So for example if we have a a wider and a narrower wire of the same length in series, since v1A1=v2A2, in the narrower wire the charges will move with a greater velocity. Since v is proportional to the electric field, this means that the electric field will also be greater in the narrow wire. Thus it will have a larger potential drop. So this is pretty trivial and directly related to Ohm's law. My question is what exactly "causes" this changing of the electric field? I'm sure this is a pretty stupid question and probably the best way to look at it is from another angle but I just can't see the intuition behind it right now.
\$\begingroup\$
\$\endgroup\$
6
-
\$\begingroup\$ At the start there's an equal electric field across the whole thing, but then the charges start moving and distort it. What you are describing is the equilibrium condition. We use a lot of equilibrium calculations in electronics. \$\endgroup\$– user253751Jul 9, 2020 at 13:53
-
\$\begingroup\$ So how does this happen in a little more detail? \$\endgroup\$– Kakukk777Jul 9, 2020 at 14:16
-
2\$\begingroup\$ There is a short moment, upon the application of an electric field, where the circuitry rapidly sets up charges (they reside on surfaces, as a rule.) For changing fields, there is a continual modification of the charge arrangements over time. I've discussed this in general terms here, here, and here. But you should read a good, readable undergrad book on physics: say, 4th edition of Chabay and Sherwood's "Matter & Interactions." \$\endgroup\$– jonkJul 9, 2020 at 16:04
-
1\$\begingroup\$ As charge re-arrangements take time (they cannot be instantaneous re-arrangements with exactly zero time), there is a delay. Adding to this is a delay whereby the changing long-range electric field forces from these momentarily re-arranging surface charges are then "felt" by charges moving in conductors (limited, at best, by the speed of light.) Some of this delay can be readily modeled as an orthogonal force -- which we call a magnetic field. \$\endgroup\$– jonkJul 9, 2020 at 16:17
-
1\$\begingroup\$ Take two (-) charges in an empty vacuum, separated by 'd'. A force is felt between them. One is moving relative to the other, which we take as "fixed." Time passes before changes in position by the moving charge affect the other. So it's not instantaneous. If we assume the electric force is instantaneous (Newtonian instead of Einsteinian), then this field effect delay gives rise to a new force (one we can construct) called a magnetic field. So Maxwell's electric and magnetic fields are "classical" or pre-Einsteinian. Full consideration of all details in solid state matter is quite "complex." \$\endgroup\$– jonkJul 9, 2020 at 16:22
|
Show 1 more comment
2 Answers
\$\begingroup\$
\$\endgroup\$
6
In the context of classical electrodynamics, it's surface charge.
You should first ask yourself "what makes the electric field follow the path and shape of the conductor?" Well, it is surface charge. The battery generates an electric field nearby it, when you place the conductor near to it without connecting it to the battery, the free charge in the conductor will feel the field and will reposition itself of the surface of the conductor in order to make the electric field inside zero. This is just plain electrostatic induction.
Let's consider a cylindrical conductor of conductivity sigma with different cross sections. When you connect the conductor to the battery, the charge will change its configuration in order to make the field inside the conductor conform to local Ohm's law j = sigma E. If you solve Maxwell's equation adding in the continuity equation and a pinch of Ohm's law in its local form, you will find that the charge density in the system will depend on gradients of conductivity and permeability:
This means that there will be rings of charge around the cylinder that will shape the field to follow the conductor, and there will be charges at the discontinuity in section to make the field 'concentrate' or 'dilute' according to cross section.
This paper has some nice pictures
Voltage and Surface Charges: What Wilhelm Weber already knew 150 years ago
(Originally published in the German Journal „Praxis der Naturwissenschaften-Physik“ (PdN-PhiS_2012_5_S_25-31)
Translation: Hermann Härtel
This is a resistor made of a material of different sigma
in this case the different field is caused by surface charge at the interface between materials
(picture from the same paper above).
In the case of a resistor made of the same material but different section you would have surface charge at the surface required to change the section's diameter. Those charges will steer the field lines inside the smaller section. Sigma is the same but both j and E will rise. When you integrate the field along the path you will find a higher potential difference.
