# Electric Potential Intuition Needed

• LikesIntuition
In summary, the voltage across parallel circuit elements is the same because they are connected to the same voltage source. This is because the difference in potential energy between two points depends only on the starting and ending points, similar to how gravitational potential energy works. The specific mechanism for establishing the voltage will depend on the type of voltage source used. The behavior of electrons in the circuit elements will also vary depending on the type of element and voltage source. Boundary conditions play a role in narrowing down the possible solutions to the differential equations that govern circuits. The setup of a field in a wire with resistance, such as in a circuit with an AA battery, can be a case of a uniform field being established, but this will also depend on the specific boundary conditions
LikesIntuition
I'm trying to understand why the voltage across parallel circuit elements is the same. I keep finding the answer "because they are both connected to the same voltage source."

Why is it the case, in a physical sense, that voltage MUST be the same across, say, capacitors that are connected in parallel to the same voltage source? How does this physically dictate the same potential difference existing across each capacitor in parallel?

I have other questions about voltage and circuits, but I will hold off on them and see where this example takes us.

Thanks for any help!

LikesIntuition said:
I'm trying to understand why the voltage across parallel circuit elements is the same.
Hi LikesIntuition, welcome to PF!

Parallel circuit elements are each connected between the same pair of conductors, and each conductor has a single voltage. So the voltage across any element is V2-V1. Since V2-V1=V2-V1 the voltages are the same.

Ok thanks! So, how exactly does that voltage come about in those circuit elements? What happens at the moment when they're connected to the conductors that have a voltage?

Edit: Also, when I put a voltage across a resistor, is a uniform field set up in that resistor? If so, why doesn't the field drop off over distance like it would in a vacuum?

Last edited:
The point is not the vacuum or lack of it. The field of a point charge drops off with distance, but here we're not dealing with a single point charge, are we? We're dealing with a large collection of charges spread over the conductor

Hi LikesIntuition!
LikesIntuition said:
I'm trying to understand why the voltage across parallel circuit elements is the same. I keep finding the answer "because they are both connected to the same voltage source."

Why is it the case, in a physical sense, that voltage MUST be the same across, say, capacitors that are connected in parallel to the same voltage source?

Voltage is (electric) potential energy per charge.

Difference in voltage is difference in (electric) potential energy per charge.

The difference in potential energy depends only on where you start and finish …

it's like gravitational potential energy, which also depends only on where you start and finish: in a playground if two slides connect the same points, the difference in potential energy between them is the same whichever slide you go down.

It doesn't matter whether it's two capacitors or any other components: if they're connected to the same two points, they have the same potential energy difference (per charge), ie the same voltage difference.

LikesIntuition said:
Ok thanks! So, how exactly does that voltage come about in those circuit elements?
You didn't specify the voltage source, so I cannot say. Different voltage sources will have different mechanisms for establishing the voltage.

LikesIntuition said:
What happens at the moment when they're connected to the conductors that have a voltage?
You didn't specify the types of circuit elements connected, so I cannot say. Different circuit elements will respond differently (i.e. with different currents) to an applied voltage step.

LikesIntuition said:
Edit: Also, when I put a voltage across a resistor, is a uniform field set up in that resistor?
Yes, approximately.

LikesIntuition said:
If so, why doesn't the field drop off over distance like it would in a vacuum?
You can have uniform fields in vacuum too. Fields don't necessarily drop off over distance, it all depends on the boundary conditions.

When I ask how voltage is set up in the circuit elements, I mean: physically, what are the electrons in the circuit doing?
Could you also explain a bit more about boundary conditions?

And thanks, tiny-tim! That explanation makes a lot of sense. So, as I cross a circuit element, I will go through some sort of change in electric potential, and that has to do with what voltage source the ends of the circuit element are connected to?

