InterviewSolution
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InterviewSolution
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A conductor with respectivityrho bounds on a dielectric with permittivity epsilon. At a certain point Aat theconductor's surface the electric displacementequals D the vectorD beingdirectedaway from theconductorand formingan anglealpha with the normalof thesurface. Find the surfacedensityof chargeson theconductor at thepoint Aand the current densityin the conductor in the vicinitity of the same point. |
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Answer» Solution :The dielectric ends in a CONDUCTOR. It is given that on one side (the dielectric side) the electric displacement`D`is as SHOWN. Within the conductor, at any point `A`, there can be no normalcomponenetof electric FIELD. For if there weresuch a field, a currentwill flowtowards despositing chargethere which in turn will set upcounteringelectricfieldcausingthe normalcomponenetto vanish. They by Gauss' theorem, we easily derive `sigma = D_(n) = D cosalpha` where `sigma` in the surface charge density at `A`. The tangential COMPONENET is determind from the circularion theorem `ointvec(E) . d VEC(R ) = 0` It must be continous across the surface of the conductor there is a tangentail electric field of magnitude, `(D sinalpha)/(epsilon epsilon_(0)) at A` This implies a current by Othm's law of `j = (D sinalpha)/(epsilon epsilon_(0) rho)` 
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