A parallel-plate capacitor with plate area 2.0 cm^2, air-gap .5mm, is connected to a 12V battery, if the air-gap is pulled to .75mm, how much work is required?
calculated Q=4.2E-11 C, V(d=0.75)=18V, delta PE=Q*delta V = 2.52E-10 J, but the answer is 1.3E-10 J, which could be a half of my answer. PE/2? why? Would you please explain this to me? Thanks
Ok, so the thing you are changing when you pull a parallel plate capacitor apart is the capacitance (it's ability to store energy).
The capacitance depends on distance like:
C=εo*(A/d)
so clearly the capacitance will decrease if the distance between the plates is increased. Let's consider the capacitance before (C), and after (C'):
C=εo*(A/d)
C'=εo*(A/d')
Ok. So now let's consider how this will effect the energy stored in the capacitor. This can be written in 3 different ways (by substituting Q=CV into the first):
PE=(1/2)QV=(1/2)CV2=(1/2)Q2/C
Which of these should we use? Well, since the plates are connected to a 12 V battery, the voltage will be the same before and after moving the plates. Note that the charge may change so we can't use the expressions with Q in them because we don't know how Q is changing (ok, yes, we could figure it out, but it would be equivalent to using the second expression, so let's not bother).
All right, so let's put it together:
W=-ΔPE=-((1/2)C'V2-(1/2)CV2)
W=-((1/2)εo*(A/d')V2-(1/2)εo*(A/d)V2)
W=-εo*A*V2/2((1/d')-(1/d))
Plugging in the numbers given in the equation above for the work done, I get: 8.49e-11J (using 12V). You also write 18V in your question, with that number I get an answer: 1.91e-10J. Sorry to muddy the waters with more answers... the method should be correct. (maybe you can recheck which numbers the question uses?)
If you haven't already, I'd suggest taking a look at the similar example (though with constant charge not voltage) posted on the main page: Ch. 17, #50.
Hope that helps. Feel free to ask questions by posting comments.
No comments:
Post a Comment