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A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hold is (sigma)/(2pi epsilon_(0))hatnwhere hatnis the unit vector in the outward normal direction, and a is the surface charge density near the hole. |
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Answer» Solution :As shown in figure, ASSUME that the hole on the CONDUCTING surface is filled by conductor. According to Gauss.s law, at A on surface, `oint vecE.dvecS =q/epsilon_(0)` `EdS cos0^(@) = (sigmadS)/epsilon_(0)[therefore sigma = q/(ds)]` `therefore E.dS = (sigmadS)/(epsilon_(0))` `therefore vecE = sigma/(epsilon_(0)).hatn` Which is electric field outside the conductor. Above electric field can be written as, `vecE = vecE_(1) + vecE_(2)`..........(2) where `vecE_(1)` = electric field at point `P_0` just outside the conductor (adjacent to point P on the surface) due to CHARGE at point P and `vecE_(2)`=electric field at point P (or at `P_0`) flue to remaining charges on the given charged hollow conductor. When hollow charged conductor does not have any hole at point P on the surface we have `vecE_(1)`and `E_2` POINTING along `hatn`and so from equation (2), we can write, `E = E_(1) + E_(2) = sigma/(epsilon_(0))`...........(3) Now consider point `P_(p)` , adjacent to point P and just INSIDE the conductor in figure (a), where resultant electric field `vecE` is `vec0`. Hence, `vecE = vecE_(1) + vecE_(2) = vec0` `therefore vecE_(1) = - vecE_(2)` `therefore vecE_(1) = E_(2)` (Taking magnitude).......(4) From eqn. (3) and (4) `2E_(2) = sigma/(epsilon_(0))` `therefore E_(2) = sigma/(2epsilon_(0))`..........(5) Now, when we have a hole at point P electric field at point P,- in the hole due to remaining charges will be, `E_(2) = sigma/(2epsilon_(0))hatn`.............(6) |
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