# Why do charges not lose potential as they travel through the circuit before reaching a resistor?

Voltage is defined as:

Given the circuit:

Why would potential not drop, from the point differentially away from the positive terminal on the battery, to the point I have labeled. I understand that charges lose energy in resistors due to inter lattice collisions, where the acceleration generated due to thermal motion and voltage bias is translated from kinetic energy to heat energy. Before the resistor the resistance is neglible so the drift velocity will be high, and collsions will be minimal. Even so as the charges move from the positve terminal the Electric field will be strong and we will be losing the potential energy as we travel in the direction of the field. With this so why dosent potential drop as we move in the direction of the field? Is the energy associated with the charge essentially translated from potential to kinetic energy, where very little of this kinetic energy is translated to heat energy, and we assume the total potential is associated with the total energy the charge has at this point (KE + PE)?

I am really looking for an answer to my question. It would be much appreciated if we address the core question I asking instead of side discussions! After the initial answer, I encourage side discussions if wanted!

• "Before the resistor the resistance is negligible so the drift velocity will be high ..." Drift velocity depends on the current density, not the resistance. If the cross-sectional area is low then, for a given current, the drift velocity will be high. If the cross-sectional area is high then drift velocity will be low. Commented Mar 3, 2020 at 0:45
• Agreed, the equation for drift velocity based on the drude model states that, that wasnt the question I was concerned about though. Commented Mar 3, 2020 at 0:50
• Ahhh, so what you are saying is that there will be an accumulation of charges on the higher voltage side of resistor, which will essentially cancel out the potential of the battery? This charge will then start to flow down the resistor, and as charges leave and potential drops then more charge is added due to the E field of the battery? Commented Mar 3, 2020 at 1:10
• @Grant - "Even so as the charges move from the positive terminal the Electric field will be strong and we will be losing the potential energy as we travel in the direction of the field." Incorrect - the electric field will be extremely low in the wire. As others have said if the field was high the current would be extremely high. Commented Mar 3, 2020 at 1:45
• @KevinWhite Wouldnt the current move to the surface of the resistor quickly because of the field, then as the charge accumulates on the surface of the resistor it works as a capacitor where the charge will fight more charge from accumulating on surface, therefore canceling the field from the anode to the surface of the resistor? Commented Mar 3, 2020 at 1:47

The resistance in a real circuit would be far higher than the resistance of the lead wires, so the bulk of the electric field would be seen across the resistor.

Now we need to interject some reality. In a schematic, the resistor is a lumped element defined by properties at its terminals. IRL, however, it is a volume of resistive material. Two of the three dimensions of this volume are typically width, and the third is length.

If you take some fraction of that length, it will have a smaller electric field across it than the whole device does. If one side is ground, different parts will show different potentials depending on how close they are to the non-ground terminal. This is, in fact, how analog potentiometers work--the center terminal moves from one end of the fixed resistor to another, changing its potential depending on its position.

So the electric field is distributed around the circuit. The trick is that most of it is distributed across the resistive element, with very little distributed around the lead wires.

• As posted in comment below. Yes but why is there little field in the conductor. I have determined the answer after doing some reading. The field at the boundary of the resistor on the conductor side and the resistor side is different. Gauss’s law explains that the difference in electric flux is due to a surface charge accumulation on the surface of the resistor. The conductor has such high conductivity that it easily moves charges from the anode to the surface of the resistor, but the buildup of charge on the resistor repels further charge making the field in the conductor differentials small Commented Mar 3, 2020 at 17:17
• As charge flows through the resistor the field on the surface diminishes and therefore the charge will be trickling from the anode to the surface of the resistor. Therefore this leads to a small electric field in the conductor before the resistOr. Thoughts on this? Commented Mar 3, 2020 at 17:18
• The amount of voltage between two points on a circuit is proportional to the fraction of the total force necessary to create the current. The higher the resistance of any segment, the more force required to push the current through. The wires/leads are made to have low resistance, the resistive element is made to provide a certain amount of resistance, usually much higher than that in the leads. Therefore, mm by mm, the electric field on the leads is much lower. BTW, resistors don't have anodes, you could say "positive terminal" of a resistor to indicate the one assumed more positive. Commented Mar 3, 2020 at 17:31
• I was talking about between the battery (anode) and surface of resistor Commented Mar 3, 2020 at 18:19

Taken from Matter and Interactions 4th Edition. This is the exact solution I was looking for @all. The battery electric field dimishes with distance but the surface charge density of the circuit rearranges itself due to feedback. Around the resistor charge builds up applying a field counteracting the field of the battery and other surface charges. Since chagres flow through the resistors slowly the surface charge distribution from the battery to the top surface of the resistor is essentially uniform, creating a very small E field.

Credit to @jonk for the tip.

