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Parallel plate capacitor's charge

  1. Feb 11, 2012 #1
    http://img94.imageshack.us/img94/4392/ppcapqs2.png [Broken]

    http://img267.imageshack.us/img267/5059/ppcapqs1.png [Broken]

    In this question the charge is calculated by isolating Q in E=surface charge density/absalon not

    My question is, the electric field of an infinite charged plane is E=surface charge density/(2*absalon naught)

    because the electric field in center of parallel plate capacitor are parallel we add

    surface charge density/(2*absalon naught)+surface charge density/(2*absalon naught)

    to give us E between the electrodes to be surface charge density/absalon not

    but when that question uses that formula to find the charge, shouldnt the charge found be divided by 2 to get the charge on each electrode since that formula we initially created by adding the E of both electrodes n this question asks to find charge on each electrode...

    so what i am really saying is that why isnt the Q in that formula (in that link) charge of both electrodes (their magnitude ofcourse) why is that Q the charge on one electrode?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 12, 2012 #2


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    As you said, the electric field due to one plate is surface charge density/(2*absalon naught) So this is equal to Q/(2A*epsilon naught) (where A is area, and Q is charge on one plate). And the total electric field is twice this (since there are two oppositely charged plates), so it is Q/(A*epsilon naught)
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