A capacitor having capacitance C is given an initial charge Q . At time t = 0 , it is connected across an uncharged capacitor of the same capacitance via a resistance R . Determine the charge accumulated on the second

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Correct answer: B. (Q)/(2)(1 - e^ -2t/RC )

Let q_1 and q_2 be the charges on the first and second capacitors at time t . By conservation of charge, q_1 + q_2 = Q . Applying Kirchhoff's voltage law in the circuit: (q_1)/(C) - iR - (q_2)/(C) = 0 Since i = (dq_2)/(dt) and q_1 = Q - q_2 , we have: (Q - q_2)/(C) - (q_2)/(C) = R (dq_2)/(dt) (Q - 2q_2)/(C) = R (dq_2)/(dt) (dq_2)/(Q - 2q_2) = (dt)/(RC) Integrating both sides with limits q_2 = 0 at t = 0 to q_2 at time t : ∫_ 0 ^ q_2 (dq_2)/(Q - 2q_2) = ∫_ 0 ^ t (dt)/(RC) -(1)/(2) ( (Q - 2q_2)/(Q) ) = (t)/(RC) ( (Q - 2q_2)/(Q) ) = -(2t)/(RC) Q - 2q_2 = Q e^ -2t/RC q_2 = (Q)/(2)(1 - e^ -2t/RC )

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