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I know that a capacitor basically stores charge, and preserves Voltage... however I'm working with a limited practice lab manual for a summer University Physics class. Yeah, it's not going very well right now.

For two different capacitors in a circuit connected in series, are their stored charges identical? Are their Voltages the same across all capacitors? And would the sum of the voltage equal the battery?

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closed as off-topic by John Rennie, sammy gerbil, heather, CuriousOne, peterh Aug 5 '16 at 16:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome to Physics.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. 4) This is not a homework help site. Check that your question follows the rules in the help center and show work. $\endgroup$ – heather Aug 4 '16 at 14:45
  • $\begingroup$ Have you tried an internet search for materials on capacitors in series/parallel? Or opened a physics text book? We expect you to make a reasonable effort to use available resources before you ask here. $\endgroup$ – sammy gerbil Aug 4 '16 at 18:47
  • $\begingroup$ Wow talk about unfriendly. You make it seem like a student isn't allowed to ask a question on a topic he can't find answers for already. The answer is -yes-. Yes I have. However my textbook does not actually spend much time describing the concepts of a compositor in a circuit. Most of the chapter is dedicated to inductors, and the next chapter is AC, followed by Optics. And any results I find online don't answer the specific question. It simply shows the equivalence of a capacitor in a circuit. $\endgroup$ – moonshineTheleocat Aug 4 '16 at 22:56
  • $\begingroup$ @moonshineTheleocat, I'd like to explain something real quick (and I mean this in the nicest possible way). We allow homework questions, but we ask that they explain what work the question-asker has done to solve the problem. I personally didn't think that you had really done that, even with the explanation in your comment, so I voted to close this question. You are free to disagree with me; I just wanted to explain the reasoning behind the close votes. I am glad you got an answer. $\endgroup$ – heather Aug 5 '16 at 0:27
  • $\begingroup$ @heather That's fine, that's a general consensus that I wasn't aware of for this portion of stackexchange. My bad. However, I'd personally suggest that you kindly inform someone of this courtesy, before doing some sort of acrobatic pirouette off the handle and into the deep end like some deranged attack dog. All I needed was a hint, or the concept. As this one problem is critical to solving the thirty something other problems. The Professor doesn't answer emails. So your're often SOLed the moment you start working on homework. $\endgroup$ – moonshineTheleocat Aug 5 '16 at 1:46
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In series, they both have the same charge. In a strictly series circuit, the current is the same everywhere along the wire, and because $I = dq/dt$, if $I$ is the same across each capacitor by definition $dq/dt$, and hence $q$, is the same.

The capacitors don't necessarily have the same voltage; voltage is given by $V=Q/C$, so if the capacitors have different capacitances the voltage will differ. Your last statement is correct because of Kirchhoff's loop rule: $\sum_iV_i = 0$, and in this case this turns to $\varepsilon + V_1 + V_2 = 0$, so the magnitude of $\varepsilon$ equals that of $V_1 + V_2$.

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    $\begingroup$ Thank you! I'd give you a plus one if I could, but it stack exchange still has that stupid rule. $\endgroup$ – moonshineTheleocat Aug 4 '16 at 22:53

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