Is Potential Constant on a Conductor?

AI Thread Summary
In a connected conductor, the electric potential at any point is uniform, confirming that for an ideal conductor, the potential is the same throughout. However, the absolute value of the potential is arbitrary and can be defined as zero at any chosen point, lacking intrinsic physical significance. In equilibrium, there is no voltage drop, but if charges are in motion, such as in a current, a voltage drop occurs along the conductor. Superconductors maintain a constant current without a voltage drop, even when current flows. Understanding these principles clarifies the behavior of conductors and superconductors under different conditions.
Sogan
Messages
3
Reaction score
0
Hi,

I consider a connected conductor. Is it right, that:

1)the potential at any point of the conductor is the same, but

2)the absolute value of the potential isn't zero in general?

I think these statements are true, but I'm not sure about it, especially with the second statement.

Thank you a lot!

Regards
 
Physics news on Phys.org
1)the potential at any point of the conductor is the same
For an ideal conductor, yes.

2)the absolute value of the potential isn't zero in general?
This value has no meaning. You can define "zero" wherever you want. There are some conventions, depending on the setup, but they do not have a physical meaning themself.
 
Sogan said:
1)the potential at any point of the conductor is the same
mfb said:
For an ideal conductor, yes.

in equilibrium :wink:

(if charges are moving, ie if there is a current, then obviously there is a voltage drop along the conductor, ie an electric potential difference)
 
Thank you very much! Now I understood.
 
tiny-tim said:
(if charges are moving, ie if there is a current, then obviously there is a voltage drop along the conductor, ie an electric potential difference)
In superconductors with constant current, you have no voltage drop, even with a current flow.
If charges are accelerating, you have an electric field and therefore a voltage drop ;).
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top