- #1
CaneAA
- 13
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Calculate the work done by a 3.0V battery as it charges a 7.8-microFarad capacitor in the flash unit of a camera.
I realize that Work = Energy. And by using U = .5CV^2 I can easily come up with the answer.
But when I first started doing the problem, I tried doing it by a more convoluted approach--which made sense to me, but I got the wrong answer. I just wanted to know why the following way of solving the problem doesn't also work:
W = U (energy)
V = 3 V
C = 7.8 x 10^-6 F
Using the equation C = Q/V, I solved for Q (charge). Q= 2.34 x 10^-5 C.
Since W = QV, I multiplied (2.34 x 10^-5)(3) = 7.02 x 10^-5 as my answer (which is wrong). I wanted to know what is wrong with my reasoning, so I don't make the same mistake again.
Thank you.
The Attempt at a Solution
I realize that Work = Energy. And by using U = .5CV^2 I can easily come up with the answer.
But when I first started doing the problem, I tried doing it by a more convoluted approach--which made sense to me, but I got the wrong answer. I just wanted to know why the following way of solving the problem doesn't also work:
W = U (energy)
V = 3 V
C = 7.8 x 10^-6 F
Using the equation C = Q/V, I solved for Q (charge). Q= 2.34 x 10^-5 C.
Since W = QV, I multiplied (2.34 x 10^-5)(3) = 7.02 x 10^-5 as my answer (which is wrong). I wanted to know what is wrong with my reasoning, so I don't make the same mistake again.
Thank you.