Conceptual understanding of capacitors

[axp107]

For some reason, I just don't get capacitors. When they store charge, do they hog it, giving less current/charge in other parts of the circuit..

So when capacitors are connected to a voltage supply, they store charge.

Q = CV ... I'm assuming V is the voltage across the capacitor plates... or is it the total voltage in the circuit

When its connected, what happens to charge q and V, voltage

What about when its disconnected?

I never really had a good conceptual understanding of capacitors and dielectric in E&M.. I only learned how to use formulas

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[BrokenGlass]

For some reason, I just don't get capacitors. When they store charge, do they hog it, giving less current/charge in other parts of the circuit..

So when capacitors are connected to a voltage supply, they store charge.

Q = CV ... I'm assuming V is the voltage across the capacitor plates... or is it the total voltage in the circuit

When its connected, what happens to charge q and V, voltage

What about when its disconnected?

I never really had a good conceptual understanding of capacitors and dielectric in E&M.. I only learned how to use formulas

Capacitors store charge (and energy) for quick delivery. When a parallel-plate capacitor is being charged, electrons from the voltage source flow onto one of its plates. These electrons repel electrons on the other plate, so the first plate accumulates negative charge and the other plate accumulates positive charge of the same magnitude. Current that flows towards the capacitor goes down with time because electrons that are already on the plate repel the newly-arriving electrons.

When the potential difference between capacitor plates equal the potentail difference of the voltage course, the capacitor becomes fully charged. Now if we disconnect the capacitor from the voltage source and connect it to say a light bulb, the capacitor will discharge very fast by providing energy for the light bulb.... This is done by first closing the circuit that chages the capacitor and
then closing the circuit that discharges a capacitor. This is how pacemakers work...

C = Q/V (ie. capacitance is the ability to store charge while maintaning a low voltage). C is capacitance, Q is the charge on the plates, V is voltage between the plates....

I am sure others will add a lot more to this and/or correct me....

 

[scgamer2000]

For some reason, I just don't get capacitors. When they store charge, do they hog it, giving less current/charge in other parts of the circuit..

So when capacitors are connected to a voltage supply, they store charge.

Q = CV ... I'm assuming V is the voltage across the capacitor plates... or is it the total voltage in the circuit

When its connected, what happens to charge q and V, voltage

What about when its disconnected?

I never really had a good conceptual understanding of capacitors and dielectric in E&M.. I only learned how to use formulas

Yes, you're correct, V refers to the voltage applied across the capacitor, which is not necessarily the voltage applied across the whole circuit.

The charge accumulates on the capacitor because it's really a break in the circuit. The electrons and "holes" attempt to move in the appropriate directions when a voltage is applied but they are unable to cross the gap in the capacitor. Eventually, so many electrons and holes accumulate on the parallel plates that no more can be added (think repulsion). In a perfect world, if you were to now remove the capacitor, the charge would remain on it. What you can do with a charged capacitor is attach its leads to a resistor and drain the accumulated voltage off of it.

A tire might make a good conceptual example. You can only inflate a tire so much, and the more you inflate it, the harder it gets to put more air into it. The same is true of trying to charge a capacitor. But, if you vent the tire, air will flow out very quickly initially and then more slowly as the pressure drops. Again, the same happens with a capacitor if you close the circuit (and when there is no external voltage source applied to it).

I can try and explain this better if you need more help.

 

[corbis11]

Don't forget about dielectrics (a non-conductor) which increase the capacitance.

 

[gm_navy]

For some reason, I just don't get capacitors. When they store charge, do they hog it, giving less current/charge in other parts of the circuit..

So when capacitors are connected to a voltage supply, they store charge.

Q = CV ... I'm assuming V is the voltage across the capacitor plates... or is it the total voltage in the circuit

When its connected, what happens to charge q and V, voltage

What about when its disconnected?

I never really had a good conceptual understanding of capacitors and dielectric in E&M.. I only learned how to use formulas

 

[gm_navy]

One way of thinking about a capacitor is thinking in mechanical terms. Think of a pipe containing water. let the pipe be blocked at a certain point by a rubber sheet or membrane. It will make sure that water can not go past or through it just as a capacitor wont allow DC through. However, when a DC voltage is applied to a circuit with a capacitor, the capacitor gets charged, in the same way as if a steady pressure be maintained on the water on one side of the rubber sheet, the rubber distends to one side to such an extent, that its elastic force of restoration becomes equal to the pressure.

However, if we were to apply a sinusoidally alternating pressure to the pipe of water, that is pressure builds up to a peak value on one side, then it falls to zero, then pressure is made to build up on the other side, and then again to zero and so on, the pipe does sustain an alternating flow of water flowing to and fro, This is exactly in the same manner as a capacitor permits alternating current to flow through it. No water penetrates the rubber membrane, just as no current moves from one plate of the capacitor to the other. The two plates get charged and discharged, just like the rubber membrane stretches first to one side and then to the other.