Understanding Charge Accumulation in Parallel Plate Capacitors

AI Thread Summary
Charge accumulation in parallel plate capacitors occurs despite them appearing as open circuits to DC voltage sources. Initially, current flows into the capacitor, allowing charge to build up on the plates until the voltage across them opposes the source voltage, at which point current ceases. While electrons flow through the circuit, they do not flow through the capacitor itself; instead, equal and opposite charges accumulate on the plates. Capacitors do not hold net charge, as the positive charges on one plate equal the negative charges on the other, resulting in a net charge of zero. The concept of "displacement current," introduced by Maxwell, explains the electromagnetic fields around capacitors despite the absence of actual charge flow within them.
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Homework Statement


My lecture notes say that no current flows through capacitors as the dielectric material is a poor conductor. It goes on to say that the capacitor appears as an open circuit to a DC voltage source. It then says that charge will accumulate on the plates with an equal amount of positive and negative charge accumulating on both of the plates.

Homework Equations


N/A

The Attempt at a Solution


What my issue is how can charge accumulate on the plates if it appears a open circuit to the DC voltage source? Doesn't open circuit mean no current flows(thus no charge)?.
 
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A capacitor appears as an open circuit only for steady, unchanging currents and voltages. Until the circuit's currents and voltages settle down to their steady-state values, the capacitor can facilitate current through it (current goes in one side and and an equal current emerges from the other side).

A common scenario is for current to flow and charge to build up on the plates until the voltage across the capacitor plates exactly opposes the voltage source supplying that current, and at that point current ceases and the accumulated charge remains fixed.
 
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NascentOxygen said:
A capacitor appears as an open circuit only for steady, unchanging currents and voltages. Until the circuit's currents and voltages settle down to their steady-state values, the capacitor can facilitate current through it (current goes in one side and and an equal current emerges from the other side).

Just in case that's confusing... It's not the same electrons that flow in one side and out the other.

Think of it from an electrostatic point of view... Perhaps take a looks at how thunder clouds build up a charge... and the opposite charge is induced in the ground or church spire below. No current flows between cloud and ground, at least not until there is a lightning strike.
 
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Actually it IS the same electrons that flow out from the + plate to the - plate thru the circuit in which the capacitor is connected. So there are actual electrons flowing thru the capacitor's leads. But there are no electrons flowing inside the capacitor. There is just equal & opposite charge buildup on the two electrodes (plates, whatever).

Furthermore, capacitors charged by a current (i.e in a circuit) never hold a net charge. The total + charges on one electrode = the total - charges on the other electrode so net charge = 0.

Maxwell, in formulating his famous equations, postulated what was later called a "displacement current" flowing inside the capacitor. It was almost an afterthought at the time, yet that term accounts for all electromagnetic radiation! It was necessitated by the fact that a magnetic field surrounds the capacitor body same as around the leads even though there is charge flow in the leads but not in the capacitor.
 
Sorry I wasn't clear. I agree it's the same quantity of electrons that flow in both leads but it's not literally the same electron.
 
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