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This is a little confusing for me. I will just ask a few questions:

First:

  • Am I correct in saying that while traversing through the circuit, charge looses some electric potential(not energy) at each point as it advances? If this is true, then why do we say that voltage drop occurs only when charge passes through a heavy resistor like a bulb?

Now, suppose I make a circuit. Now, I put a bulb very closer to the negative terminal(just the point where charge from negative terminal starts flowing). Now, the charge enters the bulb, looses a lot of its energy, and comes out of it. But, the point where it comes out of the bulb is a point in the electric field, and suppose that point is supposed to have a higher energy per unit charge(higher voltage or higher electric potential), but the charge has lost a high amount of electrical energy due to the bulb.

  • So, will the charge regain that higher energy, just because at that particular point in the field, there is a higher electric potential(not energy)? If this is true, isn't this strange, I mean because charge is loosing a lot of energy, then regaining it, without effecting the battery at all?

I request the readers to give me an intuitive explanation of this, just focusing on what is asked.

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closed as unclear what you're asking by ACuriousMind, Bill N, user36790, Sebastian Riese, Kyle Kanos Feb 28 '16 at 12:09

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Here is where I think your misconception is:

Now, the charge enters the bulb, looses a lot of its energy, and comes out of it. But, the point where it comes out of the bulb is a point in the electric field, and suppose that point is supposed to have a higher energy per unit charge(higher voltage or higher electric potential), but the charge has lost a high amount of electrical energy due to the bulb.

While you can imagine a nice electric field in some situations (like between parallel charged plates), you don't think of it that way in a wire. Instead, the electric field is created by the distribution of charges within the wire.

For ideal conductors (which is not too far from true in a simple circuit), there is no relationship between distance and the potential. It is not the case that if there is 10cm between the terminals of the battery that the potential at a distance of 5cm is something in particular.

With your described circuit, the charges move in such a way that the wire on one side of the bulb has a potential that is basically identical to the battery terminal on that side, and the the wire on the other side has the potential of the other terminal. This potential is established very rapidly after the circuit is connected.

This means that after the charge moves through the bulb, even if it is right next to the terminal, it is already at the lower potential and will stay there until reaching the battery, almost independent of the length of the wire.

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To your first point, wires do have a resistance of their own, and it would cause a drop in voltage. However, for most basic applications we can treat the wires as perfect conductors, and so ohms law tells us there is no potential drop.

To your second point, since the potential energy is related to the electric potential via $\Delta U = q \Delta V$, we see that electrons, which have $q=-e$, experience their highest potential energy at points of lowest electric potential. So at the negative terminal of your battery, the electrons have more potential energy than after they pass through the bulb, which makes sense when you think of the power emitted from the bulb as an energy loss by the charge carries.

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Voltage does drop a tiny bit across a real wire. But it's like water going just barely downhill as it flows downstream on a fairly very flat part or a river, as compared to water falling down a scary huge water fall. The water always goes downhill, but a tiny amount versus a huge amount.

But there are some misconceptions in your question. Firstly objects don't have potential energy, systems do. For instance a ball falling to earth. There is a potential energy for the earth-ball system and it depends on how far apart, $d,$ their centers are: $U(d)=-GMm/d$. And technically, the earth falls up while the ball falls down, but they feel equal and opposite forces ($GMm/d^2$ pointing towards each other) and the earth is more massive so it doesn't fall up very much. This should be a basic thing from mechanics. Learn it. It is not the case that potential energy is something a single object has, it's something the system has.

Next: electric potential isn't really potential energy per unit charge, but if things don't accelerate too much it isn't too bad.

Now let's get to a circuit. The electric potential scalar field is basically just a convenient way to specify the electric vector field at every point with less effort (and there always is an electric field at every point and sometimes there is a potential that describes it). And what happens is the field can push charges. Much like water pressure can push water. If the water pressure behind you is a bit stronger than the water pressure in front of you and the difference pushes you just hard enough to help you regain the energy you lose from rubbing against the walls of the pipe then the water can flow along. If up ahead the pipe got narrower you'd have to speed up as you reach the narrower pipe to avoid a buildup.

