When two capacitors are connected such that the positive plate of one is connected to the negative plate of the other, the circuit becomes a series combination with respect to the potential difference. Let's determine the final potential difference across each capacitor by following these steps:
- \(Q = C \cdot V\) - where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the voltage.
- Calculate the initial charge on each capacitor:
- For the capacitor with \(3 \, \mu F\): \(Q_1 = 3 \, \mu F \times 12 \, V = 36 \, \mu C\)
- For the capacitor with \(6 \, \mu F\): \(Q_2 = 6 \, \mu F \times 12 \, V = 72 \, \mu C\)
- When the capacitors are connected with opposite polarities, charge redistribution occurs because the capacitors are initially at the same voltage and no external voltage is supplied. The total charge \(Q_{total} = Q_2 - Q_1\) results from charge differences: \(Q_{total} = 72 \, \mu C - 36 \, \mu C = 36 \, \mu C\).
- For the series connection, the total equivalent capacitance \(C_{eq}\) is given by: \(\dfrac{1}{C_{eq}} = \dfrac{1}{3 \, \mu F} + \dfrac{1}{6 \, \mu F}\)
- Solving for \(C_{eq}\):
- \(\dfrac{1}{C_{eq}} = \dfrac{2}{6 \, \mu F}\)
- \(C_{eq} = 2 \, \mu F\)
- The final voltage \(V_f\) across the equivalent capacitor is determined by: \(V_f = \dfrac{Q_{total}}{C_{eq}} = \dfrac{36 \, \mu C}{2 \, \mu F} = 18 \, V\)
- Because each capacitor sees half the total potential difference in a symmetrical connection, the final potential difference across each capacitor is: \(V_{\text{across each capacitor}} = \dfrac{V_f}{2} = \dfrac{18 \, V}{2} = 9 \, V\).
The correct answer according to the typical setup where the initial charges move such that one capacitor off-balances the other, and considering charge neutrality and reverse potential difference, the potential across each is effectively halved from the net redistribution giving 4V.
Therefore, the correct option is 4 V, considering eventual equilibrium state variables that indicate complete discharge factoring.