Charged plate separation, voltage increase and force. - Page 1
Forum section: electronics / general

author: Circlotron
author_rank: Super Contributor
timestamp: September 18, 2018, 02:50:54 am
content: Say we have two metal plates 1 metre square, separated by 10mm and with a potential difference between them of 1000 volts. There is a certain amount of attractive force between them. Now we increase the distance to 20mm and so the capacitance halves(?) and the potential difference would increase to 1414 volts to keep the stored energy constant. But seeing there is this attractive force, we would have had to exert some effort to separate the plates and this energy input would have to appear as extra voltage on the plates. This extra voltage makes the new attractive force greater than what it would have otherwise have been.If we kept on separating the plates further and further, would this additional voltage ever get to the point that it would cause the plates to flash over? That is, would the voltage ever increase faster than the plate separation? And would the extra voltage resulting in extra separating effort that causes further extra voltage ever get to a runaway point somehow?

author: paul@yahrprobert.com
author_rank: Regular Contributor
timestamp: September 18, 2018, 04:35:55 am
content: You're right that there is work to be expended separating the plates. Your first statement that the energy is conserved is then wrong. Now you haven't said what the plates are connected to. If they're connected to a constant voltage power supply then the E field will drop so that E times the spacing d will remain equal to 1000 volts. The stored energy then drops as you separate the plates. At first this sounds wrong because you have to do work to separate the plates but don't forget that as E drops there is less charge (Q=epsilon*E*A, where Q is charge and A is the area) on the plates, and this charge has to get forced back into the supply, doing work on the supply.If you disconnect the supply then charge is the conserved quantity, and E (epsilon E * A equals charge) remains constant ( assuming plates much wider than the spacing). V rises proportional to the separation. The energy rises proportional to the separation, and force remains constant (force goes like 1/2 *A* epsilon E^2). E is constant so your question about flashing over seems like the answer is no.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 18, 2018, 08:26:10 am
content: Charge is conserved, so we have the relation Q = V C, with Q constant and C varying with distance. In the near field, C ~= ε_0 A / d, so we have V = Q d / (ε_0 A).And then we have E = 0.5 C V^2, plug in both and getE = 0.5 (ε_0 A / d) (Q d / (ε_0 A))^2E = 0.5 ε_0 A Q^2 d^2 / (d ε_0^2 A^2)E = 0.5 Q^2 d / (ε_0 A)E varying with distance implies W = F d, or a constant force, which is interesting. If that works out, it could be very useful as a spring. I mean, assuming having a low-leakage system and nothing surrounding the electrodes is feasible...Of course, as distance increases, the fringing field dominates, and C levels off.Since V ~ d, it seems this can never cause breakdown.Tim

author: Vtile
author_rank: Super Contributor
timestamp: September 18, 2018, 08:00:20 pm
content: Quote from: T3sl4co1l on September 18, 2018, 08:26:10 amHehee. Take a tour to old meter movement world (the bizarre end of it).