Two identical small conducting spheres P and Q carry charges of +8 and -2 respectively. When placed at a certain distance r apart in vacuum, the magnitude of the attractive electrostatic force between them is F . A

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Correct answer: C. (3F)/(4)

Initial charges on the spheres are q_P = +8 and q_Q = -2 . The initial electrostatic force between P and Q is: F = (k |q_P| |q_Q|)/(r^2) = (k (8)(2))/(r^2) = (16k)/(r^2) When the identical sphere R with charge q_R = +4 is brought into contact with P , they share the total charge equally: q_P' = q_R' = (q_P + q_R)/(2) = (8 + 4)/(2) = +6 Next, sphere R (now carrying +6 ) is brought into contact with Q (carrying -2 ). They share the total charge equally: q_Q' = q_R'' = (q_R' + q_Q)/(2) = (6 + (-2))/(2) = +2 The new charges on P and Q are q_P' = +6 and q_Q' = +2 . The new electrostatic force between P and Q at the same distance r is: F' = (k |q_P'| |q_Q'|)/(r^2) = (k (6)(2))/(r^2) = (12k)/(r^2) Taking the ratio of the new force to the initial force: (F')/(F) = ((12k)/(r^2))/((16k)/(r^2)) = (12)/(16) = (3)/(4) F' = (3F)/(4) Answer: (3F)/(4)

Two identical small conducting spheres P and Q carry charges of +8...

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