I'm trying to better understand the process of charging a capacitor with a battery. My textbook (the Halliday's Fundamental of Physics) describes this process in these terms:
When the circuit […] is completed, electrons are driven through the
wires by an electric field that the battery sets up in the wires. The
field drives electrons from capacitor plate h to the positive terminal
of the battery; thus, plate h, losing electrons, becomes positively
charged. The field drives just as many electrons from the negative
terminal of the battery to capacitor plate l; thus, plate l, gaining
electrons, becomes negatively charged just as much as plate h, losing
electrons, becomes positively charged.
So, if I understand correctly: because of a potential difference between the positive terminal + and plate h, an electric field appears and moves the electrons from h to +, then the battery does work to move these electrons from its positive to its negative terminal, and finally because of a potential difference between the negative terminal – and plate l, an electric field appears and moves these electrons from – to l.
The textbook goes on by saying:
Initially, when the plates are uncharged, the potential difference
between them is zero.
If the plates, which are isolated conductors, are uncharged, that is, there is no net charge on them, is it correct to assume that their electric potential is zero? Vh = 0V, Vl = 0V, then Vh - Vl = 0V.
As the plates become oppositely charged, that potential difference
increases until it equals the potential difference V between the
terminals of the battery. Then plate h and the positive terminal of
the battery are at the same potential, and there is no longer an
electric field in the wire between them. Similarly, plate l and the
negative terminal reach the same potential, and there is then no
electric field in the wire between them. Thus, with the field zero,
there is no further drive of electrons.
I'm ok with this, but if my previous assumption is correct (Vh = Vl = 0V), the negative terminal of the battery must have a negative potential, otherwise in the wire connecting it to plate l we would have:
- no electric field if
V- = 0V; - a reversed electric field if
V- > 0V.
The problem is that nowhere is said that the negative terminal of a battery must have a negative potential, the book says that it has a lower potential than the other terminal, so it could be zero or even positive.
So my assumption is incorrect? Why?
Best Answer
Doesn't matter.
Before the switch is closed, You know that plate h has the same potential as the
+terminal of the battery. It always will have because they are connected. You also know that the plate l has the same potential as the+terminal of the battery because the book told you so. It said "the potential difference between them is zero." That's a given initial condition.You know that the
-terminal of the battery always will have potential $V$ less than the+terminal because that's what (ideal) batteries do. They always maintain a constant potential difference between their terminals.You don't need know anything else in order to understand what will happen when the switch is closed. The whole circuit could have a positive charge (deficit of electrons,) or it could have a negative charge (surplus of electrons,) or it could be neutral; but the only things that determine what will happen when the switch is closed are the value of the capacitor, $C$, and the potential difference, $V$, across the switch.
P.S., The diagram, as drawn, is unrealistic. There should also be a resistor symbol in series with the other components. Even if the circuit was built with no actual resistor component, real batteries and real wires have resistance. If you try to analyze the trasient behaviour of the circuit when the switch is closed, you will conclude that the current is infinite (unphysical) unless you account for the resistance.