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?


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

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  • $\begingroup$ 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. $\endgroup$ – NeutronStar Jun 13 '13 at 2:23

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