InterviewSolution
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InterviewSolution
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Showthat the normal componentof electrostaticfield has a discontinuly form one sideof a charged. Surface to another given by (vec(E_(2)) - vec(E_(1))). hat(n) = (sigma)/(in_(0))where hat(n) is a unitvectornormal to the surfaceat a pointand sigma at a pointand sigma is the surface charge density at thatpoint. (The direction of hat(n) is from side 1 to side 2). Hence show that justy outsidea conductor, theelectricfield sigma hat(n)//in_(0). (b) Show that the tangential componetof electrostaticfieldis contionous from one side fo a charged surface to another. |
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Answer» Solution :PROCEEDING as in Art, NORMAL of electricfield intensity due to thin infinitie plane sheetof charge, on left side (side 1) `vec(E)_(1) = - (sigma)/(2in_(0)) hat(n)` and on right side (side 2), `vec(E_(2)) = (sigma)/(2in_(0)) hat(n)` Discotinuityis the normalcomponentfrom ONE sideto the other is `vec(E_(2)) - vec(E_(1)) = (sigma)/(2 in_(0)) hat(n) + (sigma)/(2 in_(0)) hat(n) = (sigma)/(in_(0)) hat(n) or (vec(E_(2)) - vec(E_(1))) hat(n) = (sigma)/(in_(0)) hat(n). hat(n) = (sigma)/(in_(0))` Inside a closedconductor, `vec(E)_(1) = 0``:. E = vec(E_(2)) = (sigma)/(in_(0)) hat(n)` (b) To show that the tangentialcomponentof electrostaticfield is continousfrom one sideof a chargedsurfaceto another, we usethe FACTTHAT work done byelectrostaticfield on a closedloop is zero. |
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