What happens when electrons reach the battery?Capacitor Charging and Discharging when connected to the groundVoltage and parallel-plate capacitorsTypo in physics book (capacitors)Charging a capacitor (terminals)Work done by the battery in series with capacitor with changing dielectricCan someone break down what exactly happens in this situation?What happens when a battery is connected to conductors?
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What happens when electrons reach the battery?
Capacitor Charging and Discharging when connected to the groundVoltage and parallel-plate capacitorsTypo in physics book (capacitors)Charging a capacitor (terminals)Work done by the battery in series with capacitor with changing dielectricCan someone break down what exactly happens in this situation?What happens when a battery is connected to conductors?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
We have a capacitor which is connected to a voltage source. The top plate of the capacitor is connected to the positive terminal of the source. Therefore, electrons go from the top plate to the positive terminal. What happens to the electrons when they arrive to the voltage source? I think when electrons arrive at the positive terminal the source obtains a net negative charge.
electric-circuits charge capacitance voltage batteries
edited 13 hours ago
Aaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
asked 13 hours ago
user3728644user3728644
1017 bronze badges
$endgroup$
add a comment |
$begingroup$
We have a capacitor which is connected to a voltage source. The top plate of the capacitor is connected to the positive terminal of the source. Therefore, electrons go from the top plate to the positive terminal. What happens to the electrons when they arrive to the voltage source? I think when electrons arrive at the positive terminal the source obtains a net negative charge.
electric-circuits charge capacitance voltage batteries
edited 13 hours ago
Aaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
asked 13 hours ago
user3728644user3728644
1017 bronze badges
$endgroup$
add a comment |
$begingroup$
We have a capacitor which is connected to a voltage source. The top plate of the capacitor is connected to the positive terminal of the source. Therefore, electrons go from the top plate to the positive terminal. What happens to the electrons when they arrive to the voltage source? I think when electrons arrive at the positive terminal the source obtains a net negative charge.
electric-circuits charge capacitance voltage batteries
edited 13 hours ago
Aaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
asked 13 hours ago
user3728644user3728644
1017 bronze badges
$endgroup$
We have a capacitor which is connected to a voltage source. The top plate of the capacitor is connected to the positive terminal of the source. Therefore, electrons go from the top plate to the positive terminal. What happens to the electrons when they arrive to the voltage source? I think when electrons arrive at the positive terminal the source obtains a net negative charge.
electric-circuits charge capacitance voltage batteries
electric-circuits charge capacitance voltage batteries
edited 13 hours ago
Aaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
asked 13 hours ago
user3728644user3728644
1017 bronze badges
edited 13 hours ago
Aaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
asked 13 hours ago
user3728644user3728644
1017 bronze badges
edited 13 hours ago
Aaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
edited 13 hours ago
Aaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
edited 13 hours ago
Aaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
20.6k4 gold badges35 silver badges74 bronze badges
asked 13 hours ago
user3728644user3728644
1017 bronze badges
asked 13 hours ago
user3728644user3728644
1017 bronze badges
asked 13 hours ago
user3728644user3728644
1017 bronze badges
1017 bronze badges
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
When electrons leave the top plate it then has a net positive charge which will tract electron to the bottom plate which thus becomes negatively charged.
So there is a movement of electrons from the top plate (which becomes positive) through the voltage source and onto the bottom plate (which becomes negative).
answered 13 hours ago
FarcherFarcher
55.2k3 gold badges45 silver badges119 bronze badges
$endgroup$
1
$begingroup$
But it's not the very same electrons that move from the + side to the - side of the cap. It's a shift in charge density profile.
$endgroup$
– Bill N
12 hours ago
1
$begingroup$
@BillN I am sorry of my answer gives that impression as it was never my intention to do so.
$endgroup$
– Farcher
11 hours ago
add a comment |
$begingroup$
The voltage source does work on the charges to put them onto the bottom plate until the amount of charge $Q$ on the bottom plate is given by $Q=CV$, where $C$ is the capacitance and $V$ is the potential difference between the two sides of the battery. The power source doesn't store charges like you propose.
answered 13 hours ago
Aaron StevensAaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
$endgroup$
1
$begingroup$
in voltage source direction of electric field is from top to bottom and electrons must move from bottom to top.
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
I think positive and negative poles of voltage source reduced, but voltage is constant. how?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Batteries do work against the usual forces caused by the field. That's the point of batteries. To do work on the charge. There is some process in the battery that does this. What that is depends on the type of battery.
