Question

An infinitely long round dielectric cylinder is polarized uniformly and statically, the polarization $$P$$ being perpendicular to the axis of the cylinder. Find the electric field strength $$E$$ inside the dielectric.

Answer 1

This is to be handled by the same trick. We have effectively a two dimensional situation. For a uniform cylinder full of charge with charge density $$\rho_{0}$$ (charge per unit volume), the electric field $$E$$ at an inside point is along the (cylindrical) radius vector $$\vec {r}$$ and equal to,

$$\vec {E} = \dfrac {1}{2\epsilon_{0}} \rho \vec {r}$$

$$\left (div \vec {E} = \dfrac {l}{r} \dfrac {\partial}{\partial r} (rE_{r}) = \dfrac {\rho}{\epsilon_{0}}, hence, E_{r} = \dfrac {\rho}{2\epsilon_{0}}r\right )$$

Therefore the polarized cylinder can be thought of as two equal and opposite charge distributions displaced with respect to each other

$$\vec {E} = \dfrac {1}{2\epsilon_{0}} \rho \vec {r} - \dfrac {1}{2\epsilon_{0}} \rho (\vec {r} - \delta \vec {r}) = \dfrac {1}{2\epsilon_{0}} \rho \delta \vec {r} = -\dfrac {\vec {P}}{2\epsilon_{0}}$$

Since $$\vec {P} = -\rho \delta \vec {r}$$. (direction of electric dipole moment vector being from the negative charge to positive charge.)