Clarifications about electric potential and potential difference

  • #1

greg_rack

Gold Member
363
79
Specifically, I haven't really got all the "methods" through which you could calculate or derive the electric potential and in some situations, I cannot understand how and when to apply this concept.
Is it something caused by any charge, or must there be an interaction between the two to properly talk about potential? How is a generator keeping the potential(almost, for short periods of time) constant to one end?
 

Answers and Replies

  • #2
Is it something caused by any charge, or must there be an interaction between the two to properly talk about potential?

I think you just need to be clear about the difference between potential, potential difference, potential energy, whatever. A potential ##\phi = \phi(\vec{r})## is a scalar field. Often (we're talking about statics, here) you'll start with ##\vec{E} = -\nabla \phi##, which when you take the divergence gives you$$\nabla \cdot \vec{E} = -\nabla \cdot \nabla \phi \implies \triangle \phi = -\frac{\rho}{\varepsilon_0}$$which is Poisson's equation. You need some boundary conditions. There are lots of ways to go about solving it in different problems, e.g. Green's functions. Sometimes you're lucky if ##\rho=0## in the region you're doing the integration, since you get Laplace's equation ##\triangle \phi = 0##. At this point it's really more of a calculus exercise 😜.

It is potential energies that are energies of interaction, i.e. you can't localise potential energy within a system. You should try and keep the ideas of electric potential ##\phi## and the electric potential energy ##U## of a configuration separate, although they're linked (e.g. a point charge in an external field, you say the system has potential energy ##U = q\phi##. In other cases, though, it's not so obvious how the potential energy/self energy of a system relates to ##\phi(\vec{r})##).

Some problems here:
http://teacher.pas.rochester.edu/PHY217/LectureNotes/Chapter3/LectureNotesChapter3.pdf
 
Last edited by a moderator:
  • Like
Likes greg_rack and vanhees71
  • #3
Given any collection of charges, we divide them into those charges that cause the field and those that are interacting with the field. This division is arbitrary. It is not capricious (I could have been a lawyer...) but rather made to ease the burden of calculation. For two point charges this seems silly. But for a circuit plugged into the power grid it is obviously not.
The forces on charges are the physical observable, therefore the value of the field is what matters. The field is the gradient of the potential and so any constant can be added to the electric potential (but this must be consistently done if two potentials overlap!) . The conventional practice is to set the average potential of the universe (or the planet which is why we call it ground) ) to zero. If you always talk about potential difference it doesn't matter...any consistent choice will do.


Note: Having written this already I will append it @etotheipi excellent comments above

.
 
  • Like
Likes greg_rack, etotheipi and vanhees71

Suggested for: Clarifications about electric potential and potential difference

Back
Top