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Show that the electric field at the surface of a charged conductor is given byoversetto E = ( sigma )/( 60 ) hat n, wheresigmais the surface charge density andhatnis a unit vector normal to the surface in the outward direction. |
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Answer» Solution :Consider a charged conductor whose surface charge density is ` sigma . ` To derive electric field at its surface consider a short cylinder (pill box)as the Gaussian surface about a given point . The cylinder is partly inside and PARTY outside the surface of the conductor. It has a small area of cross section ` delta S ` and NEGLIGIBLE height. Just inside the surface , the electric field is ZERO but just outside the field has a magnitudeE and is directed normal to the surface. Thus , the contribution to the electric flux comes only from the outside circular cross-section of the cylinder. ` therefore phi_in =oversetto E . oversetto (delta S) = E.delta S ` As per Gauss theorem. ` phi _in =(1)/( in_0)` (charged close)`=(1)/( in_0).(delta sigma S) ""...(II) ` Comparing (i) and (ii) we get `E =( sigma)/( in_0)rArr ""oversetto E =(sigma)/( in_0)hatn ` 
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