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Two fixed point charges +9e and +4e units are placed on a straight line AB= a at A and B respectively . Where should the third point charge be placed on AB for it to be in equilibrium ? |
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Answer» Solution :For .q. to be in equilibrium, ` F_( qq_1) =F_(qq_2) ` ie. `(1)/( 4pi in _0) (q_1q)/( x^(2)) =(1)/( 4pi in _0) (q_2q)/( (a-x)^(2)) ` ` (9e)/( x^(2)) =( 4E)/((a-x) ^(2)) ` `9( a-x)^(2) =4x^(2) "" 3(a-x) =+- 2x` ` (##JYT_AJP_AIO_PHY_XII_C01_SLV_012_S01##) ` Hence charges q should be placed at a distance of `(3A)/( 5) ` from the charge 9e on AB |
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