How to Find the Potential from a Surface Charge Density?

  • Thread starter Thread starter quasar_4
  • Start date Start date
  • Tags Tags
    Electrostatics
Click For Summary
SUMMARY

The discussion focuses on finding the electric potential generated by a surface charge density defined as \(\sigma(x,y) = \sigma_0 \cos{(ax+by)}\) on the z=0 plane. The initial approach involved a direct integration of the potential formula, which proved complex. Participants suggested using variable substitution to simplify the integration process and emphasized the importance of understanding the relationship between the surface charge density and the resulting electric field. The conversation highlighted the need for a clear understanding of the physical implications of non-uniform charge distributions.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with surface charge density concepts
  • Proficiency in multivariable calculus, particularly integration techniques
  • Knowledge of variable substitution methods in calculus
NEXT STEPS
  • Study the method of separation of variables in electrostatics
  • Learn about variable substitution techniques in double integrals
  • Explore the implications of non-uniform surface charge distributions on electric fields
  • Review the derivation of electric potential from surface charge densities
USEFUL FOR

Students and professionals in physics, particularly those focusing on electrostatics, as well as educators seeking to deepen their understanding of electric potential calculations involving surface charge densities.

quasar_4
Messages
273
Reaction score
0

Homework Statement



The z=0 plane has a surface charge density \sigma(x,y) = \sigma0 \cos{(ax+by)}. Find the potential everywhere in space.

Homework Equations


The Attempt at a Solution



Ok, so I tried to just integrate directly:

\Phi = \frac{1}{4 \pi \epsilon0} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{\sigma(x,y)}{x^2 + y^2 + z^2} dx dy

but this proves to be formidable by hand (it's a sample exam problem, so nothing but brain power can be used to integrate). My guess is this isn't the best approach to take. Can I try just using separation of variables? How do I do this knowing sigma, and not V, on the plane?
 
Physics news on Phys.org
Hi quasar_4! :smile:

have you tried simplifying the maths by changing variables, to X = (ax+by/√(a2+b2), Y = (bx-ay/√(a2+b2)) ?
 
You integral is not quite right. Your observation point has coordinates which are not the same variables as the source location used as the dummy differential variables.
 
Hm, ok, let me take a step back.

So if I understand physically, the surface charge density should induce an electric field. If the surface charge density were uniform, the field would be all in the z-direction by symmetry. I'm having a hard time picturing in my mind what's happening with the surface charge density not being uniform... seems that I can't easily just do symmetry argument here.

Here's what I *was* thinking: If I want to find the potential at some point P above the z=0 plane, I can use the vector v pointing from the origin to P (I figured if the plane's infinite, I can put the origin wherever I want) and with magnitude sqrt(x^2+y^2+z^2). Then I should be able to integrate the thing above, though I guess I should have written all the x and ys as primed variables and integrated over primed variables only. How is the integral wrong? I don't quite see it...

tiny-tim (hi!), you're right, I shouldn't have given up without trying a substitution - I just assumed automatically that it looked more complicated than it should :P
 

Similar threads

Replies
19
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
6
Views
3K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
10K