A similar thing happens in a wire. To go around a corner at a particular drift speed you'd want some charges on the outside or inside of the corner to give you a force to make that corner at that speed. If you went through a region that was stealing your energy from collisions you'd want some charge on the outside of the wire that produces a field inside that gives you a force in the direction you are trying to keep going.

That's what goes on in a resistor and a wire. There is charge on the outside making a field inside that makes you go around corners in the places you need to go around a corner and a field that replenishes your kinetic energy in places where you lose it from collisions.

So really the question is how all that charge ends up on the surface in all the right places in all the right amounts. If there weren't enough charge then you'd start to cruise out of the wire when you approached a corner so one side would start to get too many electrons and another would get too few. That ends up giving the forces you need to steer the other electrons along the propoer path. So basically it just builds up more and more as the drift velocity gets larger.

So now let's assume you have a battery and a resistor and two wires. You connect the positive terminal of the battery to one end of the resister. And you attach one end of the other wire to the negative terminal of the battery and hold it close to the resistor but don't connect them yet.

So what's the potential field look like. You could draw some equipotential surfaces. Inside the battery they are stacked up like pancakes. They come out of the battery and head over to the gap between the wire and the resistor. More rightly each is like a bubble that contains the entire disconnected wire on the inside and contains the entire other wire and the resistor on the outside and cuts through part of the battery and cuts through the empty space between the wire and the resistor.

Draw a picture if you need to, this is what us going on. As you move the wire closer to the resistor a huge number of potential surfaces are jammed into this small space, eventually the volts per meter is too huge and the air itself starts to transmit current at this point some electrons from the wire to the resistor.

Why? Because there was an electric field in the region in between, so the charges on the outside couldn't be held in and air could conduct because the energy was strong enough to rip apart the molecules into their ion parts and move the resulting ions around.

But when charges leave the wire that leaves more protons there than electrons and this creates an electric field that attracts nearby electrons which leaves a deficit next to the edge, which attracts parts next to it and so on.

But the air has effectively a huge resistance, there are lots of collisions and loss of energy so this electric field in the air in between only supports a small current. That small current has a drift velocity and that drift velocity needs some charge buildup or deficiet on the corners and outsides to move everything along the circuit. Any part of the circuit that doesn't have the right charge on the surface will start collecting will create like ups that make electric fields that discourage creating more pileup.

So it's like each part tries having different currents until it finds an current that allows each part to push the charges exactly what they need. This sounds a bit magical. But the point is that the battery can exert actual forces on the charges in it and next to it to keep the electric field a certain way inside itself. And as charges get ripped from here to there in other parts everyone steals from their neighbor but the battery actually does some work. So you can think that the more current that runs through the battery the more work it can do and so the more energy that is available to give to everything that has collisions in the air (big resistor) or your actual resistor. This equilibrium current requires some energy be given to the current going through the resistor. So one of those bubbles (potential surfaces) moves so that it isn't in the air gap any more, it's cutting through the resistor. The potential moved becasue actual charge moved to pile up extra charges and charge deficiencies on the surfaces.

So then you move the wire closer and now there is less air between the wire and the resistor and so less collisions so less energy is needed there but they we already used to be getting enough energy from the prior charge on the surfaces. So they take off faster and arrive faster at the resistor. But that leads to extra electrons on the end of the resistor so they push the electrons in from of them harder so they make the ones in front of them go faster. The faster speed leads to more collisions and more loss the speeds adjust until the pile up smooths everything out.

This leads to different charge on the surfaces and now another potential surface cuts through the resistor. And there is more current becasue now a larger current can get more power from the battery and balance it out with the losses around the circuit.

Then the air gap can be made smaller and then smaller and then smaller. And basically the current adjusts itself quickly so it goes through all of those different higher and higher currents as you close the air gap.

All those potential surfaces that started out spread through the air gap end up getting evenly spread out through the resistor instead. But they only got such a huge number of surfaces in the resistor when the resistor got up to having the right amount of current.

This answer can be unsatisfying since I basically said that air is a giant resistor and the resistance gets smaller the smaller the air gap is. But it's true. It effectively has a near infinite resistance before the breakdown field and so it's really a question of a cirucit that was always connected with a huge resistance (so all the potential surfaces cutting through that air gap part and no current flowing) and then the resistance of that air gap part getting smaller so the current in the whole circuit goes up. But that is the essential story.

