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mma

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The well-known connection between the electric field and Coulomb's law is that

These two equalities yield the [itex]F=qQ/r^2[/itex] Coulomb's law. It's important that we don't care of the electric field of q, only with of Q.

But.

In the second volume of Feynman Lectures, Feynman calculates the force acting on one of the plates of the capacitor using the whole electric field, coming from both plates together. Then he makes the following correction in the calculation:

I wonder if this logic can be applied for deriving the Coulomb's law too.

Of course we cannot apply this logic for point-charges because the electric field at the place of the charge is infinite, but we can apply to charged metal spheres, and we get the Coulomb's law as the limit when the radius of the sphere tends to zero.

So, my question is that if I calculate the common electric field of a point-charge Q and a metal sphere with charge q, and I apply the half of this electric field on the sphere, then will I get the same force as I get if I take into account only the electric field of the point charge but don't devide it by 2?

- The magnitude of the electic field at distance r from a point-charge Q is [itex]E=Q/r^2[/itex]
- The definition of E is that the force exerted the field on a point-charge q is [itex]F=qE[/itex].

These two equalities yield the [itex]F=qQ/r^2[/itex] Coulomb's law. It's important that we don't care of the electric field of q, only with of Q.

But.

In the second volume of Feynman Lectures, Feynman calculates the force acting on one of the plates of the capacitor using the whole electric field, coming from both plates together. Then he makes the following correction in the calculation:

One would immediately guess that the force acting on one plate is the charge

Q on the plate times the field acting on the charge. But we have a surprising factor

of one-half. The reason is that E0 is not the field at the charges. If we imagine that

the charge at the surface of the plate occupies a thin layer, as indicated in Fig. 8-4,

the field will vary from zero at the inner boundary of the layer to E0 in the space

outside of the plate. The average field acting on the surface charges is Eo/2. That

is why the factor one-half is in Eq. (8.18).

I wonder if this logic can be applied for deriving the Coulomb's law too.

Of course we cannot apply this logic for point-charges because the electric field at the place of the charge is infinite, but we can apply to charged metal spheres, and we get the Coulomb's law as the limit when the radius of the sphere tends to zero.

So, my question is that if I calculate the common electric field of a point-charge Q and a metal sphere with charge q, and I apply the half of this electric field on the sphere, then will I get the same force as I get if I take into account only the electric field of the point charge but don't devide it by 2?

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