Why Does Using W=QV Give the Wrong Answer for Work Done in Charging a Capacitor?

[CaneAA]

Calculate the work done by a 3.0V battery as it charges a 7.8-microFarad capacitor in the flash unit of a camera.

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.

 

[zhermes]

Good question CaneAA (and welcome to physics forums).

The problem is that the voltage and charge aren't constant, thus you would have to integrate (the differential form of) that equation to solve. If you're not familiar with calculus, I wouldn't worry about it; if you're curious however, check this out http://en.wikipedia.org/wiki/Capacitor#Energy_storage and it shows you that by integrating you end up with the same equation that you used to get the correct answer.

 

[CaneAA]

Thank you, that made sense. :)