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It was an introduction lecture to capacitors. The parallel plate capacitor has two parallel plates both charged equally and opposite in sign. They have and insulating medium in between them. When connected to an closed circuit the current will flow through capacitor for a brief time before it gets completely charged. And when the capacitor is fully charged the, the plates will have max potential difference and the circuit will be an open circuit.

What I argue is that:
How can current flow through the wire when capacitor has a insulator between them. It is such open thing to add in a circuit.

I think it is only possible if charge the capacitors enough such that the insulating medium ionise and conduct electricity, just like it happens during lightning in the atmosphere.

Ishant
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4 Answers4

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You must have understood something wrong. The current flows to and from the two sides, so you can measure it, but itn never rally flows through the capacitor. So it seems like the capacitor acts like a resistor with increasing resistance.

trula
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Capacitors, just like many other things in the world, can be understood in more than one way.

Imagine an un-charged capacitor. Imagine that you connect it to a DC power supply. What happens? The capacitor becomes charged. How? It becomes charged by a pulse of current that flows, just for a moment, pulling some electrons from one plate, adding some electrons to the other plate. No electrons flow through the dielectric that separates the plates.

Now imagine that you reverse the connections. What happens? You get another pulse of current that pulls electrons from the formerly negative plate, giving it a positive charge and, adds electrons to the formerly positive plate, giving it a negative charge.

Now, finally, imagine that you connect the capacitor across an AC power supply. The polarity of the supply is continually reversing. You get continual pulses of current. It appears as if the capacitor is "conducting AC current," when in fact, no current ever flows across the dielectric.

Electrical engineers who work with AC signals find it convenient to drop the "as if," and simply pretend that the capacitor "conducts AC current" and, "blocks DC current." It's a simplification of the underlying reality, and it is a simplification that is useful to them. It helps them to get their job done more quickly.

Literally everything we know about the universe is, on some level, a simplification. And, many things can be usefully understood on more than one level. One difference between physicists and engineers is that engineers, who are mainly interested in getting something done, prefer to look at the highest level that is not too simple to be of any use. It helps them to get the job done faster. Physicists, on the other hand, who are mainly interested in expanding our knowledge of the universe, tend to spend more time looking at the "deeper" levels.

Solomon Slow
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The charge convection current density $J$ that flows into the capacitor plate surfaces is continuing in the dielectric between the plates as a displacement current density $$J_{disp}=\epsilon \frac {dE}{dt}$$ This is what Maxwell discovered and put into his equation $$ \nabla \times H = J+ \epsilon \frac {dE}{dt}$$ In the whole circuit you have that the integral over the cross section of the current densities $$\int (J+\epsilon \frac {dE}{dt}) da=I$$ i.e., the total current $I$, is continuous.

freecharly
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AC current flows through a capacitor. No individual electron, however, passes through the capacitor. That's because the wire's electrons go to one plate, and that plate repels electrons in the dielectric, so electrons pile up next to the opposite capacitor plate, and repel electrons in that other plate out through the opposite wire. At the next half-cycle, those motions all reverse, of course.

So while there's no net movement of a charge through the capacitor, there's plenty of AC current.

The movement of charge in the dielectric isn't the only current, though: even if only the electric field in the gap (like a vacuum capacitor) is present, that AC field is a 'displacement current' term that must be deemed real, as it has the same effect as current; this is the reason that electromagnetic waves (light and radio and such) can propagate through B field action in a vacuum; the current to make the B field is that dE/dt displacement current.

Whit3rd
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