Princeton Review Physics FSQ Set 6 Question 7

[jddoc2015]

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Can anyone explain to me why in the PR explanation to this problem they are using 1/rf-1/r0 as opposed to just rf-r0. If you just use the difference in radii in the formula you get the wrong answer but I don't get where they're getting the formal they are using. Thanks!

Let W1 be the work required to move two identical point charges from a separation distance of 4 cm to a distance of 3 cm. Let W2be the work required to move the same charges from a separation distance of 3 cm to 2 cm. Which of the following is true?

A.
W1 < W2

Correct Answer

Explanation:
A. Since identical charges repel each other, the work required to move them closer together is positive and equal to the change in potential energy. Since PE = kq1q2 / r, W = ΔPE = kq1q2(1 / rf – 1 /r0). Comparing W1 and W2, we only need to look at the expression within the parenthesis. For W1, this equals 1 / (3 cm) – 1 / (4 cm) = 1 / (12 cm); for W2, this equals 1 / (2 cm) – 1 / (3 cm) = 1 / (6 cm). We can see that W1 < W2.

 

[Jepstein30]

PE = kq1q2/r
ΔPE = PEf - PEi
PEf = kq1q2/rf
PEi = kq1q2/ri
ΔPE = kq1q2/rf - kq1q2/ri = kq1q2 (1/rf - 1/ri)

You can't just use rf-ro because that's not computing ΔPE properly, as seen above.

W1 = ΔPE1 = kq1q2(1/3-1/4) = kq1q2(1/12)
W2 = ΔPE2 = kq1q2(1/2-1/3) = kq1q2(1/6)
therefore, W2 > W1 since dividing the same number (kq1q2) by a smaller number will yield a larger result.

 

[patrickmbyrne]

The closer two charges are the more work is required to bring them closer. This implies -> W1<W2 e.g. If you have two particles separated by an infinite distance then there is no force, but if you have two particles separated by almost no distance an infinite amount of work is required to bring them closer together.
edit NO MATH HERE 🙂

 

[jddoc2015]

got it, thank you both!