$endgroup$
– Aaron Stevens
13 hours ago
1
$begingroup$
Ok. that is mean when electrons arrive to the positive terminal of voltage source, voltage source push them to the negative terminal and negative terminal push them to bottom plate of capacitor?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Kind of, yes. Once charges leave the battery then the battery isn't influencing them. The battery just gives the charges energy so that there can be a charge separation between the two plates of the capacitor.
$endgroup$
– Aaron Stevens
13 hours ago
add a comment |
$begingroup$
I will rephrase your question as I interpret it. Perhaps this is not what you're getting at, but here goes regardless.
Even a small flashlight battery can push a coulomb of charge through a circuit. If you start with an even mix of protons and electrons everywhere (so there is no net accumulation of charge anywhere), and then you remove a coulomb of electrons from the negative terminal and add a coulomb of electrons to the positive terminal (which is what you're doing if you drain the battery through the circuit), now you should have +1 C and -1 C accumulated at the terminals, an inch or two apart. Thus there should be a ginormous force between the terminals, $10^12$ newtons or something, far more than the battery could possibly withstand structurally. How can this be?
The answer is that the change in charge at the terminals due to the departure and arrival of electrons is compensated by the movement of ions within the electrolyte. There is a chemical reaction at the anode that donates electrons to the negative terminal and creates positive ions in the electrolyte, and another one at the cathode in which the positive terminal gives electrons to the positive ions, and the flow of ions within the electrolyte keeps charge from building up anywhere. This may be a bit of an oversimplification of the electrochemistry involved, but I think it captures the general concept.
answered 11 hours ago
Ben51Ben51
4,1048 silver badges30 bronze badges
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When electrons leave the top plate it then has a net positive charge which will tract electron to the bottom plate which thus becomes negatively charged.
So there is a movement of electrons from the top plate (which becomes positive) through the voltage source and onto the bottom plate (which becomes negative).
answered 13 hours ago
FarcherFarcher
55.2k3 gold badges45 silver badges119 bronze badges
$endgroup$
1
$begingroup$
But it's not the very same electrons that move from the + side to the - side of the cap. It's a shift in charge density profile.
$endgroup$
– Bill N
12 hours ago
1
$begingroup$
@BillN I am sorry of my answer gives that impression as it was never my intention to do so.
$endgroup$
– Farcher
11 hours ago
add a comment |
$begingroup$
When electrons leave the top plate it then has a net positive charge which will tract electron to the bottom plate which thus becomes negatively charged.
So there is a movement of electrons from the top plate (which becomes positive) through the voltage source and onto the bottom plate (which becomes negative).
answered 13 hours ago
FarcherFarcher
55.2k3 gold badges45 silver badges119 bronze badges
$endgroup$
1
$begingroup$
But it's not the very same electrons that move from the + side to the - side of the cap. It's a shift in charge density profile.
$endgroup$
– Bill N
12 hours ago
1
$begingroup$
@BillN I am sorry of my answer gives that impression as it was never my intention to do so.
$endgroup$
– Farcher
11 hours ago
add a comment |
$begingroup$
When electrons leave the top plate it then has a net positive charge which will tract electron to the bottom plate which thus becomes negatively charged.
So there is a movement of electrons from the top plate (which becomes positive) through the voltage source and onto the bottom plate (which becomes negative).
answered 13 hours ago
FarcherFarcher
55.2k3 gold badges45 silver badges119 bronze badges
$endgroup$
When electrons leave the top plate it then has a net positive charge which will tract electron to the bottom plate which thus becomes negatively charged.
So there is a movement of electrons from the top plate (which becomes positive) through the voltage source and onto the bottom plate (which becomes negative).
answered 13 hours ago
FarcherFarcher
55.2k3 gold badges45 silver badges119 bronze badges
answered 13 hours ago
FarcherFarcher
55.2k3 gold badges45 silver badges119 bronze badges
answered 13 hours ago
FarcherFarcher
55.2k3 gold badges45 silver badges119 bronze badges
answered 13 hours ago
FarcherFarcher
55.2k3 gold badges45 silver badges119 bronze badges
55.2k3 gold badges45 silver badges119 bronze badges
1
$begingroup$
But it's not the very same electrons that move from the + side to the - side of the cap. It's a shift in charge density profile.
