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I have a hollow charged sphere and I am measuring the force between it and a grounded metal plate. How does the assumption that they are point charges (using method of image charges) become invalid for small separations?

So how is the force affected at small distances due to the charge being distributed on the surface close to the plate?

Also sidenote- how does the plate being finite affect the use of the method of image charges for this situation?

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  • $\begingroup$ Sidenote answer : Edge effects $\endgroup$ – Yuzuriha Inori Mar 10 '18 at 16:23
  • $\begingroup$ Could you please explain the effect of the edges? $\endgroup$ – user187573 Mar 11 '18 at 17:04
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    $\begingroup$ Sure! When we have a finite plate, there will be a sudden ending of the charge surface, and thus there will be accumulation of charges there that will cause bending of the electric field lines than what we would get if we had an infinite plate with no edges and no charge density problems. This change in the electric field is called edge effect, and if you make the edge very very sharp and pointed, the charges accumulate so much, there might be a breakdown near the edge (an electric discharge might be ). You might have seen this infinite plate approximation made in capacitors... $\endgroup$ – Yuzuriha Inori Mar 11 '18 at 17:08
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    $\begingroup$ ... because if there were edges, the electric field lines won't be straight, and we would have a lot of difficulty modelling a capacitor. $\endgroup$ – Yuzuriha Inori Mar 11 '18 at 17:09
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With respect to its potential and electric field outside the sphere, a charged sphere with uniform surface charge density can be replaced by a point charge at the center of the sphere. This is also a good approximation in the case of a non-uniform charge on the surface when you look at the electric field or potential at a large distance compared to the diameter of the sphere. If you consider a conducting sphere, the charge will be uniformly distributed over the surface only when it is far away from any other (charged) bodies. When the charged sphere approaches the grounded metal plate, at distances comparable to the diameter the electric field of the induced charges on the grounded plate will destroy the uniform distribution of surface charges on the sphere. Thus the charges on the sphere cannot any longer be represented by a point charge in the center of the sphere and the simple method of an image charge to obtain the electric field cannot be used when the sphere is near the conducting plate.

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  • $\begingroup$ So how is the force between the sphere and plate affected at these small distances compared to further away? $\endgroup$ – user187573 Mar 13 '18 at 16:37
  • $\begingroup$ @user187573 - When the conductive sphere approaches the plate, the charge distribution becomes asymmetric due to the electric field of the induced charges on the plate. If the charge of the sphere is positive, negative charges will be induced on the plate which attract the positive charges on the sphere and vice versa. Thus the positive charge on the sphere will move more to the side of the sphere which is nearer to the plate. The result will be that the attractive force to the plate will be stronger than in the case of the sphere charge being a point charge in the center of the sphere. $\endgroup$ – freecharly Mar 14 '18 at 1:41
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If the sphere is non-conductive and non-polarizable, it can be replaced by a point charge at any non-zero distance from the plate. Also, the image charge method is then valid.

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