Energy Change in a Capacitor by Inserting Dielectric - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-06-16T00:59:55Z https://physics.stackexchange.com/feeds/question/67791 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/67791 1 Energy Change in a Capacitor by Inserting Dielectric NeutronStar https://physics.stackexchange.com/users/25680 2013-06-12T05:30:25Z 2013-06-12T08:39:03Z <p>This is a practice Physics Subject GRE problem.</p> <blockquote> <p>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 <em>k</em>, 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 $U_0$?</p> </blockquote> <p>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?</p> https://physics.stackexchange.com/questions/67791/-/67802#67802 1 Answer by Mark Eichenlaub for Energy Change in a Capacitor by Inserting Dielectric Mark Eichenlaub https://physics.stackexchange.com/users/74 2013-06-12T08:38:31Z 2013-06-12T08:38:31Z <p>The answer you've been given is wrong. The energy after inserting the dielectric is $kU_0$</p>