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Suppose the sphere A and B in Question 1.12 have identical sizes.A third sphere of the same size but uncharged is brought in contact with the first ,then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B? |
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Answer» Solution :Since all the spheres are of same SIZE, their capacities are equaland they EQUALLY share their charges when brough in contact. On bringing 3rd sphere C in contact with ` A, q_C =q_A. =(q_A)/(2)= ( 6.5 xx10^(-7))/(2)C.` Then bringing C in contact with the B , we have ` q_C. =q_B. =(q_C+q_B) /(2)=(1)/(2)[(6.5xx 10^(-7))/(2) +6.5 xx10 ^(-7) ] =(9.75xx10^(-7))/(2) C ` ` therefore ` New forces of repulsion between A and `F=(1)/( 4pi in _0).(q_A.q_B.)/(r^(2)) =9xx10 ^(9) xx((6.5xx10^(-7))/(2)xx (9.75xx10^(-7))/(2))/((0.5)^(2) ) = 5.7xx10 ^(-3)N` |
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