$endgroup$
– Bill N
12 hours ago
1
$begingroup$
@BillN I am sorry of my answer gives that impression as it was never my intention to do so.
$endgroup$
– Farcher
11 hours ago
add a comment |
1
$begingroup$
But it's not the very same electrons that move from the + side to the - side of the cap. It's a shift in charge density profile.
$endgroup$
– Bill N
12 hours ago
1
$begingroup$
@BillN I am sorry of my answer gives that impression as it was never my intention to do so.
$endgroup$
– Farcher
11 hours ago
1
1
$begingroup$
But it's not the very same electrons that move from the + side to the - side of the cap. It's a shift in charge density profile.
$endgroup$
– Bill N
12 hours ago
$begingroup$
But it's not the very same electrons that move from the + side to the - side of the cap. It's a shift in charge density profile.
$endgroup$
– Bill N
12 hours ago
1
1
$begingroup$
@BillN I am sorry of my answer gives that impression as it was never my intention to do so.
$endgroup$
– Farcher
11 hours ago
$begingroup$
@BillN I am sorry of my answer gives that impression as it was never my intention to do so.
$endgroup$
– Farcher
11 hours ago
add a comment |
$begingroup$
The voltage source does work on the charges to put them onto the bottom plate until the amount of charge $Q$ on the bottom plate is given by $Q=CV$, where $C$ is the capacitance and $V$ is the potential difference between the two sides of the battery. The power source doesn't store charges like you propose.
answered 13 hours ago
Aaron StevensAaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
$endgroup$
1
$begingroup$
in voltage source direction of electric field is from top to bottom and electrons must move from bottom to top.
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
I think positive and negative poles of voltage source reduced, but voltage is constant. how?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Batteries do work against the usual forces caused by the field. That's the point of batteries. To do work on the charge. There is some process in the battery that does this. What that is depends on the type of battery.
$endgroup$
– Aaron Stevens
13 hours ago
1
$begingroup$
Ok. that is mean when electrons arrive to the positive terminal of voltage source, voltage source push them to the negative terminal and negative terminal push them to bottom plate of capacitor?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Kind of, yes. Once charges leave the battery then the battery isn't influencing them. The battery just gives the charges energy so that there can be a charge separation between the two plates of the capacitor.
$endgroup$
– Aaron Stevens
13 hours ago
add a comment |
$begingroup$
The voltage source does work on the charges to put them onto the bottom plate until the amount of charge $Q$ on the bottom plate is given by $Q=CV$, where $C$ is the capacitance and $V$ is the potential difference between the two sides of the battery. The power source doesn't store charges like you propose.
answered 13 hours ago
Aaron StevensAaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
$endgroup$
1
$begingroup$
in voltage source direction of electric field is from top to bottom and electrons must move from bottom to top.
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
I think positive and negative poles of voltage source reduced, but voltage is constant. how?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Batteries do work against the usual forces caused by the field. That's the point of batteries. To do work on the charge. There is some process in the battery that does this. What that is depends on the type of battery.
$endgroup$
– Aaron Stevens
13 hours ago
1
$begingroup$
Ok. that is mean when electrons arrive to the positive terminal of voltage source, voltage source push them to the negative terminal and negative terminal push them to bottom plate of capacitor?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Kind of, yes. Once charges leave the battery then the battery isn't influencing them. The battery just gives the charges energy so that there can be a charge separation between the two plates of the capacitor.
$endgroup$
– Aaron Stevens
13 hours ago
add a comment |
$begingroup$
The voltage source does work on the charges to put them onto the bottom plate until the amount of charge $Q$ on the bottom plate is given by $Q=CV$, where $C$ is the capacitance and $V$ is the potential difference between the two sides of the battery. The power source doesn't store charges like you propose.
answered 13 hours ago
Aaron StevensAaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
$endgroup$
The voltage source does work on the charges to put them onto the bottom plate until the amount of charge $Q$ on the bottom plate is given by $Q=CV$, where $C$ is the capacitance and $V$ is the potential difference between the two sides of the battery. The power source doesn't store charges like you propose.
answered 13 hours ago
Aaron StevensAaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
answered 13 hours ago
Aaron StevensAaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
answered 13 hours ago
Aaron StevensAaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
answered 13 hours ago
Aaron StevensAaron Stevens
20.6k4 gold badges35 silver badges74 bronze badges
20.6k4 gold badges35 silver badges74 bronze badges
1
$begingroup$
in voltage source direction of electric field is from top to bottom and electrons must move from bottom to top.
