InterviewSolution
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InterviewSolution
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Show that the normal component of electrostatic field has a discontinuly form one side of a charged. Surface to another given by `(vec(E_(2)) - vec(E_(1))). hat(n) = (sigma)/(in_(0))` where `hat(n)` is a unit vector normal to the surface at a point and `sigma` at a point and `sigma` is the surface charge density at that point. (The direction of `hat(n)` is from side 1 to side 2). Hence show that justy outside a conductor, the electric field `sigma hat(n)//in_(0)`. (b) Show that the tangential componet of electrostatic field is contionous from one side fo a charged surface to another. |
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Answer» Proceeding as in Art, normal of electric field intensity due to thin infinitie plane sheet of 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)` Discotinuity is the normal component from one side to 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 closed conductor, `vec(E)_(1) = 0` `:. E = vec(E_(2)) = (sigma)/(in_(0)) hat(n)` (b) To show that the tangential component of electrostatic field is continous from one side of a charged surface to another, we use the fact that work done by electrostatic field on a closed loop is zero. |
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