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Why do we have the concept of electric potential

  1. Mar 8, 2009 #1
    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.
     
  2. jcsd
  3. Mar 8, 2009 #2

    Dale

    Staff: Mentor

    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.
     
  4. Mar 8, 2009 #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?
     
  5. Mar 8, 2009 #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.
     
  6. Mar 8, 2009 #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?
     
  7. Mar 8, 2009 #6

    Dale

    Staff: Mentor

    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.
     
  8. Mar 8, 2009 #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. Mar 8, 2009 #8

    Dale

    Staff: Mentor

    What would you integrate?
     
  10. Mar 8, 2009 #9
    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.
     
  11. Mar 8, 2009 #10
    I am studying electrostatics and totally unfamiliar with the first equation.


    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 ? :)
     
    Last edited: Mar 8, 2009
  12. Mar 8, 2009 #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.
     
  13. Mar 8, 2009 #12

    Dale

    Staff: Mentor

    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.

    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.
     
  14. Mar 8, 2009 #13
    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 ?
     
  15. Mar 8, 2009 #14
    No, then you would get infinite potential in that spot.

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