How do you calculate the voltage between two charged nodes?

AI Thread Summary
To calculate the voltage between two charged nodes, one must understand that voltage is the difference in electric potential energy between them. The electric field can be approximated by summing the fields of two point charges and integrating from one surface to another. When the objects are close, their charge distributions influence each other, complicating calculations. For a two-dimensional scenario, the electric field can be calculated using the formula E = (q / (4πε₀)) * (r / r²), while in three dimensions, it becomes E = (q / (4πε₀)) * (r / |r|³). It's important to note that the charge should have a finite size for the potential to be well-defined.
Cuppasoup
Messages
6
Reaction score
0
If:
1) you know the vertex and individual charge of each node and
2) voltage is the difference in electric potential energy between two nodes

How do you calculate the voltage in a vacuum?
 
Physics news on Phys.org
Nodes where, vacuum where?
 
mfb said:
Nodes where, vacuum where?
I don't understand your question. 2 objects in space, each with an individual charge(in coulombs). How do you calculate the "tension" (in voltage)between the two charged entities?

By "where" do you mean, "what is the vertex of each node?"? If that's your question, shouldn't it be calculable with a formula?
 
Last edited:
Cuppasoup said:
I don't understand your question. 2 objects in space, each with an individual charge(in coulombs). How do you calculate the "tension" (in voltage)between the two charged entities?
That question is much easier to understand than the original post.
Calculate the electric field everywhere (in most cases, to a good approximation: the sum of the fields of two point charges), integrate from one surface to the other.
 
  • Like
Likes Cuppasoup
mfb said:
That question is much easier to understand than the original post.
Calculate the electric field everywhere (in most cases, to a good approximation: the sum of the fields of two point charges), integrate from one surface to the other.
Okay, thanks for the help! By "Calculate the electric field everywhere," do you mean to calculate 2 individual matrices of "tension" depending on the charge and shape of the object and add them together to calculate the resulting electric field?
 
If the objects are not too close, that will give a reasonable approximation. If they are close, they will influence the charge distribution of the other object and things get complicated.
 
  • Like
Likes Cuppasoup
mfb said:
If they are close, they will influence the charge distribution of the other object and things get complicated.
Oh okay, I guess that means the distinction between "objects" is then blurred.

So how do I calculate an electric field matrix?
 
Cuppasoup said:
I guess that means the distinction between "objects" is then blurred.
Two 1-meter objects with a distance of 0.5 meters between them are clearly different objects, but their charge distributions will influence each other significantly.
Cuppasoup said:
So how do I calculate an electric field matrix?
This is not a matrix, it is a vector field. If you can approximate the charge distribution as spherical, do that. Some other shapes might have analytic solutions, the general case can be treated with numerical methods.
 
mfb said:
This is not a matrix, it is a vector field. If you can approximate the charge distribution as spherical, do that.
Ok, so to start off in 2D with a circle, if I were to calculate a vector field surrounding a charged "point" in the origin, is there a formula to calculate each vector by its coordinate in relation to the charge of the point?
 
  • #10
In a two-dimensional world:
$$\vec E = \frac{q}{4 \pi \epsilon_0} \frac{\vec r}{r^2}$$
In a three-dimensional world:
$$\vec E = \frac{q}{4 \pi \epsilon_0} \frac{\vec r}{|r|^3}$$
Where q is the charge and r is the vector between charge and the point where you calculate the electric field.
Note that the charge should have some finite size, otherwise its potential is not well-defined.
 
  • Like
Likes Cuppasoup
  • #11
Excellent! Thanks!
 
Back
Top