Electric field in a circuit with a DC source

  • #1
MarcusK
8
0
So I understand that when an electric field is produced in a conductor of length L, the net electric field in the conductor will be 0 because the rearrangement of electrons in the conductor results in the production of its own electric field which cancels out the one produced initially.

However, when a conductor is connected along to opposite ends of the battery (to the positive and negative terminal respectively), the battery pumps the electrons accumulated at the boundary of the conductor from the positive to the negative terminal through its interior. Then the field inside will no longer be 0 and electrons will continue accelerating towards the positive terminal of the battery.

But this seems to contradict with the set up of a common circuit, where a wire made of conducting material, that is assumed to be ideal and whose resistance is 0 ohms, is connected to a battery (DC source) and a resistor. Technically, in an ideal conducting wire there should be no potential difference along 2 points of a wire as electrons can freely move towards the resistor and thus the potential of a charge before being driven through a resistor should be the same as the emf of the battery. And since there's no potential difference, there would be no electric field. But a battery should produce an electric field as mentioned earlier, which is confusing...
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
15,268
4,244
Yeah, this is conceptually confusing -- we all form a certain mental image for what is happening and sometimes these ideas go on the run.

A battery produces a potential difference. When there is a path from one terminal to the other, current will want to flow. Resistance determines how much or how little in the steady state case.

The very moment something is connected to the battery, the capacitance of the setup influences how much charge flows to propagate the battery potential to the setup. This charge flows until the potential difference is spread out over the entire setup -- I use the word setup, because it doesn't have to be a closed circuit: it can be a wire above a plate connected to the other terminal, or anything else that is capable to store some charge.

The point is that the amount of charge involved is minuscule because the capacity to store charge is generally very small. Number of charge carriers to charge e.g. a sphere to, say, 9 V, is relatively small.

Whereas the amount of charge that flows in a closed circuit, even with a high resistance, is humongous in terms of number of charge carriers.

Do some calculations with the definitions of current and potential and the coulomb number to get an idea.
 
  • #3
anorlunda
Staff Emeritus
Insights Author
11,199
8,599
I find it helpful to think that I choose to ignore the resistance of the conductor and the voltage drop across the conductor. That's better than saying that the resistance and drop are zero. That viewpoint reminds me that in real life, they are actually nonzero (excepting supercondutors).
 

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