Electric current flow in a circuit when a capacitor is fully charged?

AI Thread Summary
When a capacitor is fully charged, the electric field created by the charge on its plates opposes the electric field from the battery, resulting in a net electric field of zero within the circuit. This means that the electrons in the wire do not experience any force, leading to no current flow. Although the capacitor is said to be fully charged, it never truly reaches that state, as it asymptotically approaches the maximum charge. The discussion also touches on the concept that the electric field is not confined solely between the capacitor plates, as it can extend into the wires connected to the capacitor. Ultimately, the balance of electric fields prevents current from flowing once the capacitor reaches its charged state.
atavistic
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Why doesn't electric current flow in a circuit when a capacitor is fully charged?

I mean there is still the battery making the E field, why is it unable to move the electrons of the wire of the circuit?
 
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atavistic said:
Why doesn't electric current flow in a circuit when a capacitor is fully charged?

I mean there is still the battery making the E field, why is it unable to move the electrons of the wire of the circuit?
As the capacitor is being charged, electrons are building up on one of the plates. These electrons create and electric field opposing the build-up of additional charge on the capacitor. When the capacitor is fully* charged, the electric field of the electrons on the capacitor plate is equal in magnitude but opposite in direction to the electric field created by the external potential. Therefore, the net electric field inside the wire is zero and the electrons in the wire experience no external force and hence no current will flow.

*Without wanting to confuse the matter further, a capacitor is never actually fully charged. You cannot place a charge of V/R onto the capacitor plates in a finite period of time. Instead, V/R should be thought of as a limit: the amount of charge on the capacitor plates approaches the maximum value of V/R asymptotically. See http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/capchg.html" for more information.
 
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You could think of one of the wires from the battery to the capacitor, say the +ve . When fully charged the voltage at one end of the wire is the same as the other wrt to the -ve so there is current flow. No potential difference across the resistance of the wire so no current.. I = V/R.
 


@hootenanny : But isn't the E field of the capacitor confined only to the region in between the plates.

@Pumblechook: Yes but I actually wanted the reason to be in the lines of what hootenanny said.

Thnx for replies. Awaiting more.
 


atavistic said:
@hootenanny : But isn't the E field of the capacitor confined only to the region in between the plates.
No, it isn't. I can understand why you would think that this would be the case, since any diagram showing the electric field lines of a parallel plate capacitor simply show the field between the plates, but this is not the case. Consider a wire which has a 'build-up' of electrons at one end of the wire, this would create an electric field both inside the wire and one outside the wire at the end.

Do you follow?
 


But isn't the E field of the capacitor confined only to the region in between the plates.

Of course it is. The electric field inside a conductor with static charge is ZERO... You might argue about a bit of fringing at the edges..but it's irrelevant here.

Consider a wire which has a 'build-up' of electrons at one end of the wire, this would create an electric field both inside the wire and one outside the wire at the end.

Also not true..same explanation.

These seem good:
You could think of one of the wires from the battery to the capacitor, say the +ve . When fully charged the voltage at one end of the wire is the same as the other wrt to the -ve so there is current flow.
and

Therefore, the net electric field inside the wire is zero and the electrons in the wire experience no external force and hence no current will flow.

The old water pressure as voltge and water flow as current also works...a capacitor only holds a limited amount of water...when full, flow halts.
 


Well I assumed it was actually the fringe field which caused E field to cancel.
 


Naty1 said:
Of course it is. The electric field inside a conductor with static charge is ZERO... You might argue about a bit of fringing at the edges..but it's irrelevant here.
Correct, but this begs the question how does the electric field become zero?

Whilst the capacitor is charging, we have a non-static charge distribution, i.e. a current is flowing. Hence, the electric field in the conductor is non-zero. So, my question to you is, how does the electric field become zero? What causes the electric field to be zero inside the conductor, which initially had a non-zero electric field and hence, a current flowing through it?
 
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