B Electrons cannot sense electrostatic force?

Why there is not voltage or current just for 1ms if I connect a multimeter ground to the negative terminal of a DC power supply or charged capacitor? Why electrons in measure lead and DMM device cannot sense a bulk of electrons (or lack of it)? I tried with an 5kV DC power supply too. In an antenna they sense this force and move from or to the source.
 
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Why there is not voltage or current just for 1ms if I connect a multimeter ground to the negative terminal of a DC power supply or charged capacitor?
Because multimeters work on current flow and you are talking about an open circuit with zero current flow.
 
Because multimeters work on current flow and you are talking about an open circuit with zero current flow.
But an antenna also an open circuit and there is current in it.
 

sophiecentaur

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But an antenna also an open circuit and there is current in it.
This is a good observation and shows the danger of confusing steady state conditions (DC) with changing ones (AC /Step Functions / Pulses). EM can't be dealt with with simple, one-line statements. Even JC Maxwell needed Three Equations!
 
This is a good observation and shows the danger of confusing steady state conditions (DC) with changing ones (AC /Step Functions / Pulses).
Interesting but can you explain, please? I tried to use pulse DC with a push button but no effect.
 

sophiecentaur

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Interesting but can you explain, please? I tried to use pulse DC with a push button but no effect.
The title of your thread refers to Electrostatic Forces. When the forces are not Static, things are different. The reason that your DMM recorded nothing will be a combination of factors. Firstly, a Digital Analogue Converter (the heart of a DMM) takes samples and gets an average over its sample interval. The total charge that passes with an experiment like yours will depend on the Capacitance of the apparatus. The peak current could be fairly high (easily detectable by your DMM) but the average may be very small. It's the sort of thing that a 'clunky' analogue meter might show you by a tiny twitch of a needle. (Newer doesn't always mean better. :wink:)
 
... but the average may be very small. It's the sort of thing that a 'clunky' analogue meter might show you by a tiny twitch of a needle. (Newer doesn't always mean better. :wink:)
Thanks, I suspect this. And if I would use a DSO instead? Sample rate of that device is high enough I guess. Shorter time period's average could be better here.
 

anorlunda

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nd if I would use a DSO instead? Sample rate of that device is high enough I guess.
Depending on the capacitance of that wire, picoseconds or nanoseconds might be needed to see the spike.
 

sophiecentaur

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Depending on the capacitance of that wire, picoseconds or nanoseconds might be needed to see the spike.
The series R would be relevant, wouldn't it? RC time constant would be longer
 

anorlunda

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The series R would be relevant, wouldn't it? RC time constant would be longer
The way the OP stated it, is is the R of a short piece of wire.

But time constant alone does not make it visible. It is area under the curve, i.e. energy, that makes the difference between visually visible or invisible pulses.
 

sophiecentaur

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The way the OP stated it, is is the R of a short piece of wire.
Again, there is no circuit diagram to enable us all to be discussing the same situation. I assumed that a meter was in series.
 

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