Here are a few reference you might find interesting:
W. G. V. Rosser
What makes an electric current "flow"
American Journal of Physics, vol. 31 no. 11, november 1963
Bruce A. Sherwood, Ruth W. Chabay
A unified treatment of electrostatics and circuits
American Journal of Physics
(you can find it free online with a Google search. Also, Chabay and Sherwood wrote an introductory textbook that explains exactly what you want to know).
Ian M. Sefton
Understanding Electricity and Circuits: What the Text Books Don’t Tell You
(School of Physics, The University of Sydney)
Science Teachers’ Workshop 2002
and if you want to take it to the next lever, who is better than Jackson?
John D. Jackson
Surface charges on circuit wires and resistors play three different roles
American Journal of Physics 64 (7), July 1996
Do you like simulations?
Rainer Muller
A semiquantitative treatment of surface charges in DC circuits
Am. J. Phys. 80 (9), September 2012
American Association of Physics Teachers
and let's not forget the paper by Jefimenko and his oil-seed demonstrations
Oleg Jefimenko
Demonstration of the Electric Fields of Current-Carrying Conductors
American Journal of Physics 30, 19 (1962)
doi: 10.1119/1.1941887
(formatting on this site sucks!)
answered Jul 9, 2020 at 16:42
-
\$\begingroup\$ Thanks for everyone I didn't think there's this much to it and I'll get so many anwsers. I will try to understand them lol. \$\endgroup\$ Jul 9, 2020 at 17:49
-
\$\begingroup\$ The papers from American journal of physics are not available for free. \$\endgroup\$ Apr 21, 2021 at 16:44
-
\$\begingroup\$ @Sredni Vashtar can u please also explain what would happen if we connect just one piece of wire to only on end of battery say +ve terminal,,,,If there would be some net positive charge on its surface to cancel the electric field due to battery then it would mean thta the positive electrode of battery has accumulate some of its electron so it should change the potential difference of battery but all textbooks says that battery doesn't accumulate charges ,,why they are saying this? there is no proof to it,,,,,similarly why a capacitor always carry equal and opposite charges,,,what's the proof? \$\endgroup\$ May 18, 2022 at 16:37
-
\$\begingroup\$ @ArunBhardwaj the battery displaces charges at its poles thanks to its inner chemical reactions. In open circuit there is an equilibrium and the amount of charge separation does not proceed further. If you attach a piece of conductor to the posite pole, the charge that had accumulated at the pole will quickly redispose itself on the whole surface in such a manner as to make the internal electric field zero. It is only when you close the circuit on a load that current will flow - thanks also to the internal reactions in the battery itself. \$\endgroup\$ May 21, 2022 at 1:19
-
\$\begingroup\$ @SredniVashtar I still did not understand you in a clear way,,,,,,,,I mean if battery has accumulated some negative charge then it should change its overall potential difference right?? but all textbooks say that battery potential difference is always maintained because of the chemical reactions but in this case how chemical reactions can maintain potential difference ,,,the net negative charge would still have an effect on potential difference of battery as chemical reactions cannot neutralize these charges \$\endgroup\$ May 24, 2022 at 7:15
\$\begingroup\$
\$\endgroup\$
You have a choice of models.
Circuit theory will tell you what, but not how or why. Think of your wide wire as several thin wires in parallel, assume the current and voltage conditions are the same for each elemental wire, and use the resistor formulae to add them up.
The Drain Pipe Theory, aka the Hydraulic Analogy, is very intuitive, but of course it's about a totally different system, so it doesn't offer any 'how' into what's going on with electricity. It's surprising how far you can push the analogy before it breaks - potential, current, power resistance, capacitance, inductance, diodes can all be 'explained' with it, you can even build a boost SMPS with it. You only need fat and thin pipes, where the flow through and the pressure across them are proportional. A thin pipe will develop a larger pressure (aka potential) drop for the same flow.
The Drude model is inaccurate, but it's quite intuitive. It sort of works for conductors and resistors, but don't push it too far. It's a classical theory, so doesn't describe anything that needs quantum mechanics for an understanding, like conduction bands and band-gaps. It doesn't tell you how or why anything, and it's not really that good for calculating how much either. It's no better than circuit theory for adding up parallel resistances.
The fundamental theory you need is Quantum Electrodynamics, though quite how photons ping about in DC circuits is beyond me. Perhaps you'd ask this in the Physics stack. Intuitive it's not, and like all physics, it's going to tell you how to calculate the results, but not how or why.
