# Potential of dielectric cylinder with constant polarization

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1. Mar 1, 2017

### eckerm

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
σb=P⋅n

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

3. 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?

2. Mar 1, 2017

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 $P$.)