LikesIntuition said:
When I ask how voltage is set up in the circuit elements, I mean: physically, what are the electrons in the circuit doing?
As I already told you, you have to tell me the specific circuit element or voltage source before your question can be answered with any physical detail. I cannot read your mind. You can continue to ask the same question and get the same answer if you like, but a more productive approach would be to actually provide some details about the specific voltage sources and circuit elements that you are interested in.

LikesIntuition said:
Could you also explain a bit more about boundary conditions?
Laws of physics are differential equations, in the case of circuits the corresponding differential equations are Maxwell's equations and the Lorentz force law. When you solve a differential equation you don't get a single answer, you get a whole family of answers. For example, if you solve the differential equation for projectile motion you get a family of parabolas. In order to narrow down that family to a single answer you have to provide boundary conditions. For example, the initial velocity and position of your projectile.

DaleSpam said:
As I already told you, you have to tell me the specific circuit element or voltage source before your question can be answered with any physical detail. I cannot read your mind. You can continue to ask the same question and get the same answer if you like, but a more productive approach would be to actually provide some details about the specific voltage sources and circuit elements that you are interested in.

I'm sorry, I didn't understand what you meant completely the first time, and your answer wasn't the type of answer I had expected, so I thought maybe trying to reword my question might give you more insight into what I was asking. As you say, we can't read each other's mind!

So, how about a typical AA battery setting up a field in a wire that has resistance? Would this be a case of a uniform field being set up? If so, what is happening to set it up? My professor talks a good deal about electrons trying to arrange themselves to eliminate the field imposed on them, like in a conductor. Is this what's happening, but the resistance doesn't allow them to eliminate the field?

If you want a uniform field, you need a pair of large plates, parallel to each other with spacing x. The field between them, away from the edges will be a pretty good V/x volts per metre. If you want to consider the fields inside the wire, that is an entirely different matter.

LikesIntuition said:
So, how about a typical AA battery
So a battery establishes a voltage by making a microscopic push or pull on an individual electron. On the side where oxidation occurs the molecules literally push an electron onto the electrode, and on the side where reduction occurs the molecules literally pull an electron off of the electrode. The pushing and pulling do work on the electrons which is physically how the voltage is established (remember voltage is work or energy per unit charge).

LikesIntuition said:
setting up a field in a wire that has resistance?
A resistor responds to a voltage by allowing a current through it which is proportional to the applied voltage. Technically, any conductive material responds that way, and the difference between a resistor and a conductor is just the amount of current for a given voltage.

LikesIntuition said:
Would this be a case of a uniform field being set up?
Yes, approximately.

LikesIntuition said:
If so, what is happening to set it up?
The field is the negative gradient of the voltage. So where you have a voltage which changes over distance you have a field. So saying that you have set up a spatially changing voltage is the same thing as saying that you have set up a field.

LikesIntuition said:
My professor talks a good deal about electrons trying to arrange themselves to eliminate the field imposed on them, like in a conductor. Is this what's happening, but the resistance doesn't allow them to eliminate the field?
That "eliminate the field thing" only applies for electrostatic situations. When there is a current you are not in an electrostatic situation any more and you can have fields which are not eliminated in a conductor.

DaleSpam said:
The field is the negative gradient of the voltage. So where you have a voltage which changes over distance you have a field. So saying that you have set up a spatially changing voltage is the same thing as saying that you have set up a field.

That makes a lot of sense! I've never thought of it that way. So if voltage changes spatially, it means that the work done on a charge would change over that region of space, and if work is being done we have a force. Am I looking at this accurately?

Yes, that is correct. I am glad I could help!

LikesIntuition said:
So, how about a typical AA battery setting up a field in a wire that has resistance? Would this be a case of a uniform field being set up? If so, what is happening to set it up? My professor talks a good deal about electrons trying to arrange themselves to eliminate the field imposed on them, like in a conductor. Is this what's happening, but the resistance doesn't allow them to eliminate the field?
Here is an interesting explanation of the electric field inside a wire.
http://www.phy-astr.gsu.edu/cymbalyuk/Lecture16.pdf
And about the uniformity of the electric field - take a look at the generalized form of Ohms's law.
http://en.wikipedia.org/wiki/Ohm's_law
J = σE
That tells you that as long as the current density and the conductivity/resisitivity are the same, the field strength will be the same.

