# Total charge in a circuit/determining parallel or series In this image, C1 is 1 microfarad and C2 is 3 microfarads and both capacitors are charged with a potential difference of V = 100V.

S1 and S1 are closed.

The solution says this is in parallel, but I don't see how it is. It looks like it's set up in series to me. But even if it is in parallel, I thought the total charge would be 400 C, but the solution says the total charge is 200. Why is this?

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tiny-tim
Homework Helper
Hi PhizKid! The diagram show "before", with the switches open.

It shows that the two capacitors have been charged the "opposite" way round …

so what will happen to the charge when the switches are closed? I think I am having issues understanding capacitor theory. When the switches are closed, it seems to me that both capacitors hold some charges, but I'm not sure how the charges move around (how the current flows?)

tiny-tim
Homework Helper
forget that they're capacitors

the two top plates are connected by a wire, one has positive charge and the other has negative charge … what is the total? Okay, so the both the sets of top two plates and bottom two plates should each have a net charge of zero?

So since q = CV, the left set of plates each have charges of 100 C, and the right set of plates has 300 C.

So I can choose either the upper set of connected plates or the lower set of connected plates? For the case in the upper set of plates, I'd have +100 C - 300 C = -200 C. Does it matter if there's a negative?

gneill
Mentor
Okay, so the both the sets of top two plates and bottom two plates should each have a net charge of zero?

So since q = CV, the left set of plates each have charges of 100 C, and the right set of plates has 300 C.

So I can choose either the upper set of connected plates or the lower set of connected plates? For the case in the upper set of plates, I'd have +100 C - 300 C = -200 C. Does it matter if there's a negative?
100 C and 300 C are whacking great charges. Better check the units on your calculations!

But you've got the right idea. When two groups of opposite-signed charges are allowed to "see" each other through a conducting path, unless they are being held apart by some other electric force they will mix, pair up, and cancel. Whatever charges are left over constitute the final charge. Yes it can be negative if there were more negative charges than positive charges to begin with.

Okay, but if the question is asking the charge on each capacitor, how do I determine the signs for the net charge on both capacitors? I'm getting two different answers (-2 and 2, depending on which set of plates I choose to do the math with).

gneill
Mentor
Okay, but if the question is asking the charge on each capacitor, how do I determine the signs for the net charge on both capacitors? I'm getting two different answers (-2 and 2, depending on which set of plates I choose to do the math with).
Usually capacitors end up with opposite polarity charges on their plates. The phrase "the charge on a capacitor is 3 micro coulombs" implies that there is +3 μC on one plate and -3 μC on the other.

The positive polarity of the potential difference is associated with the positive charge.

Note that you'll have to determine the final allocation of the total charges on the two capacitors if you're going to say what the charge in each is. This is where the "parallel connection" point of view comes into play...