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Near the point plane surface of a uniformisotropicdielectricwith permittivityepsilon the electricfieldstrengthin vacumm is equalto E_(0) the vectorE_(0)formingan angle theta with the normalof thedielectric'ssurface (FIg). Assumingthe field to be uniformboth inside and outside the dielectric, Find: (a) the flux of the vector E througha sphere of radius R with centre located at the surface of the dielectric, (b) the circulation of the vector D around the closed path I' og length l (see fig) whose plane isperpendicularto the surface of the dielectricand parallel to the vector E_(0). |
Answer» Solution :From the previous problem, `sigma'= epsilon_(0) (EPSILON - 1)/(epsilon) E_(0) cos THETA`  (a) Then `oint VEC(E) . vec(dS) = (1)/(epsilon_(0)) Q = pi R^(2) E_(0) cos theta (epsilon - 1)/(epsilon)` (b) `oint vec(D) . vec(dl) = (D_(1t) - D_(2t)) t = (epsilon_(0) E_(0) sin theta- epsilon epsilon_(0) E sin theta) = -(e - 1) epsilon_(0) E_(0) l sin theta` |
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