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A long round cylinder made of uniform dielectric is placed in a uniform electric field of strength $$E_{0}$$. The axis of the cylinder is perpendicular to vector $$E_{0}$$. Under these conditions the dielectric becomes polarized uniformly. Making use of the result obtained in the studying problem find the electric field strength $$E$$ in the cylinder and the polarization $$P$$ of the dielectric whose permittivity is equal to $$\epsilon$$.

Solution
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By superposition the field $$\vec {E}$$ inside the ball is given by
$$\vec {E} = \vec {E_{0}} - \dfrac {\vec {P}}{3\epsilon_{0}}$$
On the other hand, if the sphere is not too small, the macroscopic equation $$\vec {P} = (\epsilon - 1) \epsilon_{\theta} \vec {E}$$ must hold. Thus,
$$\vec {E} \left (1 + \dfrac {1}{3} (\epsilon - 1)\right ) = \vec {E_{0}}$$ or, $$\vec {E} = \dfrac {3\vec {E_{0}}}{\epsilon + 2}$$
Also $$\vec {P} = 3\epsilon_{0} \dfrac {\epsilon - 1}{\epsilon + 2} \vec {E}_{0}$$.
We write $$\vec {E} = \vec {E_{0}} - \dfrac {\vec {P}}{2\epsilon_{0}}$$
using here the result of the foregoing problem.
Also $$\vec {P} = (\epsilon - 1) \epsilon_{0} \vec {E}$$
So, $$\vec {E} \left (\dfrac {\epsilon + 1}{2}\right ) = \vec {E}_{0}$$, or, $$\vec {E} = \dfrac {2\vec {E}_{0}}{\epsilon + 1}$$ and $$\vec {P} = 2\epsilon_{0} \dfrac {\epsilon - 1}{\epsilon + 1} \vec {E}_{0}$$.

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