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Theoretical electrodynamic cycle 2

  1. Jul 5, 2011 #1
    Here is a second thought experiment I designed to test if I understand certain concepts concerning electrostatics correctly. Help will be appreciated.

    There must be an error somewhere in the explanation below; I am giving the explanation as if “factual” only to explain how I understand it at present.

    Consider the setup shown in the figure for step 1. The potential difference source (cell V1) charged capacitor C1 to full capacity (voltage over capacitor equal to voltage over cell).

    In step 2 switch S1 is opened, and S3 is connected next. Work is done on resistor R2 as charges separate between the plates to establish an opposing electric field so that the potential difference between the two conductive plates of capacitor C2 is zero.

    Step 3 involves opening switch S3 and closing switch S2 afterwards. Work is done on resistor R1 as the current flows to equalize the charges on the plates of capacitor C1.

    Step 4 is to close switch S3. Work will again be done on resistor R2 as the current flows to equalize the charges on the plates. Get back to step 1 by opening switch S3 and S2, and closing switch S1. The cell will do work to charge capacitor C1.

    Assuming that there is no dielectric effect on capacitor C1 due to the charged plates of C2 in the middle of it (cannot see why there should be as the electric field due to the charged plates C2 is supposed to be zero to the left and right of capacitor C2), the work done in step 3 is equal to the work done on the cell to recharge capacitor C1 (assuming a fully efficient, reversible charging and discharging process on the capacitor). This leaves the work done on resistor R2 in steps 2 and 4 as excess from the cycle, something that is not supposed to happen.

    I will appreciate a pointer as to what I overlook.
     

    Attached Files:

    Last edited: Jul 5, 2011
  2. jcsd
  3. Jul 24, 2011 #2
    I have not received any reply to date, I assume that it has been overlooked by those who are able to answer it. Is it possible that this post is better suited for a different forum?
     
  4. Aug 18, 2011 #3
    It seems you assumed that the total charge on c1 is not influenced by the potential over c2 let met draw you a better pic of step 1 http://img40.imageshack.us/img40/7471/loledg.png [Broken].
    Now you will see 3 caps , which is exactly the equivalent of your system.

    step 2
    here you open s1 and then close s3.
    basically you discharge cap b through r2

    step 3
    here you open s3 and close s2
    you discharge the remaining energy that was stored in cap a and cap b into R1.


    Thus the total amount of energy lost is the sum of all the energies in all the caps. wich means the is no gain of energy.


    oh yeah i sent you a mail :}
     
    Last edited by a moderator: May 5, 2017
  5. Aug 18, 2011 #4
    Thank you, this makes it clear.
     
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