Assume we have a "somewhat" ideal voltage source like a DC power supply powered by mains. Take just one terminal and form a conductive path between it and earth ground. Assuming that no conductive path exists going back to the power plant ground spike, and a fuse isn't blown, wouldn't a huge amount of current flow into the earth due to the massive capacitance of the earth?

Once again, I am assuming that this voltage source can maintain its voltage indefinitely.

  • $\begingroup$ Accepting a answer within 1 hour of asking, especially when there is only 1 of them, is a bad idea. Writing a correct answer would be pointless now. $\endgroup$ – Olin Lathrop Jul 13 '13 at 22:45
  • $\begingroup$ Related: electronics.stackexchange.com/questions/75743/… $\endgroup$ – Alfred Centauri Jul 13 '13 at 23:39
  • $\begingroup$ @OlinLathrop, I don't think it's entirely pointless. Write the correct answer and the point will be that at others that read this will have the benefit of it. $\endgroup$ – Alfred Centauri Jul 13 '13 at 23:40
  • $\begingroup$ @OlinLathrop I apologize for defying the stackexchange conventions. What you say makes sense in terms of answer incentive, but I would still be very interested in your answer. I have been posing this question to many electrical engineers (both industry and EE graduates, online/offline), and I have yet to get a consistent answer. There seems to be little consensus on the finer details. I am very curious regarding how one object "grounds" itself relative to the potential of another object. $\endgroup$ – MER Jul 14 '13 at 2:34

First, the Earth's capacitance is not "huge", less than 1F, so it would not take long to charge Earth to a significant potential with moderate current. It is important, however, that Earth is conductive, so, unless the terminal has the same potential as the Earth, the current will be maintained in the circuit, as the circuit will be closed somehow through the earth and the plant equipment maintaining the terminal potential - it is difficult to assume "that no conductive path exists going back to the power plant ground spike", unless earthing is very bad either at the plant or at the contact with the terminal, in which case some limited area will be charged and the current will stop. How large the current will be, depends on such things as the resistance in the circuit, including the resistance of the earthing at the plant and at the contact with the terminal.

EDIT(07/13/2013) Actually, Earth's capacitance is not just less than 1F, it's much less - about 0.0007F

  • $\begingroup$ Ok, I hadn't realized that the Earth's capacitance was only around 1 Farad. I just want to confirm one more suspicion. Say I had a perfect insulator between myself and earth ground. I then make contact with that same single terminal. Obviously I have a body capacitance and it is quite small, but by virtue of touching just that single terminal, my body capacitance will then acquire the same voltage as the power source correct? $\endgroup$ – MER Jul 13 '13 at 20:00
  • $\begingroup$ @Gigglelot: Your body will then acquire the same voltage as the power source, yes (assuming the body is somewhat conductive). $\endgroup$ – akhmeteli Jul 13 '13 at 20:24
  • $\begingroup$ This answer is as confused as the question. The earth doesn't "have" capacitance. Capacitance is a property between two conductors. You could say there is a certain capacitance between your body and the earth, assuming you are insulated from it. That would probably be only in the 10s to 100s of pF. $\endgroup$ – Olin Lathrop Jul 13 '13 at 22:43
  • $\begingroup$ @Olin Lathrop: I respectfully disagree. The notion of capacitance of an isolated conductor is also widely used, see, e.g., web.mit.edu/8.02t/www/materials/StudyGuide/guide05.pdf : "An “isolated” conductor (with the second conductor placed at infinity) also has a capacitance." Or britannica.com/EBchecked/topic/93467/capacitance :"capacitance, property of an electric conductor, or set of conductors, that is measured by the amount of separated electric charge that can be stored on it per unit change in electrical potential." In this case the potential is with resp. to infinity. $\endgroup$ – akhmeteli Jul 14 '13 at 0:51
  • $\begingroup$ It is possible to talk about voltage with respect to infinity, and therefore capacitance too, but this is not useful for a circuit and will just confuse, as it is doing here. Note that the OP is talking about a voltage source and line power and the like. This is a circuit question. The fact that you can define the abstract concept of voltage relative to inifinity is not relevant here, and only serves to obscure the real issues. $\endgroup$ – Olin Lathrop Jul 14 '13 at 12:20

Some current will flow, but it will be not be large and it will mostly be defined via parasitic capacitance from the source to earth.


There will be some innate non-removable parasitic capacitance between source terminals (+ and -, not relevant for KVL) and there will be some parasitic capacitance from each terminal and earth.

By wiring 1 terminal to earth, you mostly bypass the capacitance on one side. But the other one will exist and it will limit the current since a low capacitance is in series with the source (the capacitance of ungrounded terminal)

If you completely neglect the resistance (non-realistic), the current will be just a dirac impulse, but the amplitude of the impulse will be proportional to the product of source voltage and total capacitance.


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