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
I think positive and negative poles of voltage source reduced, but voltage is constant. how?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Batteries do work against the usual forces caused by the field. That's the point of batteries. To do work on the charge. There is some process in the battery that does this. What that is depends on the type of battery.
$endgroup$
– Aaron Stevens
13 hours ago
1
$begingroup$
Ok. that is mean when electrons arrive to the positive terminal of voltage source, voltage source push them to the negative terminal and negative terminal push them to bottom plate of capacitor?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Kind of, yes. Once charges leave the battery then the battery isn't influencing them. The battery just gives the charges energy so that there can be a charge separation between the two plates of the capacitor.
$endgroup$
– Aaron Stevens
13 hours ago
add a comment |
1
$begingroup$
in voltage source direction of electric field is from top to bottom and electrons must move from bottom to top.
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
I think positive and negative poles of voltage source reduced, but voltage is constant. how?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Batteries do work against the usual forces caused by the field. That's the point of batteries. To do work on the charge. There is some process in the battery that does this. What that is depends on the type of battery.
$endgroup$
– Aaron Stevens
13 hours ago
1
$begingroup$
Ok. that is mean when electrons arrive to the positive terminal of voltage source, voltage source push them to the negative terminal and negative terminal push them to bottom plate of capacitor?
$endgroup$
– user3728644
13 hours ago
1
$begingroup$
@user3728644 Kind of, yes. Once charges leave the battery then the battery isn't influencing them. The battery just gives the charges energy so that there can be a charge separation between the two plates of the capacitor.
$endgroup$
– Aaron Stevens
13 hours ago
1
1
$begingroup$
in voltage source direction of electric field is from top to bottom and electrons must move from bottom to top.
$endgroup$
– user3728644
13 hours ago
$begingroup$
in voltage source direction of electric field is from top to bottom and electrons must move from bottom to top.
$endgroup$
– user3728644
13 hours ago
1
1
$begingroup$
I think positive and negative poles of voltage source reduced, but voltage is constant. how?
$endgroup$
– user3728644
13 hours ago
$begingroup$
I think positive and negative poles of voltage source reduced, but voltage is constant. how?
$endgroup$
– user3728644
13 hours ago
1
1
$begingroup$
@user3728644 Batteries do work against the usual forces caused by the field. That's the point of batteries. To do work on the charge. There is some process in the battery that does this. What that is depends on the type of battery.
$endgroup$
– Aaron Stevens
13 hours ago
$begingroup$
@user3728644 Batteries do work against the usual forces caused by the field. That's the point of batteries. To do work on the charge. There is some process in the battery that does this. What that is depends on the type of battery.
$endgroup$
– Aaron Stevens
13 hours ago
1
1
$begingroup$
Ok. that is mean when electrons arrive to the positive terminal of voltage source, voltage source push them to the negative terminal and negative terminal push them to bottom plate of capacitor?
$endgroup$
– user3728644
13 hours ago
$begingroup$
Ok. that is mean when electrons arrive to the positive terminal of voltage source, voltage source push them to the negative terminal and negative terminal push them to bottom plate of capacitor?
$endgroup$
– user3728644
13 hours ago
1
1
$begingroup$
@user3728644 Kind of, yes. Once charges leave the battery then the battery isn't influencing them. The battery just gives the charges energy so that there can be a charge separation between the two plates of the capacitor.
$endgroup$
– Aaron Stevens
13 hours ago
$begingroup$
@user3728644 Kind of, yes. Once charges leave the battery then the battery isn't influencing them. The battery just gives the charges energy so that there can be a charge separation between the two plates of the capacitor.
$endgroup$
– Aaron Stevens
13 hours ago
add a comment |
$begingroup$
I will rephrase your question as I interpret it. Perhaps this is not what you're getting at, but here goes regardless.
Even a small flashlight battery can push a coulomb of charge through a circuit. If you start with an even mix of protons and electrons everywhere (so there is no net accumulation of charge anywhere), and then you remove a coulomb of electrons from the negative terminal and add a coulomb of electrons to the positive terminal (which is what you're doing if you drain the battery through the circuit), now you should have +1 C and -1 C accumulated at the terminals, an inch or two apart. Thus there should be a ginormous force between the terminals, $10^12$ newtons or something, far more than the battery could possibly withstand structurally. How can this be?
