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Homework Help: Electrostatics point charges problem

  1. Jan 2, 2010 #1
    1. b]1. The problem statement, all variables and given/known data
    Two large metal plates are held tightly together. Two equal and opposite point charges are placed one each on either side of the plates, and equidistant from them. The line joining the two charges is perpendicular to the plane pf the plates.
    What will happen when the two plates are released?


    2. Relevant equations
    There would be an electric field at the location of each plate due to the external charges.


    3. The attempt at a solution

    The electric field at the plates due to the external charges would be equal and opposite, and hence the net field at the plates would be zero. Moreover the induced charges on the plates would be equal and opposite. Hence there should be no force of attraction or repulsion acting on the plates when they are held together. However, I am unable to figure out as to how to proceed with analysis of the scenario when the two plates are separated from each other.
     
  2. jcsd
  3. Jan 2, 2010 #2

    diazona

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    Re: Electrostatics

    Actually, it wouldn't. The plates are halfway between a positive and a negative charge. What do you know about the electric field halfway between a positive and negative charge? (in the absence of the plates) Zero or nonzero? Which way does it point?
    Think about the electric field at the surface of each plate. This electric field affects the induced charge on the plates.
     
  4. Jan 2, 2010 #3
    Re: Electrostatics

    I don't have the answer, just want to join in and try it.

    1) When the two plates are held tightly with the charge equi-distance apart, opposite charge induced on the side that are closed to the charge. Therefore the side facing the +ve charge have -ve charge induced on the surface. The side facing the -ve charge have +ve charge induced. The interface between the two plates is assumed to be held very tight to form a perfect connection and there would have NO charge on the surface. Since the plates are perfect conductors, no potentials and electric field inside the plates.

    2) The moment when plates are released, the perfect conduction boundary is broken. Since this is an electrostatic situation, no charge transfer across the boundary. Immediately upon separation, the boundary side of the plate closer to +ve charge will have +ve charge induced on the boundary surface. And there is -ve charge induced on the boundary surface of the plate closer to the -ve charge.

    3) Therefore the surface of the two plates that formed the boundary before have opposite charges induced on their surface and they should attract towards each other!!!



    Guys, I am not even convinced of what I just wrote!!!:rofl::rofl:
     
  5. Jan 2, 2010 #4
    Re: Electrostatics

    There can be no electric field inside a conductor at equilibrium, but the potential is constant, not necessarily 0!
     
  6. Jan 2, 2010 #5
    Re: Electrostatics

    I should have have said since the two charges are equal distance, the boundary is always 0 potential...... Only apply to this case.



    Well any answer? Don't leave us hanging!!!
     
  7. Jan 2, 2010 #6
    Re: Electrostatics

    I acknowledge the error in mentioning about the direction of the electric field normal to the surface of the plates, as it would obviously be unidirectional. However, since initially the two plates are in contact with each other the induced charges would be shared between them. Since positive charge means a deficit in electrons, and negative charge means a surplus of electrons, the two plates should be neutral when connected to each other as the surplus electrons from one plate would move to the other plate and neutralise the positive charges there. However, when we separate the plates, opposite and equal charges get induced on the plates and they should, therefore, attract each other. Any views on this approach?
     
  8. Jan 3, 2010 #7
    Re: Electrostatics

    Your conclusion is similar to mine except you think the plate retain the charge as they separate meaning the plate with the +ve charge of it's side will retain extra electrons after separation and the plate close to the -ve charge will retain extra +ve charges after separation.

    It is very hard for me to imagin one can separate the plates so suddenly that the electrons have no chance to return back and the plates becomes neutral. But this is a theoractical problem where anything is possible.

    My answer is in more practical realistic situation that when the plate separate, the charge return back and both plates are charge neutral. The +ve charges on the plates at the boundary are caused only by the original +ve charges that attract the electron to the side of the plate close to the charge and leaving extra +ve on the side at the boundary. The other side behave exactly the same but opposite charge. So the plate do attract to each other.



    So do you have the answer? Please post it so I can learn also.
     
  9. Jan 3, 2010 #8

    diazona

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    Re: Electrostatics

    If the charges were to return to their original plates so that each plate is individually neutral, whatever force would cause them to do so must be caused by the separation of the plates. But by the time that's happened, there is no way the charges can jump the gap between the plates. (I'm assuming the fields involved are not so high as to cause a dielectric breakdown) So the plates will remain charged after they separate - one with a positive charge, the other with a negative charge.

    My line of reasoning is that yes, the oppositely charged plates will attract each other, but each plate will also feel a force due to the point charges, and those forces tend to push the plates apart. If there's a quick intuitive way to tell which effect dominates, I'm not seeing it at the moment.
     
  10. Jan 3, 2010 #9
    Re: Electrostatics

    I know separating the plates without charges going back is theoractically possible but not practically because there is no perfect separation. The part of the surface that separate first will have the charge induced and the electrons move across the remaining contact.

    Anyway, I can go either way on that.


    About the attraction between the two plates, I think the attraction between the plates is much stronger because the distance is much shorter. In this two charge and two plate system, the charge on the surface of the plate at the boundary is equal to the the charge behind it on the other side. But the distance from the charge to the plate is much larger than the gap between the two plates if you assume the plate just separate.
     
  11. Jan 4, 2010 #10
    Re: Electrostatics

    How about another view as under:

    Scenario: Charge +Q is on the side near plate P1
    Charge -Q is on the side near plate P2
    Both P1 and P2 are thin large plates

    Initially both plates are held tightly together.

    Charge +Q induces a charge -Q on the end of P1 nearer to it and a charge -Q on the other face of P1. Similarly charge _Q induces charges of +Q and _Q on the near and far faces respectively of plate P2.
    As the plates are large in size, charge +Q does not induce any charge on plate P2 and vice versa.
    The two plates are held together by the force of attraction due to equal and unlike charges on the two faces which are touching each other.

    When the plates are separated from each other, an electric field on account of the charges on P1 would be set up between the plates as under

    Sigma/2E0 + Sigma/2E0 = Sigma/E0 (towards P2)

    where Sigma is the surface charge density on P1
    E0 represents the permittivity of free space
    This electric field would tend to cause P2 to move towards P1 when the plates are separated from each other. This should explain the phenomenon which we have been trying to understand qualitatively so far.

    I solicit views of the forum members on the above approach to the problem.
     
  12. Jan 4, 2010 #11
    Re: Electrostatics

    I think the original question said initially the two plates are held tightly together, that means the two surface are connected and there is no charge at the boundary of the two plates.

    On the point of the surface that are not connected, I can agree to your point. Charge only on the surface at the boundary that are not shorted together like a small cavity inside a metal block, it does have charge on the surface.
     
  13. Jan 4, 2010 #12
    Re: Electrostatics

    No, this isn't possible. When 2 conductors are joined, they become a single conductor.
    That means there is no boundary between the 2 sheets now.


    I believe the sheets will remain in the same position as they were.. but i contradict the statement that there would be attraction between them because there is no such force to separate them.. they still act like a single conductor.
     
  14. Jan 4, 2010 #13
    Re: Electrostatics

    To clarify: The plates are held together, NOT physically joined to each other. One of my previous posts was misleading on this score. Sorry for having created the confusion! Trust we all agree now on the approach using the concept of electric field, leading thereby to the conclusion that the two plates would attract each other when separated from one another.
     
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