Question

A capacitor of capacitance \(C\) is given a charge \(Q\). At \(t = 0\), it is connected to an uncharged capacitor of equal capacitance through a resistance \(R\). The charge on the second capacitor as a function of time is:

• A. \(\dfrac{Q}{2}\left(1 - e^{-\frac{2t}{RC}}\right)\)
• B. \(Q\left(1 - e^{-\frac{2t}{RC}}\right)\)
• C. \(3Q\left(1 - e^{-\frac{2t}{RC}}\right)\)
• D. \(\dfrac{Q}{3}\left(1 - e^{-\frac{2t}{RC}}\right)\)

Hint:

Write KVL as \(R\dfrac{dq}{dt} = \dfrac{Q - 2q}{C}\); the final equal share is \(Q/2\) and the time constant is \(RC/2\).

Answer 1

The Correct Option is A

Step 1: Set up the charge variables. Let the second capacitor carry charge \(q\) at time \(t\). By charge conservation the first capacitor holds \(Q - q\). The voltages are \(V_1 = \dfrac{Q-q}{C}\) and \(V_2 = \dfrac{q}{C}\).

Step 2: Apply Kirchhoff's voltage law around the loop. The current is \(I = \dfrac{dq}{dt}\):
\[\frac{Q-q}{C} - \frac{q}{C} = IR = R\frac{dq}{dt}.\]
Thus
\[R\frac{dq}{dt} = \frac{Q - 2q}{C}.\]
Step 3: Separate variables and integrate with \(q(0) = 0\):
\[\int_0^{q}\frac{dq}{Q - 2q} = \int_0^{t}\frac{dt}{RC}.\]
This gives \(-\tfrac{1}{2}\ln\!\dfrac{Q - 2q}{Q} = \dfrac{t}{RC}\), so \(Q - 2q = Q\,e^{-2t/RC}\).

Step 4: Solve for \(q\):
\[q = \frac{Q}{2}\left(1 - e^{-2t/RC}\right).\]
As \(t \to \infty\), \(q \to Q/2\), the equal final share, as expected.
\[\boxed{q = \dfrac{Q}{2}\left(1 - e^{-\frac{2t}{RC}}\right)}\]