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What I have since learned: (Tell me If I am wrong)

In circuits ,Batteries provide electric potential difference (voltage) ,This difference creates electric field (Over the conductor or the wire) which applies force on electrons ,This makes an flow of electrons through wires which called current.

In another way ,We can say batteries provide electrons with potential energy and this energy converts to kinetic energy and flow in the circuit ,So potential energy is used up.

I have learned also that electric potential of an electron depends on its position in electric field.

Take a look at this circuit : enter image description here

$Q$1: Since point $a$ and point $b$ are in different positions in electric field (produced by voltage of the battery) Why have they the same electric potential , Why is there no voltage drop?

$Q$2: As I have mentioned ,Potential energy of the electron converts to kinetic energy ,Now when electrons flow through a resistor ,why is there a voltage drop?

On $Q$2: I say there isn't voltage drop because when electron flow in the resistor ,Some collisions occur in the resistor ,So electrons lose some of their kinetic energy ,But due to the electric field ,Electrons accelerate again and restore their lost kinetic energy ,So why is there a drop in electric potential or voltage drop?

$Q$3: My last question about drift speed. Because electrons are in electric field ,Electric filed applies force on them ,So they accelerate ($F=ma$) ,So their drift speed $Vd$ must be higher ,hence the current $I$ must also must be higher ($I=NeVdAQe$).

Please ,I am so confused and I can't understand this.

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3 Answers 3

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A1. An ideal conductor between a and b has no resistance. The potential difference between the points is the work required per unit charge to move the charge between the two points. With no resistance no work is required is to overcome the resistance. Like a box on a frictionless surface can move at constant velocity with no force and therefore no mechanical work required. So can electrons move with constant drift velocity between points of zero resistance without doing electrical work

A2. Now think about the box on a surface with friction. To keep the box moving at constant velocity on the surface you need to apply a constant force equal to the kinetic friction force. Work is required to move the box between two points. Similarly electrical work (voltage times charge moved) is required to move the charge through the resistance.

Although electrons accelerate and decelerate on average their velocity is constant (drift velocity). But it still requires work to move them along, just like our box moving at constant velocity on a surface with friction. The electrical work per unit charge is the loss of potential energy in the form of resistance heating, Just like the work in moving the box results in friction heating .

A4. The answer A3 should answer this question. Any kinetic energy acquired by the electrons is dissipated as heat in the resistor.

Hope this helps

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  • $\begingroup$ So , Electrons gain work against resistance and lose their potential , And kinetic energy still constant in average because of accelerating and collisions , Is that right ? $\endgroup$
    – Mohammad Alshareef
    Commented Jul 14, 2019 at 22:55
  • $\begingroup$ Mohammad essentially yes except i would say electrons gain potential energy from the battery then lose it doing work against resistance $\endgroup$
    – Bob D
    Commented Jul 14, 2019 at 23:15
  • $\begingroup$ Eng.Bob D ,Just one more please .When circuit is open electrons on negative terminal are accumulated and have drift speed of zero , now when the circuit is closed ,the Battery provide electrons with potential energy and accelerate them(provide kinetic energy) ,Now how long the battery still accelerating electrons ? $\endgroup$
    – Mohammad Alshareef
    Commented Jul 15, 2019 at 11:01
  • $\begingroup$ As soon as current flows electrons alternatively gain and lose kinetic energy as they gain it from the electric field and lose it in collisions $\endgroup$
    – Bob D
    Commented Jul 15, 2019 at 12:49
  • $\begingroup$ I’ll give you a gravity analogy. A mass m has gravitational potential energy of mgh a height h above the surface of the earth. If it drops it loses potential energy while gaining kinetic energy. Suppose there were many objects in its path colliding with It such that each time it gained KE it lost an equal amount of KE in the collisions in the form of heat, so that its average velocity was constant. The end result is the loss of PE is heat dissipated, like our electron in the resistor $\endgroup$
    – Bob D
    Commented Jul 15, 2019 at 13:33
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Q1 :Since point a and point b are in different positions in electric field

The (ideal) wire connecting points a & b is an equipotential 'surface', i.e., there is no electric field inside an ideal wire. Thus, points a & b are effectively 'at the same point' in the electric field.

Q2 :As I have mentioned ,Potential energy of the electron converts to kinetic energy ,Now when electrons flow through a resistor ,why is there a voltage drop ?

