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I know that we can only read voltage across a capacitor if it's two pins are connected to the voltmeter or can't talk about any potential difference between totally different systems if they do not have a common ground. But as I think about it more, I started getting confused and can not understand the physical reason why this is the case. Let me introduce a case;

schematic

simulate this circuit – Schematic created using CircuitLab First I charge my capacitor via battery. Numbers are trivial. Then;

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simulate this circuit Then I measured the voltage to be sure that its charged. After that;

schematic

simulate this circuit Than I connected the negative probe to any universal ground(earth, a big copper plate etc.) and from common knowledge I know that I could not measure any voltages but I could not understand the physical reason of why. If voltmeter is measuring the current across its probes than why the charge carriers do not flow from the capacitor plates to ground? If voltmeter measures the electric field like an electroscope than why a plate with charge carriers and a ground do not create an electric field? Thanks for reading and sorry for trivial question.

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marked as duplicate by R Drast, PeterJ, Vladimir Cravero, jippie, Daniel Grillo Mar 7 '16 at 11:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ Current can't flow without a complete circuit - you essentially have in your last drawing an infinite resistor between the capacitor and ground. Ohms law then tells you there is no current (oversimplified). \$\endgroup\$ – Tom Carpenter Mar 4 '16 at 14:38
  • \$\begingroup\$ There is a complete circuit, completed by the capacitance between C1 (open leg) and ground. This is likely to be a few pF at best, forming a much higher impedance than your voltmeter, which will therefore not read anything useful. \$\endgroup\$ – Brian Drummond Mar 4 '16 at 15:14
  • \$\begingroup\$ @BrianDrummond That's actually still an open circuit for DC current. Basically, still a static voltage with no path, because the only way to complete the circuit is by using AC waves (in the terahertz & above regions for any meaningful xfer through that infintecimal capacitance) & the OP is describing an attempt using DC charge on a capacitor. \$\endgroup\$ – Robherc KV5ROB Mar 4 '16 at 15:24
  • \$\begingroup\$ @JRE technically, it could be argued that this one is different, because using an electroscope would show the differing static charge voltages across the capacitor, but wouldn't make that lightbulb glow. ;) \$\endgroup\$ – Robherc KV5ROB Mar 4 '16 at 15:28
  • \$\begingroup\$ Technically, an electroscope will also show the battery voltage, so it comes out the same. \$\endgroup\$ – JRE Mar 4 '16 at 15:31
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Than I connected the negative probe to any universal ground(earth, a big copper plate etc.) and from common knowledge I know that I could not measure any voltages but I could not understand the physical reason of why.

The voltmeter can only measure the voltage (potential difference) between its two terminals. It doesn't know anything about any other nodes in the circuit.

The capacitor only controls the voltage between its two terminals. It doesn't influence anything about any other nodes in the circuit.

Say you charge the capacitor to 9 V. Then you disconnect the capacitor from ground. The two terminals of the capacitor are still 9 V apart, but there's nothing keeping either one at the the ground potential. Due to a few stray electrons blowing on or off the isolated capacitor due to wind, etc., the potential relative to ground could drift by 10's or 100's of volts.

Then, when you connect the voltmeter like in your 3rd diagram, you provide a path for a small leakage from the top terminal of the capacitor to ground. Now that terminal will end up very close to the ground potential, and the other terminal will end up at -9 V, because the capacitor is still (assuming no leakage through the dielectric) maintaining a 9 V difference between its two terminals.

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Actually, I think the worst part of your question is your apology for asking a "trivial question." Your question is very valid, and quite non-trivial.

While, according to its naming convention, a "voltmeter" should show a static charge of carriers when connected to either terminal of a charged capacitor (and the charge between them, by subtracting one from the other), your multimeter is actually not a voltmeter at all.

When production multimeters are set to measure "voltage," they are internally configured to present a high (but nowhere near infinite) impedance to the circuit under test. Then, the meter measures the current across its internal load in order to estimate the open-circuit voltage of the circuit (the argument can be made that it measures voltage across the load, but the result is the same, since current/voltage across any real, non-infinite impedance are inherently linked).

Because there's no complete path for current in your circuit, the meter only "sees" a transient passage of electrons across the load, then the static potentials are balanced across the meter & no more current flows, thus no voltage is registered.

If a "voltmeter" were to actually measure "true" open-circuit voltage (like an electroscope), then your circuit would work just fine & the grounded electrode of your meter would be functioning as "reference ground," rather than its current function as "broken circuit."

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    \$\begingroup\$ Even with an electroscope, you'd be measuring the voltage between one terminal of the capacitor and ground, not the voltage difference between the two terminals of the capacitor. \$\endgroup\$ – The Photon Mar 4 '16 at 16:40
  • \$\begingroup\$ @ThePhoton If you measure 1 terminal's static potential, then the other, both referenced to the same 'other' potential (ground), then subtract the smaller result from the greater result, you could mathematically determine the charge (as already stated in my 2nd paragraph above). \$\endgroup\$ – Robherc KV5ROB Mar 4 '16 at 16:45
  • \$\begingroup\$ I don't think that point is clear at all from your 2nd paragraph. \$\endgroup\$ – The Photon Mar 4 '16 at 16:47
  • \$\begingroup\$ @ThePhoton that may be true. If you'd like, feel free to revise my wording to clarify that point. \$\endgroup\$ – Robherc KV5ROB Mar 4 '16 at 17:08
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Voltage is the potential difference between two points in circuit. If you only connect it to one side of the capacitor and the other side is disconnected then there is no circuit. It doesn't matter what kind of 'ground' the negative side of voltmeter is connected to (could be the Earth, a copper plate, a short wire, nothing). The charge on the capacitor cannot deflect the voltmeter because there is no way for it to reach the voltmeter's negative terminal.

This can be proved by looking at a real world setup in which there is a path to the voltmeter's negative terminal. Between any adjacent components there will always be some capacitance (unless they are separated by an infinite distance), so a practical circuit actually looks like this:-

schematic

simulate this circuit – Schematic created using CircuitLab

What happens in this circuit? Since we now have a complete circuit, current can flow from C1 down through the voltmeter across the ground and back up into C2, charging it up. C2 is a million times smaller than C1 so only a tiny amount of charge needs to move for the voltages on C1 and C2 to equalize. Therefore C1 will lose a tiny bit of voltage, while C2 will charge up to the same (slightly reduced) voltage.

Current will flow through the voltmeter as C2 charges up, so it will show a momentary deflection. However once charged the voltage on C2 is positive at the ground end, so from the voltmeter's point of view the voltages on C1 and C2 cancel out and it reads zero volts. Now imagine that C1 is moved further from the ground plane so that C2 is smaller and takes less charge. The initial voltmeter deflection also reduces. In a theoretical circuit with no capacitance to ground, the voltmeter will always read zero.

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