The answer is that the change in charge at the terminals due to the departure and arrival of electrons is compensated by the movement of ions within the electrolyte. There is a chemical reaction at the anode that donates electrons to the negative terminal and creates positive ions in the electrolyte, and another one at the cathode in which the positive terminal gives electrons to the positive ions, and the flow of ions within the electrolyte keeps charge from building up anywhere. This may be a bit of an oversimplification of the electrochemistry involved, but I think it captures the general concept.
answered 11 hours ago
Ben51Ben51
4,1048 silver badges30 bronze badges
$endgroup$
add a comment |
$begingroup$
I will rephrase your question as I interpret it. Perhaps this is not what you're getting at, but here goes regardless.
Even a small flashlight battery can push a coulomb of charge through a circuit. If you start with an even mix of protons and electrons everywhere (so there is no net accumulation of charge anywhere), and then you remove a coulomb of electrons from the negative terminal and add a coulomb of electrons to the positive terminal (which is what you're doing if you drain the battery through the circuit), now you should have +1 C and -1 C accumulated at the terminals, an inch or two apart. Thus there should be a ginormous force between the terminals, $10^12$ newtons or something, far more than the battery could possibly withstand structurally. How can this be?
The answer is that the change in charge at the terminals due to the departure and arrival of electrons is compensated by the movement of ions within the electrolyte. There is a chemical reaction at the anode that donates electrons to the negative terminal and creates positive ions in the electrolyte, and another one at the cathode in which the positive terminal gives electrons to the positive ions, and the flow of ions within the electrolyte keeps charge from building up anywhere. This may be a bit of an oversimplification of the electrochemistry involved, but I think it captures the general concept.
answered 11 hours ago
Ben51Ben51
4,1048 silver badges30 bronze badges
$endgroup$
add a comment |
$begingroup$
I will rephrase your question as I interpret it. Perhaps this is not what you're getting at, but here goes regardless.
Even a small flashlight battery can push a coulomb of charge through a circuit. If you start with an even mix of protons and electrons everywhere (so there is no net accumulation of charge anywhere), and then you remove a coulomb of electrons from the negative terminal and add a coulomb of electrons to the positive terminal (which is what you're doing if you drain the battery through the circuit), now you should have +1 C and -1 C accumulated at the terminals, an inch or two apart. Thus there should be a ginormous force between the terminals, $10^12$ newtons or something, far more than the battery could possibly withstand structurally. How can this be?
The answer is that the change in charge at the terminals due to the departure and arrival of electrons is compensated by the movement of ions within the electrolyte. There is a chemical reaction at the anode that donates electrons to the negative terminal and creates positive ions in the electrolyte, and another one at the cathode in which the positive terminal gives electrons to the positive ions, and the flow of ions within the electrolyte keeps charge from building up anywhere. This may be a bit of an oversimplification of the electrochemistry involved, but I think it captures the general concept.
answered 11 hours ago
Ben51Ben51
4,1048 silver badges30 bronze badges
$endgroup$
I will rephrase your question as I interpret it. Perhaps this is not what you're getting at, but here goes regardless.
Even a small flashlight battery can push a coulomb of charge through a circuit. If you start with an even mix of protons and electrons everywhere (so there is no net accumulation of charge anywhere), and then you remove a coulomb of electrons from the negative terminal and add a coulomb of electrons to the positive terminal (which is what you're doing if you drain the battery through the circuit), now you should have +1 C and -1 C accumulated at the terminals, an inch or two apart. Thus there should be a ginormous force between the terminals, $10^12$ newtons or something, far more than the battery could possibly withstand structurally. How can this be?
The answer is that the change in charge at the terminals due to the departure and arrival of electrons is compensated by the movement of ions within the electrolyte. There is a chemical reaction at the anode that donates electrons to the negative terminal and creates positive ions in the electrolyte, and another one at the cathode in which the positive terminal gives electrons to the positive ions, and the flow of ions within the electrolyte keeps charge from building up anywhere. This may be a bit of an oversimplification of the electrochemistry involved, but I think it captures the general concept.
answered 11 hours ago
Ben51Ben51
4,1048 silver badges30 bronze badges
edited 10 hours ago
answered 11 hours ago
Ben51Ben51
4,1048 silver badges30 bronze badges
answered 11 hours ago
Ben51Ben51
4,1048 silver badges30 bronze badges
answered 11 hours ago
Ben51Ben51
4,1048 silver badges30 bronze badges
4,1048 silver badges30 bronze badges
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