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(a) Two insulated charged copper spheres a and bhave theircentresseparated by a distance of 50 cm what is the nutual force of electrostatic repulsion if the charge on each is 6.5 xx10^(-7) c the radii of a and b negligible compared to the distance of separation (b) what is the forceof repuslion if each spere is charged double the above amountandf the distance between them is haved |
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Answer» SOLUTION :(a) REQUIRED Coulombian force is, `F = K(q_(A).q_(B))/r^(2)`…..(1) `=(9 xx 10^(9))(6.5 xx 10^(-7))(6.5 xx 10^(-7))/(0.5)^(2)` `F = 1.521 xx 10^(-2) N` (Repulsive because of like charges (b) New Coulombian force, `F. = k(q_(A)^(.).q_(B)^(.))/r^(2)` `=k(2q_(A).2q_(B))/(r/2)^(2)` (As per the statement) `F. = 16 k(q_(A).q_(B))/r^(2)` `therefore F. = 16 F` (From eqn (1)) `therefore F. = 1.6 xx 1.521 xx 10^(-2) = 0.2434 N` (Repulsive) |
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