• Please consider eventually accepting an answer so the question can be closed out. It is OK to accept your own answer. I am sorry for the unpleasantness in the comments to your question. Hope it doesn't turn you away from the site altogether. Commented Mar 4, 2020 at 3:30
• Was planning on it, it takes 24 hours @mkeith, you should know this.. This site has a lot of people like you who come into the comments and don’t offer any help just complain. Please leave the comments to things that has to do with the post. Commented Mar 4, 2020 at 3:38
• I didn't know that the 24 hours was enforced by the system. My apologies. It was not my intention to complain. Commented Mar 4, 2020 at 3:47
• since you are doing a master, you might want to move on from the qualitative treatmente of Chabal and Sherwood to something more quantitative. I suggest Sommerfeld's Lectures on Theoretical Physics, volume 3, p. 125, Detailed treatment of the field of a straight wire and a coil. It shows how the surface charge distribution is determined in a very symmetrical situation. I do not write more because otherwise this comment will become "unhelpful". LOL Commented Mar 4, 2020 at 16:10
• I’ll take a look into it! Commented Mar 4, 2020 at 19:05

Imagine that there was nothing in between the source and the resistor. They are still connected, but there is no "circuit" between. No resistance at all...not just negligible, but nothing. There could be no loss in potential energy, no E-field in the space between the source and the resistor, because there is no space between them.

That is the situation depicted in the schematic. We draw the voltage source and resistor a little bit away from each other because it makes the schematic easier to understand, but effectively there is an ideal conductor of zero length between the two elements. There is a strong temptation to take what you know about real circuits and impose that on an ideal circuit, but you have to understand that the schematic is just a representation of a theoretical ideal situation...without making this leap we couldn't apply rigorous mathematical analysis to the circuit.

• Actually, the OP has the right equation. But what the OP doesn't realize is that $\vec{\mathscr{E}}$ is very, very tiny for each $\text{d}\vec{x}$ along the wire. The OP's question is more about physics and I'd recommend Matter & Interactions, 4th edition, Chabay and Sherwood. It covers this exact example (Section 18.8, page 735ff.) There is also a very short period of time when such a circuit is connected up, where steady state hasn't yet been reached. That alone is worth reading about, too. (Also in the book.)
– jonk
Commented Mar 3, 2020 at 1:59
• Bought the book, exactly what I needed, thank you @jonk ! If you post this I will accept your answer. Commented Mar 3, 2020 at 2:19
• @Grant It's a very very good book and I'm sure you will enjoy it. It really attempts to bring a sense of intuitive understanding and does a very good job, I think. The best book I've seen for someone just after high school and nowhere near a graduate level physics program.
– jonk
Commented Mar 3, 2020 at 2:22
• Im a masters student, I focus on VLSI. I get alot of physics but I kind of took these answers for granted in my earlier years. I am trying to get a deeper understanding of the physics. This book already looks promising! Commented Mar 3, 2020 at 2:26
• @Grant What I like about the Matter & Interactions book is that they also walk you through the part before steady state, as a switch is engaged to complete the circuit. It's nice to have both perspectives.
– jonk
Commented Mar 4, 2020 at 3:06

Your resoning sounds logical. Consider also the fact that if the E-field in the perfect conductor were large then the current throuh it which is proportional to the product of conductivity and E-field , would be huge which is not the actual case.

• Ya, I understand the mathematical theory behind the Drude model, was just trying to piece together the physical cause of the small electric field at an atomic level. Commented Mar 5, 2020 at 1:10

The E-field in perfect conductors is zero under steady state conditions (such as your circuit). So, the integral in the equation you supply, vanishes. Thus, ΔV=0, i.e. the potential is constant along a perfect conductor.

• Yes but why is there little field in the conductor. I have determined the answer after doing some reading. The field at the boundary of the resistor on the conductor side and the resistor side is different. Gauss’s law explains that the difference in electric flux is due to a surface charge accumulation on the surface of the resistor. The conductor has such high conductivity that it easily moves charges from the anode to the surface of the resistor, but the buildup of charge on the resistor repels further charge making the field in the conductor differentials small Commented Mar 3, 2020 at 15:55
• As charge flows through the resistor the field on the surface diminishes and therefore the charge will be trickling from the anode to the surface of the resistor. Therefore this leads to a small electric field in the conductor before the resistor. Commented Mar 3, 2020 at 15:57
• The buildup of charges you have described has also been used to generate E-field of specific magnitude in the air. I can't remember the name of the phycisidt who has patented this. He has also written a very good EM textbook. Commented Mar 4, 2020 at 16:32
• @nikoschatziathanasiou Jefimenko? Commented Mar 4, 2020 at 17:12
• Yep. That's him. Commented Mar 4, 2020 at 17:19