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The electrostatic force $F$ between two point charges $q_1$ and $q_2$ changes as the distance $d$ between the charges changes as seen from Coulomb's law: $$F=\frac{q_1q_2}{4\pi \varepsilon_0 d^2}.$$

However, the electrostatic force between two parallel plates with constant charge is constant regardless of distance. How can this be?

For reference, the electrostatic force between two parallel plates of area $A$ separated by distance $d$ and holding charge $Q$ is $$ \begin{align} F &=\frac{QE}{2}=\frac{QV}{2d} \end{align},$$ where $E$ is the electric field and $V$ is the voltage difference between the plates [Ref1, Ref2]. For a parallel plate capacitor (as these two plates are), $$V=\frac{Q}{C} \text{ and } C= \frac{\varepsilon A}{d},$$ where $C$ is capacitance and $\varepsilon$ is permittivity so $$F=\frac{Q^2}{2\varepsilon A}.$$

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Those equations you are using to get the force are technically only true under the approximation that the parallel plates are infinitely big. In practical terms, they are valid for when the separation of the plates is very small compared to the size of the plates. If you have two infinitely big parallel plates, then changing the distance between them won't really affect their geometry in any relative sense and so the forces between them won't change. If you want to know the actual force between two finite sized parallel plates, you would have to worry about edge effects which are hard to calculate. But finite sized plates will indeed feel a reduced force of attraction or repulsion depending on distance between them.

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However, the electrostatic force between two parallel plates with constant charge is constant regardless of distance. How can this be?

You need to do a vector sum of all the charges on one of the plates to see how it affects a charge on the other plate.

Charges nearby are pulling directly toward the other plate. Charges far away pull less strongly, and pull considerably in a direction parallel to the plates. The pull in this direction is negated by charges in the other direction.

As you separate the plates, two things happen that exactly offset each other:

  • The charges are farther away, reducing the forces felt from them.
  • The angle to the charges is more toward the perpendicular axis.

The vector sum of the forces remains the same after this operation.

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  • $\begingroup$ Could you demonstrate your answer mathematically? I don't know enough about the behavior of electric charges to be able to agree with your explanation as currently stated. $\endgroup$ – RobotRaven Aug 29 '18 at 19:20
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Both the expression for electric field and the expression for capacitance used in the references assumes that the electric field between the two plates is uniform. This is only true if the plates are well-approximated by infinite sheets of charge (practically, this means that it's only valid in the limit where $d$ is much smaller than the size of the plates); this is because infinite uniform sheets of charge have electric fields that are constant with distance.

If $d$ is close to or greater than the size of the plates, then the electric field can no longer be approximated as uniform, because the fringing fields that were ignored before are now storing a significant portion of the potential energy. Since these fringing fields depend on distance, we must conclude that in this regime, the capacitance of widely-separated plates in a parallel-plate capacitor is dependent on distance.

Actually calculating the capacitance of this system when $d$ is close to the size of the plates is difficult, as knowledge of the fringing fields is required, and those are highly dependent on the geometry of the plates themselves. But when $d$ is much greater than the size of the plates, then the two parallel plates are actually well-approximated by point charges. In this case, we have the electric field at each plate:

$$E=\frac{Q}{4\pi\epsilon r^2}$$

so we know both the potential difference between the plates

$$V=\frac{Q}{4\pi\epsilon r}$$

and the force between the plates

$$F=\frac{Q^2}{4\pi\epsilon r^2}$$

which means that the capacitance for extremely widely-separated plates is

$$C=\frac{Q}{V}=4\pi\epsilon r$$

all of which are dependent on distance.

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