In the circuit shown, the charge on the left plate of the 20 µF capacitor is \( -50\,\mu C \). The charge on the right plate of the 12 µF capacitor is:
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In series capacitor circuits, the same charge flows through all capacitors. In parallel combinations, the voltage remains constant and charge divides in proportion to capacitance.
The configuration is as follows: a 20 µF capacitor is in series with two parallel branches, one with a 12 µF capacitor and the other with an 8 µF capacitor. These parallel branches then connect in series to a 4 µF capacitor.Because the 20 µF and 4 µF capacitors are in series with the parallel combination of the 12 µF and 8 µF capacitors, the charge across all these series capacitors is identical.Given that \( Q_{\text{20 µF}} = -50\,\mu C \), the total charge across the series equivalent is \( Q_{\text{equivalent}} = 50\,\mu C \).This same charge of \( 50\,\mu C \) flows through the entire series combination. Therefore, the net charge on the parallel combination of the 12 µF and 8 µF capacitors is also \( 50\,\mu C \). In a parallel configuration, charge distributes proportionally to capacitance.The charge on the 12 µF capacitor is calculated as:\[Q_{12\,\mu F} = \frac{12}{12 + 8} \cdot 50 = \frac{12}{20} \cdot 50 = 30\,\mu C\]Given that the left plate of the 20 µF capacitor is negatively charged, current flows from right to left. This implies that the right plate of the 12 µF capacitor (facing towards the right) will acquire an equal and opposite charge:\[Q_{\text{right plate of 12 µF}} = -30\,\mu C\]
