 in the 9th edition or 490 in the 10th edition). What would happen to the net force if q1 were negative instead of positive, but with the same magnitude of charge? The net force vector would be in the third quadrant (i.e., pointing 'southwest) The net force vector would be in the second quadrant (i.e., pointing "northwest') The net force vector would be in the fourth quadrant i.e., pointing "southeast') The net force vector would be zero because all the charges would repel away from one another

Explanation / Answer

1)Correct Ans:

When they are at the distance twice as far away, the force of attarction would be one-fourth as strong.

Reason:

F = k q1q2/r^2

F is inversely proportinal to square of distance. so if r is doubled, F becomes 1/4

2)figure required, the solution is not enough, need to see the alignment of q1,q2,q3.

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