Potential of dielectric cylinder with constant polarization

  • #1
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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?
 

Answers and Replies

  • #2
Charles Link
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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 ##.)
 
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