Extremely slight increase in length and diameter due to a miniscule amount of heat release due to resistance?physical changes that take place in wire A and wire B
The changes may not be observable, but we can measure their effects (say, with a voltmeter). So what are the theoretical changes in those wires during the discharge period?There are no observable physical changes taking place in the wires that I am aware of.
You're correct. I should have said something like, "discharge to a state of approximate equilibrum", since the exact figures are not critical to the question.Full discharge is of course an impossibility since the exponential decay to zero takes an infinite amount of time. You should phrase the proposition as: it takes an hour for the voltage to drop to 0.1% of it's fully charged value.
Possibly, but I'm thinking smaller. I was hoping for something like, "At the start of the discharge period, wire B will have ["more electrons" or "electrons that are vibrating more quickly" or "electrons further away from their protons"] than wire A".Extremely slight increase in length and diameter due to a miniscule amount of heat release due to resistance?
That may be true, but I don't think the Earth's magnetic field is relevant to the question, since I believe the circuit will behave the same in space, far away from such fields.Since the wires A and B are subjected to the Earth's magnetic field, the wires will experience two diminishing forces in opposite directions given by the cross product of current times magnetic field.
Is this school work?Given the circuit below, with the capacitor previously charged to 100 volts and the resistor large enough so that full discharge will require about an hour:
Please describe the physical changes that take place in wire A and wire B during the discharge period.
No. Well, sort of. I'm a teacher and for several years I've been trying to write an illustrated, cartoon-style book about electricity and electronics for middle- and high-school kids, but I can't seem to pin down a satisfactory answer to this question (which is fundamental to the "picture" of electricity that I want to plant in their minds). One of my inspirations is this book: https://www.amazon.com/dp/0962781592/Is this school work?
There is no state of equilbrium until infinite time.You're correct. I should have said something like, "discharge to a state of approximate equilibrum", since the exact figures are not critical to the question.
Sorry, electrons in a wire don't flow like that. Wire A and wire B will experience the same flow of electrons.Possibly, but I'm thinking smaller. I was hoping for something like, "At the start of the discharge period, wire B will have ["more electrons" or "electrons that are vibrating more quickly" or "electrons further away from their protons"] than wire A".
Then I sincerely suggest that you teach some other subject and not electricity and electronics.No. Well, sort of. I'm a teacher and for several years I've been trying to write an illustrated, cartoon-style book about electricity and electronics for middle- and high-school kids, but I can't seem to pin down a satisfactory answer to this question (which is fundamental to the "picture" of electricity that I want to plant in their minds). One of my inspirations is this book: https://www.amazon.com/dp/0962781592/
There is no physical change in the wire. Electrons flow, but the density isn't sufficient to alter the physical characteristics of the wire.Please describe the physical changes that take place in wire A and wire B during the discharge period.
Then what is physically different about wires A and B during the discharge period? How do we account for the measurable differences in potential between the two wires during the discharge period? A voltmeter connected to two different points on wire A will read very close to zero potential difference the whole time; but the same meter connected from wire A to wire B will reveal a significant potential difference at the start of the discharge period. What is the physical cause of this potential difference? There must be something different about wires A and B during the period in question. What is it?Wire A and wire B will experience the same flow of electrons.
It seems to me that a wire with electrons flowing in a given direction, and a wire with electrons that are not flowing in any particular direction, are physically different. But if the flow is the same in both wires, how do we account for the potential difference between the two?There is no physical change in the wire. Electrons flow, but the density isn't sufficient to alter the physical characteristics of the wire.
The capacitor.What is the physical cause of this potential difference?
Considering the fact that no two wires can be exactly the same, by definition, they're different.There must be something different about wires A and B during the period in question. What is it?
I hate the water analogy, but it can be illustrative in this situation...It seems to me that a wire with electrons flowing in a given direction, and a wire with electrons that are not flowing in any particular direction, are physically different. But if the flow is the same in both wires, how do we account for the potential difference between the two?
And all that happens very fast, close to the speed of light, yes? So if the flow of electrons is steady and virtually the same in both wires, how do we account for the measurable potential differences in the wires throughout the hour in question?When an electron leaves the capacitor, it displaces an electron in an atom of the metal in the wire. That causes a "chain reaction" where other atoms give up an electron and almost immediately gain one. So the net number of electrons in any atom of metal remains constant.
In your circuit, you have two electrical nets. One for the capacitor anode which includes the wire used to connect the anode to one end of the resistor. And one for the cathode of the capacitor. We don't think of the wires as extensions of anything.You also said that the "the capacitor" was the physical cause of the potential difference. If so, we could think of wires A and B as temporary "extensions" of the capacitor's plates, and the question would then become, What accounts for the potential difference between the two plates inside -- and, by extension, outside of -- the capacitor?
Okay, then what accounts for the potential difference between the capacitor's anode and cathode? In other words, what is physically different about the capacitor's anode and cathode when the capacitor is charged to 100 volts?In your circuit, you have two electrical nets. One for the capacitor anode which includes the wire used to connect the anode to one end of the resistor. And one for the cathode of the capacitor. We don't think of the wires as extensions of anything.
by Jake Hertz
by Jake Hertz
by Jake Hertz
by Duane Benson