Why does current initially rush into a capacitor?

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When a voltage step is applied to a capacitor, a large initial current flows due to rapid charge redistribution as the capacitor attempts to equalize its voltage with the source. Unlike simple wires, which have low capacitance and thus accumulate minimal charge, a parallel plate capacitor is designed to store more charge, requiring a larger current to reach the same voltage. Initially, the voltage across the capacitor is zero, causing the entire source voltage to appear across the resistor, resulting in a high current flow. As charge accumulates on the capacitor plates, the voltage across the resistor decreases, leading to a reduction in current until equilibrium is reached. This behavior illustrates that capacitors initially act like short circuits and eventually behave like open circuits as they charge.
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Hi all,

I'm trying to wrap my head around the physics behind the current that initial flows when a voltage step is applied across a capacitor. Namely, why does a large current initially flow into a parallel plate capacitor, but a large initial current does not flow if the capacitor is replaced by two simple wires (left unconnected)?

I understand that when a voltage is applied across any device, charges will redistribute (rapidly) such that the voltage across the device becomes equal to the voltage of the source.

So, what's so special about a capacitor (two metal plates close to each other) that causes this initial redistribution of charge to require such a large initial current?

Thanks!
Tim
 

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Nothing special about the plate configuration as far as that initial current. It's an uncharged conductor connected to a source of charges. Therefore the charges will move in such a way to try to equalize the potential. Same as touching any charged conductor to any uncharged conductor.

And when conductivity is high, as in a metal, that means the charges can flow quickly as they equalize, leading to a large current.

TimNJ said:
but a large initial current does not flow if the capacitor is replaced by two simple wires (left unconnected)?

Wires can be considered as a capacitance also. In fact the capacitance of wires can lead to some problems in some circuits. But it's a very small capacitance and it reaches potential V when only a small amount of charge has moved. A parallel plate capacitor is designed for charge storage, designed to require more charge before it reaches V.
 
If the capacitor were not there and you merely had the ends of two wires, there would still be a potential difference across the gap and hence some charge accumulation of positive and negative charges at each end. It's just that the amount of charge would be small because the capacitance would be low. The two metal plates have a larger capacitance, i.e. are an arrangement that stores more charge for the same voltage than the ends of the wires.
 
Thanks RPinPA. I guess that's why the peak current is simply V/R? So with normal conductors, there is a high initial current, but it is so brief that we don't even worry about it in most analysis?

In a capacitor, charge continues to flow until the voltage reaches equilibrium?
 
TimNJ said:
Thanks RPinPA. I guess that's why the peak current is simply V/R? So with normal conductors, there is a high initial current, but it is so brief that we don't even worry about it in most analysis?

Yes. I typed an update while you were responding.

TimNJ said:
In a capacitor, charge continues to flow until the voltage reaches equilibrium?

Yes, because that's what a conductor does, and capacitors are made of conductors. What the parallel plate configuration does is lower the potential by the nearby presence of opposite charges, thus requiring more Q to reach a given V.
 
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TimNJ said:
In a capacitor, charge continues to flow until the voltage reaches equilibrium?
Another way to look at it is that when the capacitor is initially uncharged, the voltage across its plates is zero therefore the voltage across the resistor is equal to the battery voltage but only initially. Current will flow through the resistor according to Ohm's law and will start accumulating on the plates with electrons moving from the negative plate to the positive. The charge accumulation on the capacitor reduces the voltage across the resistor which in turn reduces the current through it. When the voltage across capacitor matches the battery voltage, the current through the resistor goes to zero, i.e. there is no more current in the circuit. All this can be summarized with the following statement: "If an uncharged capacitor is connected to a battery through a resistor, it will behave as a straight wire (short circuit) immediately after and as an open switch (open circuit) a long time later. What constitutes "long time" depends on the time constant ##RC##.
 
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"Why does current initially rush into a capacitor?"
With a large series resistor there is no "rushing". Charge can just be leaking very slowly - and getting slower and sslloowweerrr.
 
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RPinPA said:
Yes. I typed an update while you were responding.

Thanks. I like thinking of it that way, I think. Here's what I take from that: If the source voltage is changed, then there will be a transient state, during which charge redistributes and voltages reach equilibrium. Even if there are just two unconnected wires hanging off a step-voltage source, there will be a extremely brief instantaneous "current", caused by electric field propagation down the line. In this situation, the change in voltage on a conductor is basically unopposed.

However, when you put a parallel plate capacitor at the end of those wires, the change in voltage across the capacitor is opposed by the electric field setup between the plates of the capacitor. The instant charge begins to flow into a capacitor, an opposite amount of charge flows out of the opposite plate, causing a charge imbalance, and thus electric field.

This electric field has an attractive force which "pulls" more charge onto the plate. More charge continues to flow into the capacitor until the electric field caused by an accumulation of like charges equals the electric field setup between the two plates.

Does this sound correct?

Thanks!
 
kuruman said:
Another way to look at it is that when the capacitor is initially uncharged, the voltage across its plates is zero therefore the voltage across the resistor is equal to the battery voltage but only initially. Current will flow through the resistor according to Ohm's law and will start accumulating on the plates with electrons moving from the negative plate to the positive. The charge accumulation on the capacitor reduces the voltage across the resistor which in turn reduces the current through it. When the voltage across capacitor matches the battery voltage, the current through the resistor goes to zero, i.e. there is no more current in the circuit. All this can be summarized with the following statement: "If an uncharged capacitor is connected to a battery through a resistor, it will behave as a straight wire (short circuit) immediately after and as an open switch (open circuit) a long time later. What constitutes "long time" depends on the time constant ##RC##.

Yes! From a circuit analysis perspective that certainly checks out! Thank you for that.
 
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TimNJ said:
This electric field has an attractive force which "pulls" more charge onto the plate.
Ignoring any relative potential between the circuit and Earth, there is a 'pull' on the electrons from one plate (+) and a 'push' from the other (-). The same magnitude of charge collects on each plate; no total charge is gained or lost.
 
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sophiecentaur said:
Ignoring any relative potential between the circuit and Earth, there is a 'pull' on the electrons from one plate (+) and a 'push' from the other (-). The same magnitude of charge collects on each plate; no total charge is gained or lost.

That makes a lot of sense. Thanks.

I guess I still have some confusion with respect to the different electric fields, specifically how they change with time, and how that relates to the charging behavior of capacitor. I've ordered an old edition of a book "Matter & Interactions" by Sherwood and Chabay, that I hope will have some answers to my conceptual misunderstandings.
 
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