Why do we have the concept of electric potential

In summary, the concept of electric potential simplifies calculations involving electric fields, which are a vector field. However, this sentence does not satisfactorily answer my question.
  • #1
ximath
36
0
Hi all,

I have a simple question related to electrostatics. We have the concept of electric field which is a vector field and makes sense. I am aware that it is just a virtual concept in fact, defined as force per unit charge, in other words, it doesn't exist. However, I understand the reason why we need such a concept.

We also have electric potential concept, which is totally virtual (like the electric field); defined as the potential energy per unit charge. The problem is, I am not able to see the reason why we need such a concept in electrostatics. I have searched for it and learned that since electric potential is a scalar quantity, it simplifies the calculations that involves electric field, which is a vector field. However, this sentence doesn't satisfy me at all. In my book, most of the exercises related to electric potential subject asks for the calculation of electric potential. This is rather more like a "problem" than a "tool" that helps.

In what kind of cases the concept of electric potential helps us ? Is there any example you could share with me ?

Thanks in advance.
 
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  • #2
You can say the exact same thing about the gravitational potential. At each point you can look at the gravitational field, a vector field, or the gravitational potential, a scalar field. The two are related to each other the same way that the electric field and electrical potential are related to each other. If you have a mass going down a gravitational potential you can calculate the energy that it gains independently of the path. Similarly, if you have a charge going down an electric potential you can calculate the energy that it gains independently of the path.
 
  • #3
Hi, thanks a lot.

I am in fact aware of the analogy between gravitation and electrostatics. However, although it helps clarifying a little bit, I am still not able to understand the reason why we have such concepts. How does it help us?
 
  • #4
Calculating electric fields are much harder than calculating electric potential, a potential is easy since you just have to integrate the scalar contribution made by all charges you take into consideration and you are done, while with electric fields you need to take into account the directions of the vectors and then sum those together.

And the gradient of the electric potential is the electric field, they are just two ways of describing the same thing.

Also for example when you do electrical circuits calculating electric fields would be extremely tiresome, so its a lot easier to just do the potential difference between two points. Usually you just want the line integral between two points and for that you just take the potential difference and you are done.
 
  • #5
But again, don't we still need to get our hands dirty with angles and stuff that are related to vectors ? I mean, in order to find the potential we need to integrate the dot product (E dot dl) which means we still need to deal with the angles... Is it really simpler than finding the components of the electric field directly by integration?
 
  • #6
The point is that you can find the work done on an object without doing the integration at all. You can just use the difference in potential and know that it equals the integral regardless of the path.
 
  • #7
It makes sense, supposing we know the potential difference. But if this isn't the case, in order to calculate it, don't we still need to integrate ?
 
  • #9
ximath said:
But again, don't we still need to get our hands dirty with angles and stuff that are related to vectors ? I mean, in order to find the potential we need to integrate the dot product (E dot dl) which means we still need to deal with the angles... Is it really simpler than finding the components of the electric field directly by integration?
You know the familiar I=V/R?

Volt is just the potential difference between the two sides.

Also there is no need to integrate the dot product, the potential V=q/4epir. If you have a charge distribution you just need to use that to get the potential.
 
  • #10
Klockan3 said:
You know the familiar I=V/R?

Volt is just the potential difference between the two sides.

Also there is no need to integrate the dot product, the potential V=q/4epir. If you have a charge distribution you just need to use that to get the potential.

I am studying electrostatics and totally unfamiliar with the first equation.


DaleSpam said:
What would you integrate?

As far as I understood, the potential difference is integral of E dot dl. In order to find the potential at a point, I thought we need to take this integral.

Hmm, I guess I start to understand a little bit of it now! I guess instead of doing the integration, we could simply say that [tex]V = \frac{q}{4 \pi \epsilon r}[/tex] and use this result to find the work done instead of integrating the q E dl.

This would help us, right. One more thing; by assuming [tex]V = \frac{q}{4 \pi \epsilon r}[/tex] ; do we also assume that V = 0 at r = infinity ?

