Potential of dielectric cylinder with constant polarization

Click For Summary
SUMMARY

The discussion focuses on calculating the potential V(0,0,h) for a dielectric cylinder with radius R and thickness d, oriented along the z-axis and exhibiting a uniform polarization P=pz. The key equations involve the surface charge density σb derived from the polarization, with σ equating to the polarization charge density σp at the cylinder's end faces. The challenge arises from the absence of sine or cosine terms in the surface charge density, complicating the determination of coefficients An and Bn. The solution involves computing the potential through a surface integral over the end faces, leveraging symmetry to simplify the calculations.

PREREQUISITES
  • Understanding of dielectric materials and polarization concepts
  • Familiarity with surface charge density calculations
  • Knowledge of integral calculus, particularly surface integrals
  • Proficiency in electrostatics, specifically potential calculations
NEXT STEPS
  • Study the derivation of potential for dielectric materials using surface integrals
  • Learn about the implications of uniform polarization in dielectric cylinders
  • Explore the mathematical treatment of surface charge densities in electrostatics
  • Investigate the role of symmetry in simplifying potential calculations in electrostatics
USEFUL FOR

Students and professionals in physics and engineering, particularly those specializing in electromagnetism, electrostatics, and materials science, will benefit from this discussion.

eckerm
Messages
5
Reaction score
0

Homework Statement


The dielectric cylinder is radius R and thickness d. Origin is at the center of the cylinder, which is oriented along the z-axis. It has polarization P=pzI need to calculate the potential V(0,0,h) at h>d/2.

Homework Equations


σb=P⋅n

Σ((-nR2nAn/Rn+1)-nRn-1An)sin(nφ)+Σ((-nR2nBn/Rn+1)-nRn-1Bn)cos(nφ)=-σε0

The Attempt at a Solution


I've actually done a lot of work to reach the equation above. In example problems, such as with a sphere, σ ends up with a cosφ term that can be used to set their coefficients equal to each other and solve for the An and Bn terms. In this problem, however, the normal unit vector n is parallel with the direction of polarization so σ=p. Since σ has no sin or cos terms, I can't solve for An and Bn. What am I supposed to do, or more likely, what did I do wrong?
 
Physics news on Phys.org
This is not a Legendre type problem, (unlike your problem of last week.) Instead, you get a uniform polarization charge density of ## \sigma_p=+P ## on one endface at ## z=+d/2 ## ,and a uniform polarization charge density on the other endface of ## \sigma_p=-P ## at ## z=-d/2 ##.(Note ## \sigma_p=P \cdot \hat{n} ##.) Computing the potential is simply the integral of ## V(x)=\int \frac{\rho(x')}{4 \pi \epsilon_o |x-x'|} \, d^3 x' ## where it will then become a surface integral over the two endfaces. You are only computing the potential at a point along the z-axis, so with symmetry, I think the (surface) integrals might be workable. (Note: For this problem, the polarization is considered to be spontaneous and there are no externally applied electric fields creating the polarization ## P ##.)
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 37 ·
2
Replies
37
Views
12K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
3K
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K