Laplace in electromagnetics(voltages are different in conductor?)

  • Thread starter Thread starter baby_1
  • Start date Start date
  • Tags Tags
    Conductor Laplace
AI Thread Summary
In the discussion about voltages in a conductor sphere, it is clarified that the potential inside the sphere is uniform and should be zero, as there are no electric field variations within a conductor. The equation V -> -Ez + C applies only outside the sphere, while the potential inside remains constant due to the sphere being uncharged. The solution to Laplace's equation for this scenario can be simplified without using full Legendre polynomials. An analogy with a dipole problem is suggested to further understand the potential distribution. The key takeaway is that the potential inside a conductor does not vary, reinforcing fundamental electrostatic principles.
baby_1
Messages
159
Reaction score
16
Hello
as you see this example and solution
7236109700_1397288252.jpg

2462158600_1397288252.jpg

if we assume the R of conductor sphere is 5m and check voltage in different z we obtain(for example z=1m and z=2)
gif.gif

gif.gif

and as we know in conductor we doesn't have voltage differences so this equation should be the same
gif.gif

C=0
but voltage aren't the same
gif.gif


gif.gif

gif.gif


why in the conductor sphere voltages aren't the same? or i do mistake to understand this example
 
Physics news on Phys.org
I think you're misunderstanding the example. The form of V -> -Ez + C only applies outside the sphere. The potential inside the sphere takes a different form. Since the sphere is uncharged, you can work out the potential inside it very simply - like you say, the potential should be the same at all points inside the sphere. And since it has been set to 0 at the boundary in the problem, this means that it has to be 0 everywhere inside the sphere.
 
This problem belongs in the Advanced Physics forum IMO.

The solution to Laplace's equation del2V = 0 for this case (azimuthal symmetry) involves Legendre polynomials.

You can obtain a closed-form expression for the potential everywhere outside the sphere including just outside its surface with the given boundary conditions and using just the 1st order polynomial in spherical coordinate θ.

As naz93 said, the potential inside the sphere is everywhere the same (call it zero). This is a very elementary fact of electrostatics. Any body with finite conductivity will have zero E field inside it, thus the potential does not vary inside of it.
 
Last edited:
The problem can be solved without full Legendre polynomials. As a hint, consider an alternate problem of the same uniform electric background field but with a lone +Z dipole at the origin. Find the potential everywhere in space. Is there some dipole magnitude such that the net potential on the unit sphere is a constant? What does that tell you about your sphere problem?
 
Fernbauer said:
The problem can be solved without full Legendre polynomials. As a hint, consider an alternate problem of the same uniform electric background field but with a lone +Z dipole at the origin. Find the potential everywhere in space. Is there some dipole magnitude such that the net potential on the unit sphere is a constant? What does that tell you about your sphere problem?

Dipole? Quadrupole I can see.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top