Simple Capacitor Question

[TheMightyTexan]

Though i'm sure this may have been asked before, i did not find a thread that discussed it.

I'm having a bit of difficulty understanding how a capacitor can be in a circuit and still allow electrons to flow (through/around) it to other components on the circuit.

For example, a simple circuit that contains a voltage source (battery), a light bulb, and a capacitor. The circuit is the bridge between the positive and negative terminals of the battery and it should flow through one conducting terminal in the light bulb and out the other, making its way to the capacitor, going into it and charging one of the plates...but then what? I feel like this is a really dumb thing to ask on here, but i haven't come across the answer yet and its eating up a bit of time. Thanks in advance!

Edit: Unless this the phenomenon of dielectric breakdown. However, i assumed that the capacitor is always under the same electric field or voltage source..thus it it could transfer the current with the voltage on, when the voltage is off it would immediately cross to the other plate defeating its purpose all together.

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[NextStepTutor_2]

In the situation where you have a lightbulb and a capacitor in series, once the capacitor is charged, current will not flow through it. The light bulb will therefore not light up. Typically these circuits have switches to direct the current around the capacitor or to allow for charging and discharging.

 

[TheMightyTexan]

Gotcha, thanks! My next question then is if capacitors are in series is this same scenario applied where we are supposed to assume that current is making it was down the line? One of thee TBR I believe 9.9b (the passage that points out that a large and small capacitor receive the same charge but the larger capacitor has less voltage difference) triggered my thoughts on this.

 

[TheMightyTexan]

Sorry that may have not been clear I was meaning to say, in a situation where capacitors are in series, how are we supposed to know how charge continues through the capacitor, as I could not find much information on it. (I wish I hadn't sold my physics book now) I just figured i didn't see it, but I reread ek and TBR yesterday and did not find much.

 

[TheMightyTexan]

bump

 

[JMMTB]

capacitors in series have the same charge because across the parallel plates no charged particles can the gap (or your capacitor is leaky and broken). So because of the law of conservation of charge if the positive plate of one capacitor is say +5 then the plate across from it is -5 and the plate behind it would be -5 as well.

In the picture, the longer lines represent the larger capacitor and because they have the same charge and Q=CV C1V1=C2V2. If C2 is greater than C1 then V1 must be greater than V2

 

[Hadi7183]

Don't you think the polarity of voltage on capacitors should be reversed?

 

[JMMTB]

Ya sorry about that. problem fixed now

 

[TheMightyTexan]

I get how capacitors can have the same charge, I just don't understand how current keeps going "through" them around a circuit.

Nextsteptutor's answer makes sense, i'm just wondering now if they simply omit showing this "capacitor bypass" in pretty much all circuits where something is in series with a capacitor

 

[JMMTB]

I guess I don't understand your question then. In reality current never flows through or across a capacitor, doing so would cause it to be a wire. While charging or discharging a capacitor, current flows until charge buildup is maxed out on the capacitors. Electrons are pulled off of the anode side of the capacitor into the cathode of the battery until the capacitor can not hold any more charge. On the other side the anode of the battery has electrons leave and build up on the cathode side of the capacitor.

As far as bypass capacitors, they are used to prevent noise in circuits from power sources. If they are placed in parallel to another circuit element say a resistor, current can still flow across the resistor even when the capacitor is fully charged. The difference is that Q max of the capacitor is found by finding the Voltage drop across the resistor.

Once the capacitor is fully charged we can solve for the current and voltage drops across each resistor while ignoring the branch that contains the capacitor. Once we find the voltage drop across R2 we now know the voltage drop across the C (think Kirchoff's loop rules). Q would then be that voltage drop times the capacitance.

Did that answer your question?

 

[TheMightyTexan]

Shoot now i'm really feeling ******ed lol. I should have just drawn a picture here is one, my question is simply how do electrons continue from one capacitor to another in series? More of a question to help my intuition, if they pile up on one plate, and a capacitor doesn't act like a wire, how do they get to the next? Pretty bad picture, sorry.

