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I'm trying to understand the charging of capacitors.

If we have two difference size of capacitors A & B, and a DC voltage source let say 10V. Capacitance of A is bigger than B

when we connect the voltage source to the capacitor A, it will start charging the capacitor until the voltage of the capacitor is the same as the source and it becomes a open point.

If we connect the same voltage source to the smaller capacitor, B, do we get the same voltage as the bigger capacitor? so does that mean the voltage source will charge the capacitor to the same as the source regardless of the size of the capacitor. However, the bigger capacitance will hold more charge, therefore will discharge longer than the smaller capacitor?

thanks,

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Let us consider a capacitor with capacitance C, and another one with a bigger capacitance kC, where k>1. In two separate cases, these are connected in series with a resistance R and an ideal battery of EMF V like the circuit shown.

enter image description here

Now coming to your question,

If we connect the same voltage source to the smaller capacitor, B, do we get the same voltage as the bigger capacitor?

Yes we will! (For practical purpose, we also have a resistance in series for a simple circuit, but in that case too, the steady state voltage across the capacitor will be equal to V)

so does that mean the voltage source will charge the capacitor to the same as the source regardless of the size of the capacitor.

Well.. This depends on what you mean by charging. If it is the voltage across the capacitor, then yes. But if it is the charge on a plate of capacitor, then no. In general, by charging we mean the energy stored in the capacitor in the form of electric field.

For a capacitor $C$ connected across the voltage V, this will be ${CV^2}/2$ , and similarly for $kC$, it will be ${kCV^2}/2$. And hence the capacitors are charged differently, with the "bigger" one storing more energy by a factor of k.

However, the bigger capacitance will hold more charge, therefore will discharge longer than the smaller capacitor?

The charge on a capacitor in a circuit while discharging (replacing the battery by a wire in the above circuit) can be given as $$q = Qe^{-t/RC}$$

Here, $Q$ is the charge stored in steady state with the battery which is equal to $voltage/capacitance$, and $q$ is the charge at a time $t$. As you can very well see, the charge $q$ will become 0 at $t \rightarrow \infty$. But for practical purposes, we can compare the time constant $\tau$, which according to (this)is defined as the time required to

to discharge the capacitor through the same resistor to approximately 36.8% of its initial charge voltage.

And, $$\tau = RC$$ Since the capacitance of the bigger one is more, so it will have a bigger time constant.

But only this is not sufficient to compare the discharging. We also have to consider that at the first place, the charge stored on the capacitors was different. After considering these two factors, yes the capacitor with the capacitance kC will discharge longer than the smaller capacitor.

Here I have considered the "bigger" Capacitors mentioned in the question as the one with the larger capacitance. But if it means the physical dimensions, then the capacitance will depend on if the plate area is bigger, or the plate separation is bigger.

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Assume that the voltage source is a battery (e.g., DC current is involved). The capacitance of the capacitor is given by the equation $C=Q/V$. The battery will move electrons from one plate and put them onto the opposite plate, but in doing so, the negative plate acquires more and more negative charge while the positive plate acquires more and more positive charge. The negative charge repels the next electron that the battery tries to add to the negative plate, and the more charge there is on the negative plate, the stronger the repulsion becomes. This means that there is a point where the battery cannot add another electron to the negative plate, and this point occurs when the voltage across the capacitor is equal to the battery emf. Based on the equation for capacitance given above, this means that the larger the capacitance, the more charge the battery can add to the negative plate, but once again, only until the capacitor voltage drop equals the emf of the battery.

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  • $\begingroup$ i understand that. So the smaller capacitor will hold less charge but the voltage is the same as the bigger capacitor correct? Only the charge is different with the bigger vs smaller capacitor. $\endgroup$ – Ace8888 Jun 13 '17 at 2:55
  • $\begingroup$ Yes, you understood it correctly. $\endgroup$ – nasu Jun 13 '17 at 4:38
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You can think of a capacitor $D$ with a very small capacitance $c$ being charged with a charge $q$ and potential difference $V$ across the plates with $q=cV$.

Then another $N-1$ identical capacitors are charged up with the same charge and the same voltage.

Now connect all N capacitors up in parallel and you have a capacitor with effective capacitance $C=Nc$ storing charge $Q=Nq$ with potential difference $V$ across the plates.

So for a given potential difference across the plates more capacitance leads to the storage of more charge.

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