I know that a capacitor has positive and negative charge distribution on either of its plates. But saying that net charged provided to it by the connected battery is zero doesn't seem to be correct. I understand that the two plates have opposite charges to create the necessary electric field between the plates to store the electric energy,but then the very statement that "it has received net charge=0 from the battery" confuses me a little
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1$\begingroup$ Disconnect the cap from it's charging source on both ends and then short the leads together. Using a (very) sensitive meter, measure the voltage between the shorted leads and ground. That will give you a sense of the net charge on the capacitor (which is going to be very, very close to zero). $\endgroup$– Hot LicksDec 15, 2019 at 3:05
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1$\begingroup$ are we talking physical capacitor? or theoretical capacitor, and which design of capacitor? unbalanced charges can accumulate of the external surface of a capacitor same as for any other conductor. $\endgroup$– JasenDec 16, 2019 at 0:46
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7 Answers
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but then the very statement that "it has received net charge=0 from the battery" confuses me a little
The battery doesn't deliver charge to the capacitor. It moves charge from one plate of the capacitor to the other leaving one plate with a net positive charge and the other plate with a net negative charge. It takes energy to move the charge between the plates. That energy is stored in the electric field of the capacitor as electrical potential energy and equals $\frac{CV^2}{2}$. The battery provides that energy.
The overall net charge on the capacitor is zero both before and after charging. The charge has just been redistributed.
Hope this helps.
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$\begingroup$ IIRC it's possible for capacitors connected in series with each other to individually have non-zero net charges, since they effectively function as one capacitor with a bigger distance between the plates, right? $\endgroup$ Dec 15, 2019 at 0:30
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1$\begingroup$ @nick012000 Not quite sure what you mean by “net” charge. Capacitors in series have the same amount of positive and negative charge on their plates and each capacitor has the same amount of negative and positive charge for an overall net charge of zero for each capacitor. $\endgroup$– Bob DDec 15, 2019 at 12:30
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$\begingroup$ "The overall net charge on the capacitor is zero" unless it has been electrostatically charged anyway. You can have charge on a balloon, no reason you couldn't get one on a capacitor. Just not by connecting battery to it. $\endgroup$– MołotDec 16, 2019 at 13:04
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$\begingroup$ @Mołot I would not consider an electrostatically charged balloon in isolation to be a capacitor, which is normally defined as two conductors separated by an insulator (or space). The balloon is a single charged surface. A charged capacitor has an equal amount of positive positive charge on one plate and negative charge on the other. If you add them together you have zero net charge. $\endgroup$– Bob DDec 16, 2019 at 13:26
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1$\begingroup$ @BobD Maybe I didn't make myself clear. Pretty much anything can have net charge just the way baloon can have it, right? So capacitor can too. Thus, "The overall net charge on the capacitor is zero" may be true, but may be false. $\endgroup$– MołotDec 16, 2019 at 13:28
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A capacitor whose terminals are not connected to anything can hold a net charge, just as a balloon or a bit of dust can hold a net charge.
However, a capacitor whose terminals are attached to the terminals of a battery will have no net charge induced by the battery because the battery will pull electrons from one plate of the capacitor and push the same number of electrons onto the opposite plate.
answered Dec 14, 2019 at 15:32
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It doesn't have to always be zero, but in this case, when an uncharged capacitor is connected to a battery in series, the net charge on the capacitor will be zero. The key point here is that batteries provide energy to components, not charge. Batteries have an internal mechanism that ensures that the net charge of the battery stays constant.
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The voltage across a capacitor only depends on the charge difference. The total charge added by connecting it to a voltage source is zero unless of course the two plates are not identical.
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$\begingroup$ Does the voltage source add charge to the capacitor, or does it supply energy to move charge from one plate to the other of the capacitor? Just food for thought. $\endgroup$– Bob DDec 15, 2019 at 22:08
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yes it's near zero. any unbalanced charges will repel and migrate to the outer surface of the capacitor thus limiting the amount of unbalanced
charge to the amount which the battery could have placed on a similarly sized conductor.
Adding 6 positive charges to the left plate without drawing any from the negative plate causes charges to be distributed on the external surface of the capacitor, three of the positive charges go into the gap and draw negative charges from the external surface of the negative plate leaving a positive charge there too.
unless the battery itself has an electrostatic charge it will not be able to place a significant unbalanced charge on the capacitor.
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Consider a 1000 microfarad capacitor charged to 10 volts. It has a charge of 10 millicoulombs.
Consider another similar capacitor 10 centimeters away.
Now, if there was no equal opposite charge in two plates and the capacitor would truly have the 10 millicoulomb net charge, what would the force be?
The force according to Coulomb's law would be 90 meganewtons, or in other words, about 9174 tons of force.
The two quite ordinary capacitors would smash into each other with a huge force. You would need 9174 tons of force to keep them separated.
This simple thought experment demonstrates why the two capacitors cannot have their charge as purely negative or positive. They must have equal and opposite charges in their plates, cancelling the net charge.
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$\begingroup$ Why doesn't the same thing happen with the two separate plates? $\endgroup$– anonDec 16, 2019 at 12:56
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1$\begingroup$ @GS-ApologisetoMonica That would be the subject of another very good question, but my understanding is that the area of the plates is very large (much larger than the external area of the capacitor due to rolled plates) so that the force per unit area is not so big after all... But go ahead, and ask the other question! $\endgroup$– juhistDec 16, 2019 at 12:57
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You are right the voltage won't be zero.
but the thing is the plates of a capacitor (as used in electronic devices) are very close together and have a large area and the vast majority of the charge put into the capacitor is matched by a couter-current out through the other lead.
however some of the charge couples to the rest of the universe instead of to the other plate.
If you were to charge a capacitor from a battery and then remove it and discharge it without introducing any other charge to the capacitor. measurement with a sensitive electrometer would show the capacitor potential as being somewhere between that of the two terminals of the battery (dependent on how the capacitor was constructed - some of them connect one terminal to the case of the capacitor in which case that terminal would dominate the result)
If the battery was atop the output dome of a VanDeGraf generator (and somehow connected) the measured potential could be substantial with respect to ground.
However in electronic circuits this external electrostatic charge is usually ignored because its effect is small and only the internal electrostatic charge of the capacitor is considered.