As far as I know, we could select any arbitrary point to have zero potential -- hence which means this equation is only valid when V = 0 at infinity. Or am I totally wrong ? :)
 
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  • #11
Calculations gets much simpler if we say that a charge free space have 0 potential, and that also feels more natural.

And the first equation is just the simple electric current equals electromotive force divided by resistance, you should have seen it in your first physics class ever.
 
  • #12
ximath said:
As far as I understood, the potential difference is integral of E dot dl. In order to find the potential at a point, I thought we need to take this integral.

Hmm, I guess I start to understand a little bit of it now! I guess instead of doing the integration, we could simply say that [tex]V = \frac{q}{4 \pi \epsilon r}[/tex] and use this result to find the work done instead of integrating the q E dl.
Exactly, you jumped right ahead and understood what I was hinting at. In textbook problems you are often given E and asked to find V, which is a pain. However, in many more realistic scenarios you either explicitly know V (i.e. your wall socket provides 120 V) or you have a charge distribution from which you can directly calculate V without having to integrate E. In fact, in the latter case it is easier to calculate V than E anyway since it is a scalar instead of a vector.

ximath said:
This would help us, right. One more thing; by assuming [tex]V = \frac{q}{4 \pi \epsilon r}[/tex] ; do we also assume that V = 0 at r = infinity ?

As far as I know, we could select any arbitrary point to have zero potential -- hence which means this equation is only valid when V = 0 at infinity. Or am I totally wrong ? :)
That is correct. You can add an arbitrary constant to your voltage field without changing anything, so you have to pick some convention. The usual convention is that V=0 at infinity. The same convention is used with gravitational PE. It is just a convention, but if you don't want to use it you have to add an offset term to your above expression for V.
 
  • #13
Klockan3 said:
Calculations gets much simpler if we say that a charge free space have 0 potential, and that also feels more natural.

And the first equation is just the simple electric current equals electromotive force divided by resistance, you should have seen it in your first physics class ever.

You're right, I had seen it, but I am trying to figure the things out one by one :)

Thanks a lot all, for your help and clarifications, I guess I understand it now.

One last thing, just curiosity, if I selected V = 0 at r = a; what would be the corresponding equation for V ?

I guess I would have (r-a) instead of r in the denominator... Is it correct ?
 
  • #14
ximath said:
One last thing, just curiosity, if I selected V = 0 at r = a; what would be the corresponding equation for V ?

I guess I would have (r-a) instead of r in the denominator... Is it correct ?
No, then you would get infinite potential in that spot.

Instead you just put V=+q/4pier-q/4piea
 

1. Why do we need the concept of electric potential?

The concept of electric potential is crucial for understanding and analyzing the behavior of electrically charged particles in an electric field. It helps us to calculate the work done on a charged particle as it moves through the electric field and the resulting change in its kinetic energy. Without this concept, we would not have a quantitative way to measure the potential energy of charged particles in an electric field.

2. How does electric potential differ from electric field?

Electric potential is a scalar quantity that represents the amount of potential energy per unit charge at a given point in an electric field. It is a measure of the electric field's intensity, but it does not convey information about the direction of the field. On the other hand, electric field is a vector quantity that represents the force per unit charge acting on a charged particle at a given point in an electric field. It indicates both the magnitude and direction of the force experienced by a charged particle.

3. 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. It is directly proportional to the electric potential difference between two points, as given by the equation U = qΔV, where U is the electric potential energy, q is the charge, and ΔV is the potential difference. In other words, electric potential is a measure of the potential energy per unit charge at a specific point in an electric field.

4. Can electric potential be negative?

Yes, electric potential can be either positive or negative. It depends on the relative positions of the charged particles and the direction of the electric field. A positive electric potential indicates that a positively charged particle would experience a repulsive force, while a negative electric potential indicates an attractive force towards the source of the field.

5. How is electric potential measured?

Electric potential is measured using a device called a voltmeter. It consists of two probes connected to a power source and a display screen that shows the potential difference between the two points. The unit of measurement for electric potential is volts (V). In an electric circuit, the potential difference between two points is equal to the work done per unit charge in moving from one point to the other, and this can be measured using a voltmeter.

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