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.
Let’s analyze the configuration:
- The 20 µF capacitor is in series with two parallel branches:
- Top branch: 12 µF
- Bottom branch: 8 µF
These two branches then connect to the 4 µF capacitor.
Since 20 µF and 4 µF are in series with the parallel branch (12 and 8 µF), the charge across all series capacitors must be the same.
Given:
\[
Q_{\text{20 µF}} = -50\,\mu C \Rightarrow Q_{\text{equivalent}} = 50\,\mu C
\]
The same charge flows through the entire series, so the net charge on the equivalent of 12 µF and 8 µF (which are in parallel) must also be \( 50\,\mu C \). In parallel, charge divides proportionally to capacitance.
\[
Q_{12\,\mu F} = \frac{12}{12 + 8} \cdot 50 = \frac{12}{20} \cdot 50 = 30\,\mu C
\]
Since the left plate of the 20 µF is negatively charged, current flows from right to left, meaning the right plate of the 12 µF capacitor (towards right) gets an equal and opposite charge:
\[
Q_{\text{right plate of 12 µF}} = -30\,\mu C
\]
