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The electric field outside of any capacitor plates is zero.If we take the negative plate to be at ground potential then the positive plate's potential will be $V=E_xd$ ($E$ is the electric field and d is the distance between the plates).The potential beyond the positive plate will remain constant at $V$ since the Electric field is Zero beyond it. Does this mean that if I take a positive charge and place it a little distance above the positive plate the potential at that point where the charge is placed is higher than that of the positive plate.By inference the charge should go towards the positive plate when released and if able to move.This would imply that the charge wiuld move to the positive plate no matter what amount of charge the positive plate carries. Would the potential beyond the negative plate be fixed at zero following the same reasoning? Could someone please clarify as this doesn't sound right. How can the charge build up indefinitely on the positive plate in this manner?

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First, note that the electric field outside of any capacitor is not zero. It is zero only for the ideal case of a perfect infinite parallel plate capacitor.

Your inference about the movement of the positive charge is wrong. Yes the potential is higher there than it is at the other plate, but that is not enough to cause a force on the charge. What is needed is a gradient of the potential. $$F=-\frac{\mathrm{d}U}{\mathrm{d}x} = -q\frac{\mathrm{d} V}{\mathrm{d}x}$$ If the potential is strictly zero, then the gradient is also zero, and there is no force.

Update after comment

A charged particle does not respond to its own field, and hence it does not feel its contribution to potential. The positive particle will feel only the fields due to the capacitor plates. The field outside is zero, and the potential that the charge feels is constant.

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  • $\begingroup$ If I take an external positive charge outside the positive plate then the lines of force would start from the external charge and end on the positive plate.The potential just out side that charge would be the sum of the potentials due to the capacitance plates and the ecternal charge which would be positive.If I took a charged conductor and touched it to the positive plate it should flow from the conductor to the plate. $\endgroup$ – Chappy May 9 '17 at 17:03
  • $\begingroup$ Is this the question: The presence of a positive test charge just outside the positive plate raises the potential in the half-space outside of the positive plate to a value higher than the potential of the positive plate itself. Therefore the positive test charge will be attracted to the positive plate (which is at a lower potential)? $\endgroup$ – garyp May 9 '17 at 18:42
  • $\begingroup$ @garyp..yes.I'm trying to figure out what this would ultimately lead to. $\endgroup$ – Chappy May 9 '17 at 19:11
  • $\begingroup$ @Chappy I've updated my answer to address the clarification. $\endgroup$ – garyp May 10 '17 at 1:13
  • $\begingroup$ But it does mean that I can build up a huge reservoir of charge by moving the external charge under an external force to the positive plate? $\endgroup$ – Chappy May 10 '17 at 2:29

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