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Two identical point charges ,q each , are kept 2 m apartin air. A third pint charges Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q. |
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Answer» Solution :Let two CHARGES ,q each , are kept at points A and B respectively separated by a distance 2m. Let a third point charges Q be placed at point C SITUATED at a distance x m from A . For equilibrium of system the net force on any charges must be zero. Considering net force on charge Q, we have ` ""oversetto (F_(CA)) +oversetto (F_(CB)) =oversetto 0or |oversetto (F_(CA))|=oversetto (|F_(CB)|) ` ` therefore ""(1)/(4 pi in _0) (qQ)/(x^(2))=(1)/(4pi in _0).(qQ)/((2-x)^(2)) ` `rArr "" x= 2 -x or x=1 m ` Again considering net force on charge q situated at point A, we have ` ""oversetto (F_(AC)) +oversetto (F_(AB)) =oversetto 0 ` `rArr ""(1)/(4 pi in _0).(qQ)/((1)^(2)) +(1)/(4 PIIN _0) .(q^(2))/((2)^(2)) =0 ` `rArr "" Q +(q)/(4) = 0 or Q= -(q)/(4)` 
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