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I Is a symmetric charge distribution the lowest potential

  1. Sep 11, 2017 #1
    Is the potential energy of a symmetric planar (x,y) charge distribution lower than any non symmetric distribution ? from the discussion on Gauss's law and symmetric charge distributions I would think so because the electric field could only be normal to the (x,y) plane in the symmetry case but would have an additional component parallel to the x,y plane due to the non-symmetric component of the distribution.
     
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  3. Sep 12, 2017 #2

    mfb

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    In general there is no well-defined minimum, and there will always be many non-symmetric distributions with an energy lower than various symmetric distributions. Are there any constraints on the charge distribution?
     
  4. Sep 12, 2017 #3
    Given a specific symmetric planer charge distribution in x,y at z=0 and add a small non-symmetric addition at z=0 to this originally symmetric distribution - then this new charge distribution should have a higher potential energy. The symmetric distribution is defined for x,y at z=0. The distribution is symmetric for rotations in the x,y plane. Symmetry implies that the distribution and associated field are invariant for a rotation thus this Es field must not be a function of x,y , only of Z , thus the Es field must be perpendicular to the x,y plane. Consider the modified distribution with a small non-symmetric addition. This non-symmetric addition will have an En field at z=0 that is a function of x,y . The total e field at z=0 Es + En will be greater than Es
    I am interested in perturbations of existing symmetric distributions. It seems to me that perturbations away from symmetric distributions will always lead to a higher energy. Thus removing the non-symmetric component will lead to a lower energy.
     
  5. Sep 13, 2017 #4

    Nugatory

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    Are you adding charge or just moving the initial charge around? Any addition, regardless of the symmetry or lack thereof, will increase the potential energy by the amount of work required to bring the additional charge in from infinity. On the other hand, if you're just moving the existing charge around then it is easy to reduce the potential energy while breaking the symmetry - for example, remove some of the charge to infinity in a particular direction.
     
  6. Sep 13, 2017 #5
    Keep the charge of the system the same. Alter the distribution of the charge by adding area over which the charge is distributed. NewTotal area = original symmetric distribution area + new area that is not symmetric, but charge on the system unchanged. Now original symmetric area has less charge having shared some charge with new non-symmetric area. Is the energy of the system unchanged? The electric field now has a gradient that it did not have before in its symmetric shape.
     
  7. Sep 14, 2017 #6

    mfb

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    No, and the change will depend on the details of your distribution.

    Start with a uniform distribution over the unit disk. Now take 10% of this charge and distribute it in some asymmetric way in a large area around x=1000. The total potential energy went down by about 19%.

    Start with a uniform distribution over the unit disk. Now take 10% of this charge and distribute it in a disk with radius r=0.001 somewhere (doesn't matter where). You now increased the total potential energy by about a factor 100.
     
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