- #1

stunner5000pt

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Griffith's problem 4.13 page 173

[tex] \vec{E}(\vec{r}) = \frac{a^2}{2 \epsilon_{0} s^2} [2(\vec{P}\bullet\hat{s})\hat{s} - \vec{P}] [/tex]

I m wondering if this is in any way similar to that of a sphere in that we can find the potnetial (from the solution of Laplace's equation) and from there we can find the electric field both inside and outside.

Is there a shorter method, though?

[tex] \sigma_{b} = \vec{P} \bullet \hat{n} [/tex]

[tex] \rho_{b} = - \vec{\nabla} \bullet \vec{P} [/tex]

wel ok

[tex] \sigma_{b} = (P_{x},P_{y},P_{z}) \bullet (n_{x},n_{y},n_{z}) [/tex]

and we know that

[tex] \frac{\partial P}{\partial n} = \vec{P}\bullet\hat{n} [/tex]

but the thing is does this hold??

[tex] \frac{\partial P}{\partial n_{x}} = \frac{\partial P}{\partial x} [/tex]??

**A very long cylinder of radius a carries a uniform polarization P perpendicular to its axis. Find the electric field inside the cylinder. Show taht the field outside the cylinder can be expressed in the form**[tex] \vec{E}(\vec{r}) = \frac{a^2}{2 \epsilon_{0} s^2} [2(\vec{P}\bullet\hat{s})\hat{s} - \vec{P}] [/tex]

I m wondering if this is in any way similar to that of a sphere in that we can find the potnetial (from the solution of Laplace's equation) and from there we can find the electric field both inside and outside.

Is there a shorter method, though?

**problem 4.14****When you polarize a neutral dielectric, charge moves a bit, but the total remains zero. Thi fact should be reflected i nteh bound charges simga b and rho b. Prove from the given equations taht the total bound charge vanishes**[tex] \sigma_{b} = \vec{P} \bullet \hat{n} [/tex]

[tex] \rho_{b} = - \vec{\nabla} \bullet \vec{P} [/tex]

wel ok

[tex] \sigma_{b} = (P_{x},P_{y},P_{z}) \bullet (n_{x},n_{y},n_{z}) [/tex]

and we know that

[tex] \frac{\partial P}{\partial n} = \vec{P}\bullet\hat{n} [/tex]

but the thing is does this hold??

[tex] \frac{\partial P}{\partial n_{x}} = \frac{\partial P}{\partial x} [/tex]??

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