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Let's just say you run a 5V current from the battery into a circuit. From what I understand, this means that I am essentially pushing high potential electrons into the circuit. These electrons would be able to do work through a resistor like a bulb by dropping from higher potential to lower potential and thus transferring energy to the bulb in terms of light, heat, whatever... so far so good. The problem comes in when I read in the textbook about capacitance.

  1. First problem comes in when I read that if I connect a higher volt battery - let's say 10V, I will get more charge. I thought it would be the same amount of charge (group of electrons) but this charge would be at higher potential. Is there a principle here that I am missing that says something about higher potential electrons and increasing the number of them?

  2. I understand you can increase the capacitance of a capacitor by increasing the area, having a good material between the plates or decreasing the distance between two plates. Does this mean the potential difference you can create between the two plates can be bigger than 5V that are originally supplied, so a capacitor in itself is becoming some sort of battery (not in chemical reaction creating higher potential sense) but some sort of a source for creating voltage, higher or lower than originally supplied?

  3. How does the capacitor discharge this net voltage between the plates? One side of the plate is supplied with extra high potential electrons creating a net negative charge and the other side has repelled regular electrons creating a net positive charge. This creates an electric field and the distance between them creates a potential (voltage). But how exactly can this be passed on through the current? The higher potential electrons are all on the other side of the plate, so who exactly transfers this energy (carried out by high potential electrons) if there are not high potential electrons on the other side of the plate to begin with? I am visualizing all this in my mind and textbook is of no help whatsoever.. Thanks for taking the time to read this lengthy post...

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    $\begingroup$ Please see meta.physics.stackexchange.com/q/6413/50583 for how to write good question titles and also note that you shouldn't include any "thanks" into your question , see meta.physics.stackexchange.com/q/360/50583. $\endgroup$ – ACuriousMind Dec 23 '16 at 16:02
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    $\begingroup$ Welcome to Physics Stack Exchange. This is a wonderful site for physics questions and answers, but it only works when we follow some basic guidelines about writing good questions. Each post should ask a single focused question. Failure to do this makes it impossible for anyone to write a good answer, and so your post will be closed. If you find yourself writing a numbered list of questions, that is almost surely telling you that you are asking too many different things. Second, use proper punctuation etc. For example, put a question mark at the end of questions. $\endgroup$ – DanielSank Dec 23 '16 at 16:06
  • $\begingroup$ "5V current" doesn't make sense. Please clarify that. $\endgroup$ – DanielSank Dec 23 '16 at 16:07
  • $\begingroup$ current = flow of group of electrons (charge)/time - so the flow of electrons that carry 5V potential - was that confusing? I thought this is standard lingo. I said from the getgo, I am not a Physicist. $\endgroup$ – Harsh Bhavsar Dec 23 '16 at 16:21
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The confusion reflected in your post seems to be a result of the lack of clear idea of some of the essential definitions.

Voltage of a Battery It is the amount of energy provided to a unit positive charge as it travels through the battery - from the negative pole to the positive pole.

Capacitor It is an object with two places to be connected to the external world and the external world will always give $q$ amount of charge to one of the connections and $-q$ charge to the other, producing a certain potential difference between the two connection points which is proportional to the absolute value of $q$.

Capacitance Capacitance of a capacitor is the amount of charge supplied to one connection of the capacitor and to be taken from the other in order to produce a unit potential difference between the two connection places.

