Force between parallel plates of set voltage

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SUMMARY

The discussion centers on calculating the electrostatic force between two parallel plates with a voltage difference of 2.5kV, where one plate is at -10kV and the other at -7.5kV. The formula used for this calculation is F = ε0 AV² / 2d², which is applicable as the force depends solely on the voltage difference, not the absolute voltages. The energy stored in the capacitor is expressed as E = (C V²)/2 = (ε0 A V²)/(2 d), confirming that the voltage difference is the critical factor in determining the electrostatic force.

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Hi PF,

I have a setup with two plates that are connected to two different voltages and separated by vacuum. In my case, one plate is set at -10kV, and the other plate is set at -7.5kV. I know the area of these plates, the separation of the plates (which is much smaller than the plate area), and the voltage difference between the two plates (2.5kV) and I am trying to set up a way of calculating what the electrostatic force between these plates would be under these conditions, but I am having trouble conceptualizing the problem.

I am used to thinking about this problem in terms of capacitors, where a voltage is applied across the parallel plate capacitor and each plate is imparted with opposite charges ±Q. In that situation, I know I can calculate the force between the plates using the following equation:

F = ε0 AV2 / 2d2

But I don't know if this equation still holds in my situation, where I have two parallel plates in space, each connected to a different voltage. Can I just insert the 2.5kV voltage difference between my plates into the equation above for V and get an accurate result? I tried it and got a reasonable result but don't exactly trust it.

Thanks for the help.
 
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Yes, what you have done should be correct. The common mode voltage doesn't make any difference, it is the voltage difference that causes the force. It's easy to come up with the formula you quoted. The energy stored in the capacitor is E = \frac{C V^2}{2} = \frac{\epsilon_0 A v^2}{2 x}, where x is the plate separation. The force will just be -\frac{dE}{dx} = \frac{\epsilon_0 A V^2}{2 x^2}.
 
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Thanks, phyzguy!
 
Only the voltage difference matters, not the absolute voltages.
 
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