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Total charge in a capacitive circuit

  1. Nov 17, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.

    I am confused about an example in my textbook related to charge stored by a capacitor

    A 100 uF capacitor and a 700 uF capacitor are placed in series, along with a cell of 3V.
    Calculate the p.d. across the 100 uF capacitor.

    The book states the solution as follows:
    "Using the reciprocal formula, the capacitance of the whole circuit is 87.5 uF.
    This means the total charge stored Q = VC = 3 * 87.5e-6 = 2.63e-4

    The total charge stored is equal to the charge stored by the 100 uF capacitor, which is equal to the charge stored by the 700 uF capacitor.
    So V = Q/C = 2.63e-4/100e-6 =2.6V"

    I understand all of this apart from the last section. I understand that the charge stored on both capacitors will be the same, but I do not understand how they can both be equal to the total charge - surely the total charge is the sum of the charges stored on both capacitors?

    Any help much appreciated.
  2. jcsd
  3. Nov 17, 2016 #2

    Doc Al

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    Staff: Mentor

    Try thinking of the arrangement as being one giant capacitor instead of two. One plate of that giant capacitor is the left plate of C1, the other the right plate of C2. What are the charges on those plates?
  4. Nov 17, 2016 #3


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    Staff: Mentor

    The phasing of the paragraph is not the best, perhaps, and glosses over some technical details regarding how capacitors "store" charge. As you point out, each of the capacitors ends up with what we call a charge Q of the same size as the total charge. It gets even more problematic if you consider that what we call a charge on a capacitor consists of equal and opposite charges on its plates, actually summing to zero net charge on the capacitor as a whole.

    In this case the total charge stored is taken to mean the charge moved by the voltage source and placed on the effective (equivalent) capacitance.


    So in the above circuit the voltage source V moves a total charge Q "onto" the capacitance comprised of C1 and C2 in series. But the net charge for the two is in reality zero, the voltage source having also collected Q via its negative terminal while supplying Q from its positive terminal.

    So when we say that a capacitor has a charge Q on it, we really mean that a charge Q has been removed from one plate and placed on the other, generally via some external path (wiring) and EMF doing the moving.
  5. Nov 18, 2016 #4
    Thanks very much - that explained it perfectly.
    V grateful :)
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