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From what I understand, a resistor causes a voltage drop. So it would seem that the end of the capacitor attached to the resistor would have the same voltage as the end attached to the battery. If there is no voltage change, then how would the capacitor become charged?

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If there is no voltage change, then how would the capacitor become charged?

I don't follow your logic here. It's true that there is a voltage drop across a resistor if there is non-zero current through the resistor.

If there is zero volts across the capacitor (at some time), then all of the battery voltage is dropped across the resistor (if you don't see this, stop here and think about it some more).

That is, when there is zero volts across the capacitor, there must be non-zero current through the resistor (assuming the battery voltage is non-zero). Since the resistor and capacitor are series connected, there is non-zero current through the capacitor which necessarily means that the voltage across the capacitor is changing.

As the voltage across the capacitor changes, the voltage across the resistor must change which implies the series current is changing. The capacitor is 'fully charged' when the voltage across the capacitor is (effectively) the same as the battery voltage. In that case, the voltage across the resistor is (effectively) zero and so there is zero series current.

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  • $\begingroup$ So in the equation $C=Q/V$, the $V$ is referring to the voltage across the capacitor when it has some charge $Q$? And the charge getting there is the result of the current through the capacitor? $\endgroup$ – Vityou Oct 9 at 3:00
  • $\begingroup$ @Vityou, it's better to write $Q=CV$ so that it's clear that the capacitance is the constant of proportionality between the charge $Q$ (on one plate, the other plate has charge $-Q$) and the voltage across the capacitor. Yes, a current through the capacitor implies a changing $Q$. $\endgroup$ – Alfred Centauri Oct 9 at 10:23
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In fact, it is the voltage across the capacitor that stops the charging process.

There is no problem running current through the capacitor when the voltage across it is zero. When the switch is thrown (not shown), there is a potential difference between the battery terminal and the capacitor. Charges will try to accumulate on the capacitor plate. Without the other plate present, the charging would stop in a very short amount of time, picoseconds or shorter I suppose (but I'm not sure). But the presence of the other plate changes things. For every charge carrier entering the positive capacitor plate, one is removed from the negative and returned to the battery. There's practically no resistance to this process.

But as the charge builds up on the plates, an induced electric field develops in the wire in the direction that opposes the charging of the plate. Eventually that field is equal to the field applied by the battery. At that point the current ... and the charging process ... stops entirely.

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  • $\begingroup$ If there is no problem running current through a capacitor when the voltage across it is zero, why is the capacitance described in terms of applied voltage: $C=Q/V$? $\endgroup$ – Vityou Oct 9 at 2:48
  • $\begingroup$ $C=Q/V$ makes no mention of current. It's a relationship between the charge stored on the plate and the voltage across the plates at any point of time. $\endgroup$ – garyp Oct 9 at 10:40

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