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Two identical metallic sphere's having unequal opposite charges are placed at a distance 0.9 m apart in air. After bringing them in contact with each other they are again placed at the same distance apart. Now the force of repulsion between them is 0.025 N. calculate the final charges on each of them. |
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Answer» SOLUTION :As two metallic spheres are idential, after bringing them in contact charges on both will be same i.e., `q_1 =q_2 =q` (say ).As R= 0.9 m and F = 0.025 N. Hence from the relation ` F= ( 1)/( 4pi in _0).(q_1 q_2)/(r^(2)) =(1)/(4pi in _0).(q^(2)) /(r^(2))`, we have ` ""q= sqrt ( 4 pi in _0. F . r^(2)) = sqrt ((0.025xx ( 0.9)^(2))//( 9xx 10 ^(9))) = 15 mu C .` |
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