InterviewSolution
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InterviewSolution
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A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is((sigma )/(2in_0))hatn, where hatnis the unit vector in the outward normal direction, andsigmais the surface charge density near the hole, |
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Answer» Solution :Let us consider a hollow charged conducting BODY, P having a tiny hole H cut into its surface, If `sigma ` be the surface density of charge near the hole , then we know that the electricfield at a point B outside the conductor and near the hole =`oversetto E= (sigma )/(in_0)hatn, ` and electric field at a point A inside the hollow conductor =0. Obviously electric field at point A as well as B may be considered as the VECTOR sum of electric field due to the ramaining conductor `oversetto (E_1)`and electric field due to hole portion of the conductor ` oversetto (E_2) ` As shown in at point A, we have ` |oversetto(E)| =|oversetto (E_1)+ oversetto(E_2)| =oversetto(E_1)-oversetto(E_2)=0 and ` it leads to ` |oversetto(E_1) = |oversetto(E_2)|` Again at point B, `|oversetto(E_1)=|oversetto(E_1) +oversetto(E_2) | =oversettoE_1+oversetto(E_2)=2 E_1 =(sigma )/(in_0) ` ` rArr ""oversetto(E_1) = ( sigma )/( 2in _0) hatn , `where `hatn `is unit vector in the outwordnormal direction. ` (##U_LIK_SP_PHY_XII_C01_E02_005_S01.png" width="80%"> |
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