1) Two charges are a distance of 0.3 m apart. Charge #2 is to the right of charg

Question

1) Two charges are a distance of 0.3 m apart. Charge #2 is to the right of charge #1. Charge #2 (which is twice as large as charge #1) experiences a force the size of 23 pN due to charge #1. Assume that there are no other interactions present.
If the charges are instead separated by a distance of 0.6 m apart, then calculate the size of the force that charge #1 experiences. (include 3 significant figures in your answer)

2)Two charges are a distance of 0.2 m apart. Charge #2 is to the right of charge #1. Charge #2 (which is twice as large as charge #1) experiences a force the size of 18 pN due to charge #1. Assume that there are no other interactions present.
If the charges are instead separated by a distance of 0.6 m apart and the size of charge #1 is doubled, then calculate the size of the force that charge #1 experiences. (include 3 significant figures in your answer)

Explanation / Answer

The formula that applies here is F = kqq/r2

Part 1)

Since we are doubling the distance, the force will be reduced by a factor of four

F = 23/4 = 5.75 pN

Part 2)

In this case, the distance is triples, so 3 squared is 9. Charge 1 is doubled.

The net change is 2/9, so...

F = (2/9)(18) = 4.00 pN

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