sArGe99 said:
Homework Statement
In an RC circuit under charging, does the current actually flow across the plates of a parallel plate capacitor?
No, but if you hook up an ammeter to the circuit, it will detect a decaying current as charges accumulate on the plates.
Does charge emanate from both terminals of the battery and move to the plates of a capacitor or does charge only emanate from positive terminal and get stored in one of the plates and manage to polarize the other? How is it that both plates have equal magnitude of charge?
The entire circuit is neutrally charged and remains so
at all times. As soon as you connect the capacitor, the positive terminal of the battery is at higher potential than the plate to which it is connected. The free electrons in the conductors will move from the positive plate, through the battery and onto the negative plate. When the positive plate plus connecting wire is at the same potential as the positive terminal, electrons stop flowing. The electrons on the negative plate leave an absence of electrons on the positive plate because the whole thing is neutral. That is why the charges o the plate are equal and opposite.
Under steady state, does no current pass through the wires connected to the capacitor even if a resistor is connected in series with it? In such case how do we find the current through the circuit,do we take into account the resistor connected in series with capacitor or just the other ones?
Remember these simple rules:
1. When an
uncharged capacitor is suddenly connected to a battery, it
initially (that means an infinitesimal amount of time after the connection) behaves as if it were a short (straight wire).
2. When a
fully charged capacitor is suddenly connected to a resistor, it
initially behaves as a battery of voltage Q/C.
3. A long time after a capacitor is connected to a circuit (steady state), it behaves as if it were an open switch.
By "behaves" I mean that, if one of the conditions as described above is met, you can replace the capacitor in the given circuit with a short or battery or open switch and analyze the new circuit instead as far as currents in resistors, etc. are concerned.