Will the Voltage Split Equally Between Two Different Sized Electrodes?

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When two electrodes of different sizes are connected to a 100V voltage generator, the voltage does not split equally due to their differing capacitances. The larger electrode has a lower charge density and, consequently, a lower potential compared to the smaller electrode. If one electrode is grounded, the potentials relative to ground will be 0V and 100V, regardless of the electrode sizes. The surface charge density is greater on the smaller electrode, resulting in a higher electric field strength. Therefore, the potential difference remains 100V, but the individual potentials differ based on the size and capacitance of the electrodes.
Fmhood
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I have been thinking about the following situation, without getting the physical understandig straight:

Two electrodes, one twice the size (area) of the other, are connected to a voltage generator to obtain a voltage difference of 100V between the electrodes. The question is if the voltage will be split symmetrically between the electrodes? (i.e. is the voltage difference between each electrode and ground the same?).

My argumentation goes as follows: The voltage generator makes sure that the electrodes have opposite charges in equal amounts. This means that the large electrode will have half the charge density compared to small electrode. The potential of the electrode is however a superposition of the potential of each charge. Since the product, charge density times "size" is the same for both electrodes, their potential difference with respect to ground must be the same.

Am I making a mistake here?

thanks...
 
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Say the surface area of electrode E1 is 2x area of E2.
Assuming that the same ratio also applies for the capacitances of both electrodes wrt Earth then you could make an equivalent circuit with 2 capacitors in series connected to 100V. The potential of E1 wrt Earth is then 33V and that of E2 is 67V. So for the same charge the lower capacitance has the higher voltage.
 
As long as a 100V source is connected across the electrodes, the difference in potential is 100V.

The definition of potential is the work W required to move a positive test charge q+ from one equipotential surface to another. Where q+ = e is the elemental charge.

V2 - V1 = dW/q+

You have defined V2-V1 = 100 by inserting a voltage source. If the plates are perfect conductors they will always have a difference of 100V regardless of where you place the ground.
 
Fmhood said:
their potential difference with respect to ground must be the same.

Am I making a mistake here?

thanks...

I thought this was the question?
 
Sketch attached of my understanding of the question. Choice of ground node is arbitrary. There is an E-field in the gap between the plates which may or may not be a linear function from 0 to 100V.
 

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  • 100vsketch.png
    100vsketch.png
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Thanks for the answers.

My apologies for not stating the question clearly. I was thinking of the situation that Per Oni is referring to. In light if the capacitance of the electrodes, it makes sense that equal amount of charge corresponds to different potentials when the capatitances are different (if area is linearly dependent with capacitance).

However whether a voltage generator actually is constructed the way SystemTheory has sketched it, is a relevant. If I understand it right, then if one electrode is grounded, the potential of the electrodes will be 0V and 100V, respectively, no matter the size of the electrodes... ?
 
Fmhood said:
However whether a voltage generator actually is constructed the way SystemTheory has sketched it, is a relevant. If I understand it right, then if one electrode is grounded, the potential of the electrodes will be 0V and 100V, respectively, no matter the size of the electrodes... ?
Yes, in that case that is their potential wrt ground.

System Theory can you tell us how to generate and print those pictures?
 
I use the "paint" program in the Windows accessories folder under the Start menu. The tools are sufficient for making a quick rough sketch with some practice. Use black and white and Save As *.png or *.gif seems to produce a small file size for resource economy.

The capacitance of a system without symmetry is typically measured rather than calculated. I think you are correct that the surface charge density will be greater on the smaller plate.
 
SystemTheory said:
I use the "paint" program in the Windows accessories folder under the Start menu. The tools are sufficient for making a quick rough sketch with some practice. Use black and white and Save As *.png or *.gif seems to produce a small file size for resource economy.
Thanks. I'll try sometime.
The capacitance of a system without symmetry is typically measured rather than calculated. I think you are correct that the surface charge density will be greater on the smaller plate.
With your picture in mind and with Q the same on both surfaces then surface density D=Q/A which means that D is higher for the smaller surface. Since field E=D/epsilon zero also E is higher for the smaller surface, something which becomes clear when you draw the field lines.
 
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