What Is the Force Between Plates of a Parallel-Plate Capacitor?

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SUMMARY

The force between the plates of a parallel-plate capacitor can be derived using the relationship between electric field and charge density. The force exerted by each plate on the other is given by F = (σ^2 A)/(2ε0), where σ is the surface charge density and ε0 is the permittivity of free space. Doubling the distance between the plates results in a change in voltage, specifically V' = V/2, where V is the original voltage. The work required to double the distance is calculated using the formula W = F * d, where d is the change in distance.

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Homework Statement



Imagine you have a parallel-plate capacitor with area A and charge Q = [itex]\sigma[/itex]A, separated by a distance x, which may vary while the charge stays fixed in the plates.

a. The force that each plate exerts over the other can be expressed as _______.
b. How much work is required double the distance that the plates had between them originally?
c. If the original voltage is V, once the distance between the plates is doubled, the voltage between the plates will be _________.

Homework Equations



C = (ε0 A)/d

E = σ/A

σ= Q/A


The Attempt at a Solution



I really don't know where to start this one. Since it asked for force I thought about using Coulomb's Law but that took me nowhere. I'm guessing the equations I put above are vital in all of this, but I can't seem to figure the relationships out. Any help at all would be appreciated. Thanks.
 
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Thanks. I'll check that out then.
 

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