Recent content by awstinf

Alright I've figured it out. What was tripping me up is I was thinking that the overall voltage must change because the overall separation changed. But if I treat it as a equiv capacitor then the voltage is the same because its one equiv capacitor and the voltage is supplied by the battery and...

Sigh... Again. Can someone explain to what happens to charge and voltage in the scenario I listed.

Ah well. Someone help me get closer to understanding the answer?

"Three capacitors are connected in series across a DC power source that produces a constant potential difference. A fourth capacitor is now added in series with the other three. As a result of adding this capacitor, the total energy stored in the system of capacitors has B. decreased."...

If you double the area the capacitance increases by 2 voltage would not change but charge would increase by a factor of 2 I'm guessing now. If you double the separation while keeping voltage constant the charge would decrease? the only proof of that I can think of is Q=CV V stays the same and C...

Directly related? Higher voltage higher charge? via qk/r^2*d

I guess more of my question is what happens to charge and voltage. I'll do that but it's difficult for me to gain intuition that way without someone explaining it to me. It seems like if they are added in parallel the charge would increase and the capacitance would increase but again I am not...

Homework Statement Three capacitors are connected in series across a DC power source that produces a constant potential difference. A fourth capacitor is now added in series with the other three. As a result of adding this capacitor, the total energy stored in the system of capacitors...