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Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but unchanged 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 : INITIALLY Coulombian force exerted between `q_(A)` and `q_(B)` `F = K(q_(A).q_(B))/r^(2)`……(1) Final charges on A and C, `q_(A)^(.) = q_(C )^(.) = (q_(A) + 0)/2 = q_(A)/2` Most final charges on C and B, `q_(C)^(..) = q_(B)^(.) = (q_( C)^(.) + q_(B))/2 = (q_(A)/2 + q_(B))/2 = (q_(A) + 2q_(B))/4`........(2) Final Coulombian force between A and B, `F. k(q_(A)^(.).q_(B)^(.))/r^(2)` (If SEPARATION between A and B remains same) `THEREFORE F. = k(q_(A)/2)(q_(A) + 2q_(B))/4)/r^(2)` `therefore F. = 1/8 xx k xx (q_(A))(q_(A) + 2q_(B))/r^(2)` `=1/8 xx k (q_(A))(3q_(B))/r^(2), (therefore q_(A) = q_(B))` `=3/8 k (q_(A).q_(B))/r^(2)` `=3/8 F` `=3/8 xx 1.521 xx 10^(-2)` `therefore F. = 5.704 xx 10^(-3) N` (Repulsive) |
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