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I'm reading the capacity chapter of Serway's book, and I had a question about the charging of a parallel plate capacitor. Let's assume the following situation with a modification of the circuit in the figure: we connect the negative terminal of the battery and one of the capacitor plates to ground. The positive terminal connects directly to the plate as in the figure. I understand that the negative terminal and the plate will be at the same potential (gnd) and there should be no flow of electrons. This is so?

Physics for Scientists and Engineers 6th. By Raymond A. Serway, John W. Jewett

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At the first instance when the capacitor is hooked up, there will be a brief response whereby the plates are charged, after which the system reaches equilibrium and there will be no flow of current; only a set of capacitor plates with a potential difference as per the indicated battery. Connecting first to ground makes no difference in the scenario. The bottom line is that the capacitor will store the amount of charge on it's plates according to the potential difference it is subjected to, i.e. $C=Q/V$. The only way to avoid any current flow whatsoever, even at the first, would be to connect a capacitor that has been precisely pre-charged to the amount corresponding to the battery voltage, and that the potential of the negative plate is identical to the that of the negative terminal; whatever it may be. If this were exactly the case then there would never be any electric field in the wires, since the electric field depends on a potential difference as per: $\vec E=-\nabla\phi$. This is why the author specifies the connecting of an uncharged capacitor, so as to allow for a problem that is completely specified by the given quantities, viz. capacitance C, battery voltage V.

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  • $\begingroup$ "...The only way to avoid any current flow whatsoever, even at the first, would be to connect a capacitor that has been precisely pre-charged to the amount corresponding to the battery voltage...". If the battery negative terminal is connected to ground and the plate too, wouldn't the plate be discharged and at the same potential as the negative terminal? $\endgroup$
    – Gabriel Diez
    Commented Feb 15 at 21:32
  • $\begingroup$ I am getting at the point that if the battery terminal and the negative plate are at the same potential, i.e. "ground" and that the difference of potential between the plates is the same as that of the positive terminal and ground, if it were pre-charged in this way, then there would be equilibrium, therefore no discharge when you hook it to the battery. $\endgroup$
    – Albertus Magnus
    Commented Feb 15 at 21:37
  • $\begingroup$ Thanks for the reply. I'm ashamed that I still don't understand. We start from the initial situation with the plates discharged, when the cables are connected to the terminals (this means positive terminal to a plate, negative terminal to ground, and the remaining plate to ground), why do electrons flow towards the ground plate? if this plate is at the same potential (0) as the negative terminal. $\endgroup$
    – Gabriel Diez
    Commented Feb 16 at 1:51
  • $\begingroup$ @GabrielDiez Don't be ashamed! We have all been there, and most certainly will be there again. Well, right before you hook up the negative plate, it is not at the ground potential (it is at whatever potential); thus when you hook it up, the potential shifts to that of ground via the equalizing charges. The act of connecting the cables to the terminals brings the system together electrically as one. $\endgroup$
    – Albertus Magnus
    Commented Feb 16 at 2:07
  • $\begingroup$ Thanks, I got it! And once the ground plate is neutralized, how does it start to gain electrons? Could it be due to the electric field generated by the other plate? $\endgroup$
    – Gabriel Diez
    Commented Feb 16 at 2:34

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