# Q & A: energy, charge, etc. in capacitors

Q:
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
A:

First, a slight correction to Scenario 1. It's not really the potential energy that goes up by a factor of ĸ, 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.