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Two identical small metallic sphere having arbitrary charges are separated by a distance ' r ' in air. They are then touched by two more identical metal spheres respectivelyand seperated After- wards A and B are brought to r/(2)distance. Show that initial interaction is same as final interaction- |
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Answer» Solution :`F=( 1)/( 4piin _0) (q_1q_2)/( r^(2) ) `where, `q_1` and `q_2`are the arbitary charges in A and Brespectively . As and A and B are touched by identical spherethe charges are RESPECTIVELY reduced by ` (q_1)/( 2 ) and ( q_2)/( 2 ) ` Since they are then placed at `(r)/(2) `the new FORCE is ` F.=(1)/( 4pi in _0) ((q_1)/(2) xx (q_2)/( 2) )/( (r)/(2) ^(2)) = (1)/( 4pi in _0) (q_1q_2)/( r^(2)) ((1//4)/(1//4)) ` ` =(1)/( 4pi in _0) (q_1q_2)/( r^(2))` F, hence result. |
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