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Suppose that the positive terminal of a battery is connected to one of the plates of the capacitor, and the other plate isn't connected anywhere. The other terminal of the battery is connected to a conducting wire. The positive terminal of the battery would then attract some of the electrons from the plate and thus leave a positive charge. The other plate would then become negatively charged. The capacitor would then be charged. Why doesn't this happen in practice?

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  • $\begingroup$ Why would the other plate become negatively charged? Where does that negative charge come from if the plate isn't connected? $\endgroup$
    – John Doty
    Commented Jul 14, 2022 at 19:10
  • $\begingroup$ @JohnDoty Wouldn't the negative charges be attracted to the positively charged plate, therefore resulting in polarization. $\endgroup$
    – Piksiki
    Commented Jul 14, 2022 at 19:13
  • $\begingroup$ Again, where would those negative charges come from if the plate's not connected? $\endgroup$
    – John Doty
    Commented Jul 14, 2022 at 19:36
  • $\begingroup$ @JohnDoty the charges within the plate $\endgroup$
    – Piksiki
    Commented Jul 14, 2022 at 20:15

2 Answers 2

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Connecting one plate to the positive terminal of a battery and leaving the other plate disconnected will simply result in the existing free charge (electrons) on each of the two plates to redistribute themselves within each plate in response to the field of the positive battery terminal. The net charge on each plate will still be zero, i.e., the capacitor does not become "charged" because there is no net charge on each plate.

In order for the capacitor to become charged there needs to be a means to remove charge from one plate and deliver that charge to another. Although the positive battery terminal attracts electrons pulling them to one side of the plate, it can't remove them and deposit them on the other plate, which is how a capacitor gets "charged". That requires the negative terminal be connected to the other plate to complete the circuit.

See the diagrams below. Notice that although the free electrons move within each plate, each plate remains neutral.

Hope this helps.

enter image description here

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  • $\begingroup$ why can't the battery remove the negative charges from the plate, at least until charge buildup occurs on the positive terminal? $\endgroup$
    – Piksiki
    Commented Jul 14, 2022 at 20:53
  • $\begingroup$ The electrons that moved to the right side of the plate connected to the battery terminal are kept from coming off the plate by the attraction of the positive charge it left behind on the plate. It's similar to the situation with fully charged capacitors in series. Although a negatively charged plate is directly connected to a positively charged plate electrons don't move between the plates, because they are bound by the charge on the other plate. $\endgroup$
    – Bob D
    Commented Jul 14, 2022 at 21:07
  • $\begingroup$ won't the plate connected to the positive terminal eventually reach the potential of the positive terminal? If there is no charge movement from the plate, how is this potential set up? $\endgroup$
    – Piksiki
    Commented Jul 16, 2022 at 20:57
  • $\begingroup$ @Piksiki Did you think about two series charged capacitors? A negatively charged plate of one capacitor is connected to the positively charged plate of the other capacitor. What is necessary for the two plates to be at the same potential? $\endgroup$
    – Bob D
    Commented Jul 16, 2022 at 21:55
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The answer is that there is a very tiny capacitance between the "dangling" lead of the capacitor and the other terminal of the battery. The full circuit is then Batter -> real capacitor -> weak parasitic capacitor.

You can then solve the two capacitors in series to determine how much charge will build up on the two terminals of the real capacitor. If you include the internal resistance of the battery you can calculate the time constant for how long it will take the capacitor to charge up.

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