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Electric field and distance in Parallel-Plate Capacitor

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Jul10-14, 08:08 AM
P: 13
In the Image attached, we have two tubes of glass each filled with charged metal and they are in space. As we know, the electric field is only between them and negligible in other places.
If we moved an electron worth of charge from - to + we have execute some work on it.
In case 1 we have to execute double the amount of work as in case 2.
I see that the electric field on the plates are the same, so where is the increase in electric field force that we have to work against? Can you describe this to me on the picture?
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Jul10-14, 08:11 AM
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Quote Quote by Amerez
What I fail to understand is if we move the electron along a path "not" between the plates but "around" the plates. Since there is no field around the plates, where is the force countered responsible for the work done on the electron (or the potential energy increase of the electron)?
The charged capacitor analysis cannot have it both ways. Either the capacitor is connected to a charge transfer circuit such as a resistor or a battery, or it is open circuit with a fixed charge. If one electron moves from one plate to the other, then the plates are connected by that electron's path, and the charge is changing.

If an electron starts at one of the charged plates and travels to the other plate by any route whatsoever, the electron will change it's energy. The force on the electron will be countered by the electric field and therefore, in turn, by the physical capacitor's plates.

Since the oppositely charged plates attract, the dielectric insulator that keeps the plates apart counters the force due to electric charge attracting the plates. Moving an electron between the plates changes the charge and therefore the force of the plates on that dielectric structure.

Any conductor connected to a plate is part of that plate. There really is an electric field outside the inner volume. The simple assumption, (for the purpose of analysis), that the field is entirely within the restricted volume between the physical plates, is totally invalidated by considering the real electric field that must be outside the plates.

If the electron starts away from the capacitor and circulates once around the capacitor, in a closed path, it will see no net change in energy. If it is not in the capacitor's field, then it is irrelevant to the capacitor analysis.
Jul10-14, 08:32 AM
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Quote Quote by Amerez
In the Image attached, we have two tubes of glass each filled with charged metal and they are in space.
If there is charge difference between the tubes then there must be an electric field across the two dielectric glass envelopes and the space between them.

What stops the electrically attracted glass tubes from accelerating towards each other, then crashing, breaking the glass and neutralising the charge?

You appear to be hypothesising a capacitor with charge, but without an electric field.
How are you going to move your test electron through the glass dielectric?

C = Q / V. A charge difference implies a voltage difference V = Q / C, where C is the capacitance between the different charges.

If the charges are the same in the two cases you present, and only the separation is different, then the voltages will be different because the capacitance is different.

Power = voltage * current, half the voltage, for the same electron current, implies half the power.
jim hardy
Jul10-14, 09:36 PM
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I still didn't get where the electric field we are working against exists from the explanation
it's between the plates. And that's where the energy in a capacitor is stored. Halving the distance between the plates halves the volume there.
Jul14-14, 08:01 AM
P: 13
Thank you Jim and Baluncore

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