Energy, Charge, Etc. in Capacitors

I'm studying to become a physics teacher, and as part of that effort, I'm reviewing The Princeton Review's "Cracking the AP Physics 2 Exam 2017 Edition." There's a section on pages 178-179 in the chapter on Direct Current Circuits that is very confusing in the last part of the explanation. I really would like your help in clarifying this. Here is the scenario and question. Thank you so much if you are able to help me.Scenario 1: You charge up a capacitor without a dielectric. You then insert a dielectric into a capacitor while still connected to the battery. The following happens: Capacitance C increases by a factor of ĸ, the dielectric constant.Charge Q increases by a factor of ĸ.Voltage V stays the same.Electric field E stays the same.Potential energy of the capacitor UC increases by a factor of ĸ.Scenario 2: You charge up a capacitor without a dielectric. You then disconnect the battery, and then add the dielectric. The following happens:C increases by a factor of ĸ.Q stays the same.V decreases by a factor of ĸ.E decreases by a factor of ĸ.UC decreases by a factor of ĸ.Question to you: What happens to C, Q, V, E and UC if the battery is then reconnected to the capacitor in Scenario 2 above? In other words: capacitor charged up without dielectric → battery disconnected → dielectric added → after everything in Scenario 2 has occurred, the battery is then reconnected to the capacitor. What then happens to C, Q, V, E and UC. Thank you in advance for your very kind reply to this question. Please note: After I pasted this question on to your website, I realized that the symbol for potential energy of the capacitor looks like "UC." The "C" is really a subscript following the capital letter "U." Thanks.
- Fred Moldofsky (age 61)
77598-2628

First, a slight correction to Scenario 1. It's not really the potential energy that goes up by a factor of Ä¸, it's the Helmholtz free energy, U-TS, where T is absolute temperature and S is entropy. For many typical high-ĸ materials, the change is mainly a reduction of S as electric dipoles line up in the field.

Now for your question:

Reconnecting the battery just brings all the values back to what the would have been in scenario 1, assuming that everything reaches thermal equilibrium. That seems to be the assumption throughout, which makes things simple.

Mike W.

(published on 09/04/2017)