Will Separating Capacitor Plates Cause a Spark?

AI Thread Summary
Separating capacitor plates after charging them can lead to a significant decrease in capacitance due to increased distance, which may not necessarily cause a spark if the charge remains stable. The discussion highlights that achieving a capacitance of 100 microfarads with a plate area of 5 cm² is impractical due to physical construction limitations. Even under ideal conditions, the electric field strength would exceed the dielectric strength of common materials, making such a configuration unfeasible. Theoretical calculations suggest that the attractive force between the plates would be drastically reduced, potentially by a factor of 100 million. Overall, the proposed scenario faces significant practical and theoretical challenges.
error404
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yet...more electrical attraction :)

hello everyone
i posted a question erleair about electrical attraction, and the answers were really helpful, i hope i can help me with this too

here's the situation, i have a circiut with a battery of 100 volts and a capacitor having "100 Micro farad, 100 volts", i'll leave the circuit to charge the capacitors 100% after charging is done, i want to open the circiut, so that the cahrge is trapped on the plates (1e-2 coulomb) of the capacitor, after the charge is trapped on the plates, i want to separate the plates using some insulator (wood or whatever) so that the space between the 2 plates is 1 meter, and the chrage is still on them.

the area of each plate is 5 Cm2.


there are 2 things i want to know acutally
1- the capacitance law is C= E.A/X
when i pull the 2 plates far from each other, the "x" is going to be increased and the capcitance will be decreased. a high charge on a low capacitance, would the plates give a spark and get rid of that big charge?? or would it keep the charge? what else would happen?

2-if everything worked perfect, and i could take the 2 plates away from each other, a whole meter, would the attraction ofrce be hundreds of Newtons? (apporximatley, according to coulomb's law)

thanks
Error404
 
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I feel for 1) the capacitance is decreased, but distance between the plates is increased, so it won't be a problem since same charges are far apart.
 


I think I know where you’re coming from Error404, but you do need to learn a little about how large value capacitors are physically constructed in order to really understand the practicalities (or otherwise) of the questions you’ve been asking. You probably should also read up on Poissons Equations (http://en.wikipedia.org/wiki/Poisson's_equation) to gain some insight into the difficulties of making precise calculations of charge and field distributions for non trivial spatial configurations.

Let me just say that 100uF at 5 cm^2 is not achievable with the type of planar (non electrolytic) construction that would allow you to separate the plates in the manner you desire.

As an example consider a parallel plate capacitor with A=5E-4 m^2 and a Titanium-dioxide dielectric (relative dielectric constant =100). Even with this dielectric we need a plate separation of about 4E-9 (4 nm) to achieve 100uF. This is a spacing of around 10 or 15 molecules thick. Now even if we could make the dielectric this thin and perfectly uniform and flat, and even if we could attach the electrodes with near perfect contact on an Angstrom scale without them being metallurgically bonded (both pretty big ifs) then a voltage of 100 volts would give a field strength of 2.5E10, exceeding the dielectric strength of the TiO2 by about 10000 times. In other words, even if this capacitor could be constructed near ideally, it’s rated working voltage would be only about 0.01 volts. So sorry to rain on your parade yet again, but this will decrease your force by a factor of 1E8 (ten to the power of eight).
 
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