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Two point charges +q and -2qare placed at the vertices B and C of an equilateral triangle ABC of side 'a' as shown in Obtain the expression for The magnitude , (b) The direction of the resultant electric field at the vertex A due to these two charges. (c) |
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Answer» Solution :As shown in electric field at vertex A DUE to charge +q placed at vertex B is ` ""E_B =(1)/( 4 pi in _0) . (q)/( a^(2)) ` along BA and electrical field at vertex A due to CHARGES -2q placed at vertex C is ` "" E_C =(1)/( 4 pi in _0)(2q)/( a^(2)) ` along AC Obviously ` oversetto (E_B) and oversetto (E_C) ` are inclined at an angle ` THETA = 120^(@)` from the another . Hence Magnitude of net electric field at A `"" E= sqrt( E_B^(2) + E_C^(2)+2E_BE_C cos 120 ^(@))= sqrt(E_B^(2) +E_C^(2)-E_B E_C ) ` ` rArr ""E= ( 1)/( 4 pi in _0).(sqrt3q)/( a^(2)) ` (b) If the net electric field `oversetto E ` is directed at angle `beta ` from the direction AC, then ` tan beta = ( E_Bsin 120^(@) )/(E_C+ E_B cos 120 ^(@)) =(E_B.(SQRT3//2))/(E_C-(E_B//2))=(E_B.(sqrt3//2))/(2E_B-(E_B//2))` `rArr "" tan beta = (1)/(sqrt3) or beta= tan ^(-1)((1)/(sqrt3)) =30^(@) `  `<br> <img src=)
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