I Electric Potential in circuit

Click For Summary
The discussion focuses on the application of the electric potential formula in circuit models. The formula for electric potential at a point suggests that higher positive charge concentrations should exist where potential is higher, which contradicts the neutrality of points in a simple circuit. It is clarified that the formula applies to voltage due to point charges, not to continuous charge distributions found in circuits. The presence of a continuous distribution of surface charge complicates the application of the formula, as there are no solitary point charges in the circuit. Thus, the traditional electric potential equation does not accurately represent the behavior of circuits.
eyeweyew
Messages
35
Reaction score
6
TL;DR
Electric potential at a point equation for circuit and net charge
I reviewed some of the fundamental physics and I looked back at the equation for Electric potential at a point p:
$$V(p) = k \sum_{i} {\frac {q_i} {r_i}}$$
where

- p is the point at which the potential is evaluated;
- ri is the distance between point p and point i at which there is a nonzero charge;
- qi is the charge at point i

and I still find it's kind of contradicting with the simple circuit model such as the one below. Both point a and point b should be neutral with no net charge so their electric field is 0 and the voltage is flat on the graph according to Gauss law. I understand the electric potential of point b is ε higher than that of point a (i.e. V(b)-V(a)=ε) means it takes ε work to move a +1 test charge from point a to point b along the circuit.

But according to Electric potential formula at a point, should that also imply there are higher positive net charge concentration around point b than point a so how can they both neutral with no net charge? Does that mean the equation for Electric potential at a point does not apply in a circuit model but if so, why?

electric_circuit_voltage_plots-001.png

image reference: https://tikz.net/electric_circuit_voltage_plots/
 
Last edited:
Physics news on Phys.org
Is that formula supposed to be the voltage at a point or the voltage due to a point charge? Read the surrounding text carefully
 
Dale said:
Is that formula supposed to be the voltage at a point or the voltage due to a point charge? Read the surrounding text carefully
It is voltage at a point due to other point charges. I edited my post to clarify it. Thanks!
 
eyeweyew said:
It is voltage at a point due to other point charges. I edited my post to clarify it. Thanks!
So that formula doesn’t really apply. There are no solitary point charges in that circuit. There is a continuous distribution of surface charge along all the conductors. That distribution doesn’t have a nice closed form expression.
 
Thread 'Why higher speeds need more power if backward force is the same?'
Power = Force v Speed Power of my horse = 104kgx9.81m/s^2 x 0.732m/s = 1HP =746W Force/tension in rope stay the same if horse run at 0.73m/s or at 15m/s, so why then horse need to be more powerfull to pull at higher speed even if backward force at him(rope tension) stay the same? I understand that if I increase weight, it is hrader for horse to pull at higher speed because now is backward force increased, but don't understand why is harder to pull at higher speed if weight(backward force)...

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
Replies
16
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
6
Views
2K
Replies
3
Views
462
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 10 ·
Replies
10
Views
1K