- #1
stunner5000pt
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Homework Statement
Griffiths Problem 4.13
A very long cylinder, of radius a, carries a uniform polarization P perpendicular to its axis. Find the electric field inside the cylinder. Show that the field outside the cylinder can be expressed in the form
[tex] \vec{E}(\vec{r}) = \frac{a^2}{2\epsilon_{0}s^2} \left[2(\vec{P}\cdot\vec{\hat{s}})\hat{s}-\vec{P}\right][/tex]
2. The attempt at a solution
Now i was wondering if i could use Laplace's equation here... since there is not volume charge.
If i were to use this equation to solve for the potnetial
[tex]V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \oint_{S} \frac{\sigma_{b}}{r} da' + \int_{V} \frac{\rho_{b}}{r} d\tau'[/tex]
Suppose we were solving for points inside then bound surface charge is zero and the volume charge cannot be zero
SInce the polarization is perpendicular to the axis, then the polarization is +P for theta from 0 to pi
but for theta from pi to 2pi then the polarization is -P correct??
But how can the bound volume charge be non zero if the polarization is constant??