Now, I will try to resolve all the three of the issues employing the above definitions properly with the basic rules of Physics:

  1. How much charge will flow is not generically fixed in a circuit. How much charge will flow is determined by the circuit elements and their specifications. For example, in a circuit consisting of a single resistor and a single battery, the law that decides how much charge will flow (per unit time) is the Ohm's law, namely, $V=IR$. This implies that for a given resistance, more the voltage difference provided by the battery, more amount of charge will flow per unit time - not that only a unit charge will be driven by a higher potential but the total number of charges will also increase. This fact is in a way reflected in the formula for power consumed by a resistor with resistance $R$ connected to a battery of voltage $V$ being quadratically rather than linearly. So, there is no reason to think that higher voltage should just mean the same number of electrons at a higher potential. In the case of a capacitor, we know that in a capacitor, the amount of charge needed to produce a certain voltage is directly proportional to the voltage. Thus, greater the voltage, greater the amount of charge accumulated on each of the plates of the capacitor.
  2. By changing the capacitance of a capacitor (via whatever method), you can't change the potential difference between its plates as long as the battery connected is fixed. But, there is another interesting case in which capacitor acting as the source of potential difference can be realized. One can charge a capacitor up to a certain potential with the help of a battery and then disconnect the capacitor from the circuit. This will not discharge the capacitor but will preserve the amount of charge on each of the plates that were accumulated during the charging process. Now, this capacitor plates can be pulled apart to increase the potential difference between them and then can be used in another circuit as the source of a potential difference which is higher than the actual potential difference of that battery that was used to charge the capacitor in the first place. This is really interesting but there is nothing unusual here. In the sense that although this might seem like some extra energy is available, it not really the case. The increased potential difference is woing to the work done in order to pull the plates apart.
  3. There is no discharge between the plates. Negative pole of the battery pulls the charge from one plate and the positive pole pushes towards the other plate of the capacitor. One can be sure that there is enough charge on both the plates for this process to take place because there is a really huge number of free electrons in the metals - so huge that it can be taken to be infinity for all the practical voltages and currents.
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  • $\begingroup$ Thank you so much for 1 and 2. You are awesome! I understand it well now! But I am still confused on 3. Can you please expand. From what I understand, positive end of the battery attracts the electrons (hence low potential) and pushes them towards negative end of the battery where they dont want to be (hence higher potential electrons) - but this is opposite of what you stated - can you please explain? $\endgroup$ – Harsh Bhavsar Dec 23 '16 at 21:44
  • $\begingroup$ @HarshBhavsar Thank you for your appreciative gesture. I have written that the negative pole pulls charge from one plate and sends it towards the positive inside the battery and finally pushes out of the positive side. I haven't made it explicit but I was talking about the positive charge. It is true that in an actual circuit the movable charges are indeed electrons that are negatively charged but still qualitatively, it makes no difference, in this case, to discuss the mechanism in terms of the positive charge instead of the negative charge. $\endgroup$ – Dvij Mankad Dec 24 '16 at 12:46
  • $\begingroup$ Ahh ok makes sense... I find the actual movement of electrons to be much more helpful but I can see how you stated in terms of conventional current flow (thanks, Ben Franklin!)... $\endgroup$ – Harsh Bhavsar Dec 27 '16 at 16:33
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  1. Electrons will literally come out of the battery when it is connected in a circuit. If these electrons flow onto a capacitor, then you're right that the potential on a capacitor is related to the number of electrons on the plate. If you look at the expression for the potential energy stored in a capacitor, notice that it is a function of charge for the very reason you mentioned

$$ U = \frac{1}{2}CV^2 = \frac{q^2}{2C},$$

so more energy comes from more charge.

  1. No you can not make the potential difference larger than the difference of the battery connected. This is related to your first question, the thing you modulate by changing the capacitance is the amount of charge carriers you get loaded up on the plates due to a specific potential difference. The battery is the thing that drives electrons onto the capacitor.

  2. So long as a battery is connected there won't be any discharge. It's only when you remove it and there is no longer a potential difference due to the battery that the electrons will flow back around the circuit the other way and discharge the capacitor. Also sometimes you can get dielectric breakdown and the charge will jump across the plates through the air, but this is not the type of discharge you were asking about.

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  • $\begingroup$ Your third point is debatable. If you disconnect the battery and there is not any remaining complete pathway (imagine a simple battery/capacitor series circuit), then the capacitor will remain charged until the charge slowly leaks into the air. $\endgroup$ – Bill N Dec 23 '16 at 17:08

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