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I'm using a 2nd hand Elenco scope In good shape I picked up with 10x probes. I don't want to drag on but do you see my point? With a resistor or an inductor attached to a voltage the other terminal is still physically in connected a way. After all in inductor is just a piece of wire. But with a capacitor the second plate is not connected to the charges in any way. If the cap was initially uncharged there wouldn't be any electric field generated. So how is the voltage jumping from one terminal to the other? They are physically separated and a dielectric is non conductive.
DC bias on amplifier outputs
- Thread starter petewh
- Start date
electrostatic field
The oscilloscope completes the circuit, creating a path for current flow in the external circuit - the 'scope with a 10:1 probe places a 10 megohm resistor between the probe tip and the circuit common. With a voltage source in the circuit, an electrostatic field can be produced between the electrodes of the capacitor. For each electron added on the negative side an electron is "driven out" of what becomes the positive side. Electrons accumulate on the negative side and are removed from the positive side until the the strength of the field is sufficient to oppose the applied voltage and prevent any more current in the external circuit - the capacitor is fully charged for the circuit under consideration. If there is no current flow in the external circuit there can be no voltage appearing across a resistor in that circuit.
Surely all of this is described in detail in dozens or hundreds of places on the web, including Wikipedia.
The oscilloscope completes the circuit, creating a path for current flow in the external circuit - the 'scope with a 10:1 probe places a 10 megohm resistor between the probe tip and the circuit common. With a voltage source in the circuit, an electrostatic field can be produced between the electrodes of the capacitor. For each electron added on the negative side an electron is "driven out" of what becomes the positive side. Electrons accumulate on the negative side and are removed from the positive side until the the strength of the field is sufficient to oppose the applied voltage and prevent any more current in the external circuit - the capacitor is fully charged for the circuit under consideration. If there is no current flow in the external circuit there can be no voltage appearing across a resistor in that circuit.
Surely all of this is described in detail in dozens or hundreds of places on the web, including Wikipedia.
I didn't think of the probe as a circuit element completing the circuit. As far as the capacitor then being in circuit and the electrostatic field I do understand that. Thanks for clearing that up.
I was stumped. I do remember it's an old puzzle for me I never resolved much on my own.
Lately I have been thinking of probe use as transparent action. It's been a couple of years that I used probes on 1x and loaded down a 555 timer with a long time constant with resistors in circuit near 1 Meg.
Thanks again.
I was stumped. I do remember it's an old puzzle for me I never resolved much on my own.
Lately I have been thinking of probe use as transparent action. It's been a couple of years that I used probes on 1x and loaded down a 555 timer with a long time constant with resistors in circuit near 1 Meg.
Thanks again.
