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Homework Help: Capacitor problem from Irodov (3.122)

  1. Dec 3, 2015 #1


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    Gold Member

    1. The problem statement, all variables and given/known data
    What charges will flow after shorting of the switch (Sw) in the circuit below through sections 1 and 2 in the direction indicated by the arrows?


    2. Relevant equations
    Kirchoff's laws

    3. The attempt at a solution
    Initially, both capacitors will have a charge of $$Q_I=\frac{C_1C_2}{C_1+C_2}E$$
    Finally, after shorting the switch, potential difference across ##C_1## becomes zero. So final charge on ##C_1## is zero and final charge on ##C_2## is ##C_2E##.
    So the charge that flows through section 1 is $$C_2E-\frac{C_1C_2}{C_1+C_2}E$$ But answer given is ##C_2E##.


    Please check if I am correct.
    Some charge has to flow through section 1 to make the charge on right plate of C1 zero. That charge is $$-\frac{C_1C_2}{C_1+C_2}E$$ since initially, the right plate had a positive charge of magnitude ##Q_I##.
    Now, to make charge on left plate of C1 zero, a charge of
    $$\frac{C_1C_2}{C_1+C_2}E$$ will flow from upper plate of C2. (Assuming the charge can't flow through the battery). So this makes C2 lose all its initial charge and after shorting the switch, ##C_2E## flows in section 1. Hence I got the given answer.
    But is my assumption correct?
    Last edited: Dec 3, 2015
  2. jcsd
  3. Dec 3, 2015 #2


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    Staff: Mentor

    You've come to the right result. You can play bit with sources of charges when you are balancing the circuit for some end result, and you can imagine negative charges moving, too if you want. So long as you always conserve charge the math will always work out and give you the correct result. Here's how I would implement the strategy that you've given.

    When I do these I usually label each capacitor plate with its initial charge, and only "move" positive charges just like conventional current (you can create a negative charge on a plate by taking a positive charge away from the plate). So if we let Q be the initial charge that you calculated for the two capacitors, each capacitor will have +Q on one plate and -Q on its other plate. So the setup looks like this:


    Now when the switch closes you know that C1 should end up with no charge and C2 with ##C_2 E## as a charge. Well clearly we can "steal" the Q required to cancel the -Q on C1 from the top plate of C2. But that means pulling Q onto the bottom plate of C2 at the same time, leaving C2 uncharged, too (don't worry, we're not done!). This has the benefit of satisfying C1's requirements and current has only had to flow in the right hand loop, i.e. through point "2" (see the right hand loop and red indications in the figure below).

    Next, move ##C_2 E## onto C2 using the left hand voltage source and loop. That will be the only current that had to flow through "1".


    All that's left is to note the charges that flowed past points 1 & 2, being sure to note their directions of travel.
  4. Dec 3, 2015 #3


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    Homework Helper

    No, this is the total change of charge on C2. And charge flows through two ways on it.
    The battery behaves as a short for the current. current does flow through a battery.
    After shorting the switch, the whole charge of C1 flows off according to the red arrow,. C2 gains charge by the blue arrow and losses charge by the red arrow.


    Edit: gneill was much faster!
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