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(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E_(2)-E_(1)).n = sigma/epsilon_(0) where hatn is a unit vector normal to the surface at a point and sigmais the surface charge density at that point. (The direction of hatnis from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is sigma hatn //epsilon_(0). (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.] |
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Answer» Solution :Electric field NEAR the surface of a charged plane PLATE, `E = (sigma)/(2in_(0))` If electric field is considered to be vector then the unit vector `hatn` is one side and - `hatn` is other side of plane.  Electric field from side (1) to side (2), `vecE_(2)= (sigma)/(2in_(0)) . hatn` Electric field from s.ide (2) to side (1), `vecE_(1)=-(sigma)/(2 in_(0)). hatn` Electric field to side (1) `(vecE_(2)-vecE_(1))=[(sigma)/(2in_(0))-(-(sigma)/(2in_(0)))]hatn` `= (2sigma)/(2 in_(0))hatn` `= (sigma)/(in_(0)) hatn` where `sigma` is the surface charge density and unit vector `hatn` is from side (1) to side (2), Since `E_(1)` and `E_(2)` are opposite to each othe so electric field has discontinuity on plat and the electric field DISAPPEAR inside the conductor hence `vecE_(1) =0` So, electric field outside the conductor `vecE=vecE_(2) = (sigma)/(in_(0))hatn` |
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