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Potential Difference between two parallel charged plates.

  1. Oct 28, 2015 #1
    How do we find out the potential difference between two equal and opposite charged (conducting) parallel plates mathematically? Let the charge on a plate be 'Q', Total area of a plate be 'A', the distance between the plates be 'd'.
    I need a direct mathematical solution please, I've come across various indirect solutions involving the product of the electric field and distance 'd'.

    The solution may include use of integration explained in detail.

    Please help me with it. My mind has blown off searching books and internet, always getting indirect solutions.
     
  2. jcsd
  3. Oct 28, 2015 #2

    Dale

    Staff: Mentor

    Could you point to one of these indirect solutions? From what you describe it sounds like you just divide by d to get the direct solution.
     
  4. Oct 28, 2015 #3
    I am not sure what you mean by an indirect solution. Usually the solution is derived from the definition of potential. Here are the steps:
    1. Take a point charge from one plate to the other, calculate the work done by the electric field, which you know, since you know the charge and the area and the separation.
    2. The negative of the work is the change in potential energy.
    3. Divide the change in potential energy with the charge that you moved.
    That gives the potential difference.
     
  5. Oct 28, 2015 #4
    By indirect solution i meant,
    V= E*d
    Where E is the electric field between the two plates.
    And d is the distance of seperation of plates.
    And by a direct solution I meant finding the potential difference just by using the variables, charge Q on the plate, the distance of seperation d between the plates and Area A of the plate.
    And yeah it may include the use of Integration.
     
  6. Oct 29, 2015 #5
    You are supposed to already know that E = Q/ε0A. So there is no difference between your so called indirect and direct solutions. The integration is a trivial part of the solution, which arises when you calculate the work or potential energy. It is trivial in this case because the electric field is uniform.
     
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