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A long round cylinderical 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)`. The axis of the cylinder is perpendiucular to vector `E_(0)`. Under these conditions the dielectrics becomes polarized unifromly. making use of the result obtained in the foregoing probem, find the electric field strength `E` is the cylinder and the polarization `P` of the distance whose permittivity is equal to `epsilon`. |
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Answer» As in 3.98, we wire `vec(E) = vec(E) = vec(E_(0)) - (vec(P))/(2epsilon_(0))` using being the result of the foregoing problem. Also, `vec(P) = (epsilon - 1) epsilon_(0) vec(E)` So, `vec(E) ((epsilon + 1)/(2)) = vec(E_(0))`, or, `vec(E) = (2 vec(E_(0)))/(epsilon + 1)` and `vec(P) = 2epsilon_(0) (epsilon - 1)/(epsilon + 1) vec(E_(0))` |
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