Before the wire and resistor started to generate current you effectively had some extra negative charge on wire that touched the battery but not the resistor and had some deficiency of positive charge on the resistor and the wire connecting the resistor to the battery. That's how they were held at those different potentials. But the charge on the outside was arranged to have no electric field inside the parts.

Once the current started to flow the charge in the outside needed to be the kind that makes electric fields inside the parts. Small fields inside the wires since little energy is lost for a given current (wires effectively has super tiny resistance) and comparatively larger fields inside the resistor since comparatively more energy is lost for a given current. Those fields developed because of actual imbalances. It's just the imbalances existed for a very short time. The current might have electrons drift at 1cm/s but the field associated with a charge imbalance is a lightspeed broadcast. So the other parts of the circuit react to a charge imblance somewhere long before he charges themselves get there. But a region of too many electrons has a field due to it that pushes every other electron away (so ones ahead will start clearing out of the way and ones behind will slow down thus bunching up there and sending less into the region which relieves the build up there a bit. And then does the same thing in the place behind).

So the imbalances actual diffuse themselves. But they diffuse themselves at the lightspeed rate. This is still a simplification since you can have a current in a region that is completely balanced between charges, that is even what happens inside. Basically you could imagine that when the electric field is pointing in the direction of the circuit, that a current can flow and if it is strong enough to match how much energy is lost to collisions at that speed then the speed can be steady. But if it hits an edge of a wire instead of going along the wire then it contributes to a charge deficiency there, or going backwards it springs from a place where charge pileups are created. But those pileups or deficiencies make the fields in those directions weaker. So there is a race. When you moved the wire closer the fields due to the other charges is still what it is (those charges are far away) and when the field didn't line up with the wire then you develop a charge imbalance on the surface, then other parts of the surface react to the fields from those pileups and so create their own pileups.

But if there is a way to configure the potential surfaces so the fields point along the wires with the steady state size then that field doesn't change. And the other fields all react in a way to weaken themselves over time. So you can imagine that the field settles into that equilibrium configuration.

You can make a model of the material and write Maxwell as a series of vector equations and actually see the equilibrium develop in the situations it does. But I think it's never clear without seeing the full model.

A complete model would include that energy is stored in electric fields and in magnetic fields and that it flows in a direction orthogonal to both fields so it flows out of that battery from the sides, along those potential surfaces and into the sides of the resistor and works its way from the outside of the resistor to the inside of the resistor.

And that's where the energy always was, never in some potential, but in the fields and it transfers from fields to current in places with both.

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  • $\begingroup$ Your answer has produced a number questions which I would like to ask you. You imply that your equipotential bubbles extend outside the conducting parts of the circuit ie into the air not just at the gap. Is this true of all of the bubbles or are some contained within the conductors? If there are these bubbles outside the conductors then does it follow that there an electric field outside the conductors and if there is of what strength is it compared with the electric field inside the conductors? $\endgroup$ – Farcher Feb 28 '16 at 6:54
  • $\begingroup$ @Farcher Since the wires are actually resistors of very tiny resistance the bubbles do extend into them from the sides just like for the resistor. A certain number of bubbles cut through the battery and a they are like percentiles if 10% of the lower has been consumed by the time you get to a certain portion in the circuit then that bubble cuts through that part of the circuit. Not exactly especially when things are changing in time. The electric field outside the circuit elements is actually how the energy flows from the battery to the circuit elements. $\endgroup$ – Timaeus Feb 28 '16 at 7:18
  • $\begingroup$ @Farcher Most of the energy flows in the thin layer right outside the edge of the circuit elements. $\endgroup$ – Timaeus Feb 28 '16 at 7:19
  • $\begingroup$ "the bubbles do extend into them from the sides just like for the resistor" and once inside the bubble surface is approximately perpendicular to the conductor surface? If two current carrying conductors of the same area but differing mobile charge density are joined together the electric field in each conductor is different? Does that mean that local to the junction there is a redistribution of charges? If yes is this due to a local distortion of the lattice ions/atoms? $\endgroup$ – Farcher Feb 28 '16 at 8:06

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