Last edited by a moderator:
LikesIntuition said:
And thanks, tiny-tim! … So, as I cross a circuit element, I will go through some sort of change in electric potential, and that has to do with what voltage source the ends of the circuit element are connected to?

sorry, i don't like that description at all

it depends not only on the source but also on everything else in the circuit

any particular circuit element only gets the voltage difference that is left over after the voltage difference across everything else in series has been subtracted

you should think of a voltage (a potential) as existing at every point, and then the difference in voltage between two points tells you the work done per charge moving between them along any path

(of course, often there's only one path)

LikesIntuition said:
DaleSpam said:
The field is the negative gradient of the voltage. So where you have a voltage which changes over distance you have a field. So saying that you have set up a spatially changing voltage is the same thing as saying that you have set up a field.
That makes a lot of sense! I've never thought of it that way. So if voltage changes spatially, it means that the work done on a charge would change over that region of space, and if work is being done we have a force. Am I looking at this accurately?

yes, but I'm not sure that that's helpful in most cases

yes you do need the spatial change of voltage to find the voltage across eg the gap in a capacitor (you are using work done = force times distance)

but no in most parts of most circuits the spatial change is unused and unhelpful (you are using work done per time = volts times charge per time = volts times current, with no mention of force)who cares what the electric field (force per charge) is? … you can find everything using voltage current etc!

(this is different from the gravitational comparison …

in gravity, W = Fd only,which involves the potential

but in electricity, W = Fd and W = Vq, and the potential is only involved in the latter, which is not explicitly spatial)​

Last edited:
gracy
DrZoidberg said:
Here is an interesting explanation of the electric field inside a wire.
http://www.phy-astr.gsu.edu/cymbalyuk/Lecture16.pdf
And about the uniformity of the electric field - take a look at the generalized form of Ohms's law.
http://en.wikipedia.org/wiki/Ohm's_law
J = σE
That tells you that as long as the current density and the conductivity/resisitivity are the same, the field strength will be the same.

Thanks! The diagrams in the pdf were exactly the type of thing I was looking for!

Last edited by a moderator:

## 1. What is electric potential?

Electric potential is the measure of the amount of electrical potential energy that a charged particle possesses at a certain point in space.

## 2. How is electric potential different from electric field?

Electric field is a vector quantity that represents the force that a charged particle will experience at a given point in space. Electric potential, on the other hand, is a scalar quantity that represents the potential energy of a charged particle at a given point in space.

## 3. How is electric potential measured?

Electric potential is measured in units of volts (V) using a voltmeter. It can also be calculated by dividing the work done by an electric field on a charged particle by the charge of the particle.

## 4. What is the relationship between electric potential and electric potential energy?

Electric potential energy is the potential energy that a charged particle possesses due to its position in an electric field. The electric potential at a point is equal to the electric potential energy per unit charge at that point.

## 5. How does distance affect electric potential?

As distance increases between a charged particle and a source of electric potential, the electric potential decreases. This is because the electric field becomes weaker with increasing distance, resulting in a decrease in the electric potential energy per unit charge.

• Electromagnetism
Replies
7
Views
1K
• Electromagnetism
Replies
21
Views
2K
• Electromagnetism
Replies
4
Views
1K
• Electromagnetism
Replies
14
Views
880
• Electromagnetism
Replies
16
Views
1K
• Electromagnetism
Replies
1
Views
995
• Electromagnetism
Replies
3
Views
1K
• Electromagnetism
Replies
7
Views
1K
• Electromagnetism
Replies
14
Views
3K
• Electromagnetism
Replies
11
Views
2K