Mathematically, wouldn't the loss of electrons increase voltage?

In summary, the conversation discusses the concept of voltage and how it relates to the flow of electrons between objects with a difference in charge. The potential difference between two objects is what is expressed as volts, and when a conductive wire is connected between them, electrons flow from the object with excess electrons to the object with fewer electrons. The conversation also touches on the relationship between voltage, joules, and coulombs, and how this affects the flow of electrons. There is also a discussion on the difference between electrostatic and electrodynamic problems, and how the movement of electrons can affect the voltage within a conductor. Overall, the concept of voltage is defined as the energy lost or gained by a charge as it moves in an electric field, and
  • #1
ptownbro
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Dumb question, but something I don't get.

I see this expressed in many basic examples of what voltage is. You have 1 object that has too many electrons and 1 object has too little. The potential difference between them is what is expressed as/in volts. If you connect a conductive wire between them, the electrons will flow from the object with the excess electrons (negative) to the object with lower amounts of electrons (positive). Here's my (lame) attempt at drawing it here =) (Forgive the extra "." only way I could get the spacing).

______________ ........_______________
|.....|........|......|
|...6 electrons...|---------------------------->|...2 electrons...|
|_____________|........|______________|

There's a difference in electrons (assume coulombs of electrons. I wrote electrons to help visualize it in my drawing) of 4 coulombs creating voltage. 2 coulombs would need to flow from object 1 to 2 to equalize the electrons between them, bringing the difference to zero and eliminating the voltage.

My question:

If voltage = joules / coulombs, then as electrons flow from object 1 to object 2 and is losing electrons, wouldn't that increase the voltage in that mathematical equation? Your denominator is going down, so mathematically, wouldn't voltage increase?
 
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  • #2
if you spend a work of 50 joules , this means you either spent the whole 50 joules on one electron and created a P.D. of 5o volts , or you spend them on two electrons and create a p.d. of 25 volts .

but by the way its not about electrons flowing from where the number is high to where the number is low , its about overcoming the potential difference even if the number of electrons is the same( more electrons usually mean more potential difference but that's not always the case ) .
in case of batteries , its not about equalizing the number of electrons , its about equalizing the potential difference across the poles of a battery , also i find it better not to think of the electric current in terms of electrons , think of it in terms of charge . electrons are weird little things !
 
  • #3
Think of it as a two-step process.

1. The charge distribution creates an electric field, which is equivalent to voltage difference between the sides.

2. In the presence of this electric field, the charges experience a force which makes them move from higher voltages to lower voltages.

There is an interdependency between the charge positions and the electric field. One depends on the other, which depends on the other...


Now voltage is DEFINED so that its downhill gradient points in the direction of the electric field (so 1 to 0 means the electric field points left to right). As the electrons move, the electric field that they create changes, and the voltage changes. Eventually, with 4 on each side, the voltage will be constant throughout the conductor. This is the steady state.

Voltage can be thought of in terms of the amount of Joules one Coulomb loses when going from e.g. 1 V to 0 V, but this isn't particularly useful when the voltage depends on where the Coulomb of charge is! Your case here, where the electric field generators (the charges) are moving, is an electrodynamic problem, so you need to be more careful when using this formula. Most of the time the electrostatic approximation is fine, charges either move so fast you assume they're always in a stationary state (e.g. in perfect conductors), or they don't move at all (e.g. in perfect insulators).

Your case is an electrodynamic problem because you're asking for the details of the electrons' movements as they equilibrate within a conductor. That means you can't assume the voltage distribution is constant.

Logically, try to think of it this way:

Charge distribution causes electric field/voltage
Electric field/voltage causes forces to act on charges
As these charges move in this field, they lose or gain energy.

The energy they lose/gain is the voltage times their charge.

So calculate the voltage, then the energy, not the other way around.
 
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  • #4
Hmmm...
1. How can you have potential difference without a difference in electrons?
2. Better not to think of electric current in terms of electrons? I've just started this, but everything I've read/seen describes it as a flow of electrons. You saying it's not?

Setting those aside for a moment, let me stick with the traditional definitions as they pertain to my question... Using my example, again, if there are 6 coulombs in object 1 and 2 coulombs in object 2, as electrons flow from object

Assume joules = 100. As the difference decreases I would have expected the voltage to go down (assuming constant joules).

Volts = joules / coulombs (shown as differences below)
Volts = 100 / 4 = 25
Volts = 100 / 3 = 33
Volts = 100 / 2 = 50
Volts = 100 / 1 = 100

What am I doing wrong here?
 
  • #5
ptownbro said:
If voltage = joules / coulombs, then as electrons flow from object 1 to object 2 and is losing electrons, wouldn't that increase the voltage in that mathematical equation? Your denominator is going down, so mathematically, wouldn't voltage increase?
Yes, it would, but that is only very indirectly related to the scenario you were talking about.

In the equation V=E/Q, if you reduce Q but keep E constant then V will indeed increase. In other words, if you want to store the same amount energy in a smaller amount of charge then you need a larger amount of voltage.

However, in the scenario you described above E is not constant. As the electron goes from one plate to the other it can do work. So both E and Q are going down. When both numerator and denomintor are going down the effect on the ratio is ambiguous and depends on which is going down more, which you have to determine from other factors.
 
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  • #6
Ah. Ok. That answers it. I didn't realize joules would also naturally go down as well. Consider my question solved! =)

Thanks!

Though I still don't see how you have potential difference without difference in electrons or how current is not a flow of electrons?
 
  • #7
ptownbro said:
Though I still don't see how you have potential difference without difference in electrons or how current is not a flow of electrons?
Current could be a flow of electrons or protons or ions. Similarly, a potential difference could be due to a difference in protons or in ions rather than electrons. You also can get potentials due to changes in magnetic fields even in vacuum with no charges whatsoever.
 

1. What is the relationship between the loss of electrons and voltage?

In simple terms, voltage is the measure of electrical potential difference between two points. When electrons are lost, the number of charged particles in a system decreases, resulting in a higher potential difference and therefore an increase in voltage.

2. How does the loss of electrons affect the overall electrical system?

The loss of electrons can have a significant impact on the overall functioning of an electrical system. It can cause disruptions, such as power outages or malfunctions, and can also alter the behavior of other components in the system.

3. Does the loss of electrons always result in an increase in voltage?

Not necessarily. The relationship between the loss of electrons and voltage depends on various factors, such as the type of material, the size of the system, and the presence of other charged particles. In some cases, the loss of electrons may not have a significant impact on the voltage of a system.

4. Can the loss of electrons be reversed?

Yes, the loss of electrons can be reversed through processes such as recharging or repositioning. In some cases, external sources of electrons can also be introduced to balance out the loss and restore the system's equilibrium.

5. How does the loss of electrons relate to the concept of current?

The loss of electrons is closely related to electric current, which is the flow of charged particles in a system. When electrons are lost, a current is generated as the remaining charged particles try to balance out the charge difference. This current can then be used to power electrical devices or perform other tasks.

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