There is a voltage drop (potential difference) across the resistor because the charge density on one terminal is different from the charge density on the other terminal. It must be this way because, otherwise, the flow of charge into the resistor would not equal the flow of charge out of the resistor.

Q3 : My last question about drift speed. Because electrons are in electric field ,Electric filed applies force on them ,So they accelerate (F=ma)

True but it's also true that they collide with the lattice of the resistor material giving up some KE. This is why a resistor warms up when there is a current through.

I say there isn't voltage drop because when electron flow in the resistor

But, empirically, there is a voltage drop regardless of what you say.

Some collisions occur in the resistor ,So electrons lose some of their kinetic energy ,But due to the electric field ,Electrons accelerate again and restore their lost kinetic energy ,So why is there a drop in electric potential or voltage drop ?

Right, the electrons leave the resistor with the same kinetic energy (on average) as they enter with. But why do you (1) acknowledge the existence of the electric field through the resistor but (2) question why there is a potential difference?

Isn't the potential difference given by the line integral of the electric field through the resistor?

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  • $\begingroup$ $A$ do you (1) acknowledge the existence of the electric field through the (2) question why there is a potential difference? , I mean that there is a voltage because of the battery , and there is an electric field because of the battery , but What I don't understand why there is a voltage drop because of the resistor. $B$ :Could you please explain why there is a potential drop ? I know there is a potential drop BUT WHY ? where , why and how electrons lose electric potential ? Please include pictures if you can. $\endgroup$
    – Mohammad Alshareef
    Commented Jul 14, 2019 at 22:01
  • $\begingroup$ @MohammadAlshareef, I'm afraid I don't understand why you don't see this. Since you agree that there is an electric field through the resistor, it follows that there is a difference in potential through the resistor. You even state, in your question, that there should be a potential difference at different points in the electric field so I'm honestly at a loss why you believe there should be a potential difference between the ends of an ideal wire (through which there is no electric field) but no potential difference between the ends of resistor (through which there is an electric field). $\endgroup$
    – Alfred Centauri
    Commented Jul 14, 2019 at 22:47
  • $\begingroup$ Haha , I think you have got angry , Just I am stupid . Yes I believe there is a potential difference but I am asking why ? likewise I believe that water fall from high but why ? because of gravity . $\endgroup$
    – Mohammad Alshareef
    Commented Jul 14, 2019 at 22:51
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Okay, I'll try to shed some light on your questions. However, I don't know what things you know, so I'll try to give simple answers.

Q1: Since point a and point b are in different positions in electric field (produced by voltage of the battery) Why have they the same electric potential , Why is there no voltage drop?

The reason is that we consider that they are ideal wires. That means, they are perfect conductors with 0 resistance.

If they were real wires, there would be a tiny voltage drop. You could calculate its value:

$V=I\cdot R$.

But, as their resistance is 0, the voltage drop is also 0. That's why working with ideal wires is good.

In fact, this means that it doesn't matter how long the wire is, so you can use longer wires, shorter wires, or bend them, etc. It will not affect the circuit (as long as they are ideal).

Nevertheless, when you set up a real circuit, they're real wires, so the longer they are, the more voltage drop they experience. Use shorrt wires.

Real wires are harder to deal with. That's why we usually model real wires as ideal wires + a resistance, and that extra resistance accounts for the losses of the real wire.

Q2: As I have mentioned ,Potential energy of the electron converts to kinetic energy ,Now when electrons flow through a resistor ,why is there a voltage drop?

So, if the battery inserts a voltage difference, and wires do not drop it, the drop must occur in the resistor.

On Q2: I say there isn't voltage drop because when electron flow in the resistor ,Some collisions occur in the resistor ,So electrons lose some of their kinetic energy ,But due to the electric field ,Electrons accelerate again and restore their lost kinetic energy ,So why is there a drop in electric potential or voltage drop?

You're right: kinetic energy is lsot due to collisions, that's why resistances heat up.

This does not mean there cannot be a voltage drop. There is.

Roughly speaking, the potential energy becomes kinetic energy. But, due to collisions, instead of accelerating a lot, they speed down, until they reach a "steady" velocity, which defines a uniform current. Thsi is the product of the equilibrium between accelerating force and collisions.

Check that this is also coherent with conservation of mass and charge: there cannot come out more than what comes in.

This also answers your $Q3$, I think. I hope this helped. You can tell me anything in the comments.

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