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Surfaces charges in a circuit

Hello ,This image represents surface charges in a circuit caused by the battery's electric field, if you were to place an other electric field source of the same magnitude (but not connecting it to the circuit (on top of the battery or under it)) and placing so it has the same direction of the battery's electric field for me it would do 2 things :

First, since they are in the same direction their power will add up resulting in a doubled potential difference (e.g voltage) (well not totally since the distance at least a bit more), this will translate to the surfaces charges as a double of their density everywhere to maintain the same electric field inside the wire.

Second, The battery will get used more quickly due to the higher current.

However I know that this is wrong because :

This requires if I place the other electric field source to counter act the electric field of the battery, or if the other electric field is stronger will imply many weird reactions(after thinking about it you need a really strong second battery) .

The evenements I describe don't seem to really happen in the real life,

So if someone could show me the error in my reasoning,Thanks. TLDR: In a circuit why dosen't the electric field of other external things even help a tiny bit when stacking with the battery's electric field.

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marked as duplicate by John Rennie electromagnetism Jul 23 at 5:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It actually did awnser my question $\endgroup$ – mohamed azaiez Jul 22 at 9:49
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Whatever static external fields that may be present in the vicinity of the circuit only affects the charge distribution on the surfaces of the wires. There is no "power" associated with these, assuming they are not changing, or they area changing very slowly.

These surface charges will "shield" the external fields. The field distribution within the wires/battery, and hence the operation of the circuit, will still depend as usual on what components make up the circuit, in accordance with Kirchhoff's Voltage Law (a consequence of Faraday's Law) and Kirchhoff's Current Law (a consequence of conservation of charge).

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  • $\begingroup$ Are you talking about The surface charges used in sheilding from electric fields. But i am talking about the surface charge distribution in an electric circuit tu-braunschweig.de/Medien-DB/ifdn-physik/ajp000782.pdf or i did not understand your awnser and you're indeed talking about surface charge distribution in an electric circuit, please confirm $\endgroup$ – mohamed azaiez Jul 8 at 5:29
  • $\begingroup$ I'm talking about the surface charges induced in the components in the circuit by the external fields you were asking about. There will be two components to the surface charge distribution: the one set up by the circuit itself without external fields (and the one the paper you linked is talking about), and the one induced by the external fields. The latter component will "shield" the circuit from these external fields, and the circuit will operate as usual. $\endgroup$ – Puk Jul 8 at 5:43
  • $\begingroup$ "Assuming you mean you have a full electrical circuit and you're wondering what an external field does to it: The external electrical field would change the voltage, " link physics.stackexchange.com/questions/237426/… Who should i trust now ); $\endgroup$ – mohamed azaiez Jul 8 at 5:52
  • $\begingroup$ i don't understand how can they sheild the external field without changing the surface charge distribution thus changing the voltage could you provide more details ? $\endgroup$ – mohamed azaiez Jul 8 at 5:56
  • $\begingroup$ The answer you linked to is not inconsistent with mine. Yes, if you bring some charge near one of the plates of a capacitor, it will induce charges and change the potentials on the plates. However the fact remains that any surface charge has the two components I mentioned earlier, which can be superposed. The surface charge due to external fields won't impact the circuit operation. The I-V relation of a capacitor is $I = C\frac{dV}{dt}$. Since the voltage difference due to the external field is not time-varying (or is slowly-varying) the circuit will operate as if there were no external field. $\endgroup$ – Puk Jul 8 at 6:16

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