Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This is a practice Physics Subject GRE problem.

A parallel-plate capacitor has plate separation $d$. The space between the plates is empty. A battery supplying voltage $V_0$ is connected across the capacitor, resulting in electromagnetic energy $U_0$ stored in the capacitor. A dielectric, of dielectric constant k, is inserted so that it just fills the space between the plates. If the battery is still connected, what are the electric field $E$ and the energy $U$ stored in the dielectric, in terms of $V_0$ and $U_0$?

I was able to figure out $E$ fairly easily. Since the voltage stays the same (by virtue of it being connected to a battery), $E = V/d$. I am having difficult understanding why the answer for $U$ is what it is though. The answer is $U = k^2U_0$. What equation(s) would give this result? I know $W = \frac{1}{2}VQ$, and with $V$ staying the same $Q$ must be the variable being multiplied by $k^2$. But $Q = CV$ so $C$ should also be increasing as $k^2$ but for a parallel plate capacitor $C$ only increases by a single $k$ when inserting a dielectric. How to resolve this discrepancy, or what am I missing?

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

The answer you've been given is wrong. The energy after inserting the dielectric is $kU_0$

share|improve this answer
    
I see now, I actually was looking at the wrong "correct" answer. The correct answer is indeed kUo. And that is the correct answer because of what I said above, that W = 1/2 VQ and Q = CV, so increasing C by a factor k increases W by a factor k. –  Joshua Jun 13 '13 at 2:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.