2
$\begingroup$

Say you have a battery, with a wire connecting the negative and positive terminal. Initially, (the transient state) the electric field is not uniform and is perpendicular to the surface of the cross section of the wire.

This forces the surface charges to rearrange, and the surface charges keep moving until the electric field in the wire is uniform and at any point, the electric field is perpendicular to the cross section of the wire at that point. This is the steady state, and is when current is flowing.

Now consider a circuit that consists of a battery and a capacitor connected to the both the terminals. Could someone walk me through how the capacitor is charged using the concepts of surface charge redistribution discussed above?

Please note I am not looking for a general explanation, I'm looking for an explanation that shows how the surface charges redistributing themselves during the transient state to ensure the electric field is perpendicular to the wire results in the capacitor charging.

$\endgroup$
  • 1
    $\begingroup$ The first three Google results have more than satisfactory explanations. You should check this, this and this out $\endgroup$ – Pranav Hosangadi Feb 19 '14 at 1:13
  • $\begingroup$ @PranavHosangadi No, they don't. I'm specifically asking for an explanation that utilizes the concept of surface charge. None of the links do that. $\endgroup$ – dfg Feb 19 '14 at 1:20
  • $\begingroup$ Downvoter, care to explain? $\endgroup$ – dfg Feb 19 '14 at 1:21
  • 1
    $\begingroup$ dfg, almost any physical capacitor is, by virtue of its construction, in essence, an open circuited transmission line. I'm not interested in "walking you through this". Take that walk yourself; it will be much more rewarding. $\endgroup$ – Alfred Centauri Feb 19 '14 at 1:39
  • 1
    $\begingroup$ I'm with @AlfredCentauri on this. This is the same physics that you get in any other piece of conductor, which means that the details are tedious and boring unless you can motivate the desire for a detailed microscopic analysis of this geometry in terms of something interesting to be learned from it. Nor does the pure number of questions you have asked on the same physics help. I imagine that people would like you to generalize what you have been taught so far; take some responsibility for running with it. $\endgroup$ – dmckee Feb 19 '14 at 3:13
1
$\begingroup$

Initially when you attach the capacitor to the battery, said battery will act to create an electric field within the wire. On the side of the negative terminal this field will point perpendicular to the cross section of the wire toward the terminal of the battery (electric field points toward negative charge). On the side of the positive terminal the field will point perpendicular to the cross section of the wire away from the terminal of the battery. This is because the battery arranges the electric charge in such a way that the negative terminal is electron rich and the positive is electron deficient.

Now, this electric field will cause charges to move. On the side of the negative terminal, the electric field will push the electrons away from the battery and toward one side of the capacitor. On the side of the positive terminal, the field will pull electrons from the capacitor and toward the battery. When this happens we see that the side of the capacitor attached to the negative terminal of the battery is accumulating electrons (becoming negative), and the side of the capacitor attached to the positive terminal of the battery is losing electrons (becoming positive). But what is this motion of charges doing to the field? With electrons accumulating on the negative side of the capacitor, the field within the wire is decreasing (because now there is less of a potential difference between the negative terminal of the battery and the negative side of the capacitor due to the accumulation of the charges). Electrons will continue to be pushed toward the negative side of the capacitor, however, until this field decreases to zero. This will happen when enough charges accumulate on the negative side of the capacitor to bring the negative capacitor plate to the same potential as the negative terminal of the battery. By the same argument, electrons will continue to be pulled away from the positive side of the capacitor until the field is brought to zero when the positive capacitor plate achieves the same potential as the positive terminal of the battery.

From reading the comments above it seemed to me that this is the type of explanation you are, perhaps, looking for. I hope it is helpful.

$\endgroup$
  • $\begingroup$ Thanks! If I understand correctly, the charge builds up on the capacitor to neutralize the field from the battery, resulting in the field in the wire being zero. But there are other possible distributions of charge which causes the field in the wire to be zero. $\endgroup$ – dfg Feb 24 '14 at 2:15
  • $\begingroup$ For instance, you could have a buildup of charge at the terminals of the battery (positive charge buildup at the negative terminal and negative charge terminal at the positive terminal) - this would also result in the field being zero. So how do we know charge building up on the capacitor plates is always the charge distribution that is created to neutralize the field? $\endgroup$ – dfg Feb 24 '14 at 2:16
  • $\begingroup$ I didn't mean in the battery, I mean't in the wire - at the edge of the wire that connects to the terminal. I was under the impression that the battery only maintains a negative and positive charge in the battery, but doesn't have control over what's outside it. $\endgroup$ – dfg Feb 24 '14 at 2:29
  • $\begingroup$ Sorry, I don't follow. Why would the positive charges be pushed away from the negative terminal? The positive charges would only have to overcome the 9V if it was at the positive terminal? Wouldn't it be attracted to the negative terminal? $\endgroup$ – dfg Feb 24 '14 at 2:51
  • $\begingroup$ I see. But then there wouldn't be a problem with the charge building scenario I gave right? Because there would a positive charge buildup at the negative terminal, and negative charge buildup at the positive terminal so no charge would have to overcome any potential. Am I mistaken? $\endgroup$ – dfg Feb 24 '14 at 2:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.