 

[JMMTB]

No that's an excellent question. So when the capacitor is first connected to the battery it is neutral. Thus there is already charge imbalances. The anode of the battery is more negative than the neutral plate so electrons move from the anode to the capacitor to distribute the charge. By doing so it repels electrons on the other side of the capacitor as an electric field begins to form. Theses electrons move over to the next capacitor plate (C2). Although electrons never jump the gap, electrons still move along the wire between the capacitors in series because of the differences in charge on the plates.
You could make an analogy to chemistry with induced dipoles. The molecules are not becoming ionic with the loss or gain of an electron, but the electron density is redistributed by the influence of another molecule.

 

[TheMightyTexan]

Ok, i knew this would bite me in the ass. I thought that concept made enough sense, but i just ran into a Ek 1001 circuit questions (Physics #836) where the answer said to ignore the loop that had a capacitor in it because it is fully charged and no current flows through it. These two scenarios, where capacitors in series receive current, and a charged capacitor that doesn't allow current to flow past it, appears to be contradicting.

 

[TheMightyTexan]

No that's an excellent question. So when the capacitor is first connected to the battery it is neutral. Thus there is already charge imbalances. The anode of the battery is more negative than the neutral plate so electrons move from the anode to the capacitor to distribute the charge. By doing so it repels electrons on the other side of the capacitor as an electric field begins to form. Theses electrons move over to the next capacitor plate (C2). Although electrons never jump the gap, electrons still move along the wire between the capacitors in series because of the differences in charge on the plates.
You could make an analogy to chemistry with induced dipoles. The molecules are not becoming ionic with the loss or gain of an electron, but the electron density is redistributed by the influence of another molecule.

Perhaps i just didn't fully understand this, let me see if i can restate it.

Capacitor 1 in series has a build up of current (moving electrons) on the closest plate to the battery forming a (-) charge. This creates repulsion of (local electrons?) on the plate next to it and as those electrons leave that plate then becomes (+). The electrons that are pushed away from the (+) plate on capacitor 1 move down series to capacitor 2, where it charges a plate creating a (-) charge and repeating the cycle until there are no more capacitors and then at the end of the chain the e- imbalance i guess would either have to congregate or travel back to the cathode.

Gahh, my understanding of circuits must be pretty far off still :/

 

[JMMTB]

So for EK 836 it is important to note that the capacitor is in parallel with the other circuit elements (yes it is in series with one resistor but the important thing is that it is in parallel to the rest of the circuit). Just like you said, after fully charged, no more current flows because the plates have reached the max charge they can hold - their capacitance. The diagram for questions 843-846 is a simpler showing of this.

Your understanding of capacitors in series sounds correct. Even after a long time, a capacitor in series will impeded current and it will stop.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html#c1

This shows how as the Q increases I falls respectively.

 

[TheMightyTexan]

Wow, my friend just explained this to me today, just wanted to throw it in here to seal the deal in case it might help someone in the future.

My conceptualization of capacitors was off. It should have been obvious since i knew about how resistance works. I knew each capacitor got charged, but it guess it was throwing me off cause i was thinking since they did, they were getting fully charged or something. Some of the concepts i knew, but wasn't able to throw together in a practical blend. The capacitance of a capacitor will vary, meaning my visualization of a "fully charged" capacitor was wrong. Since items in series receive the same current the voltage drop is different, the total capacitance LOWERS with each capacitor added in series. 1/Ctotal = 1/C1 + 1/C2 + 1/C3. . . if you added an infinite amount of capacitors, it only weakens how much capacitance each capacitor has. Combining this with the equation Q=CV if the capacitance is lowered and the voltage received is lower, then the charge must be lower. This paired with JMMTB's earlier explanation was what i was missing. @JMMTB thanks for the help buddy!

 

[JMMTB]

No Problem! Good luck on the test!