Capacitors in Parallel and in Series — Question 9

Question 9

Opposite initial polarities can reverse one capacitor during parallel reconnection

C1=2.00 nF initially has signed voltage +10.0 V; C2=3.00 nF initially has signed voltage -4.00 V. Like-named terminals are joined with no source attached. Assume transients settle and no charge escapes the two joined nodes.

1. Find the common final voltage.

2. Find both final positive-reference terminal charges and the charge change on capacitor 2.

Question 9 - Solution

Opposite initial polarities can reverse one capacitor during parallel reconnection

(a) Method and calculation

$$Q_{\mathrm{node,i}}=C_1V_1+C_2V_2=20.0-12.0=8.00\,\mathrm{nC}.$$$$V_f=\frac{Q_{\mathrm{node,i}}}{C_1+C_2}=1.60\,\mathrm V.$$

(b) Interpretation and further calculation

$$q_{1,f}=3.20\,\mathrm{nC},\quad q_{2,f}=4.80\,\mathrm{nC}.$$$$\Delta q_2=4.80-(-12.0)=16.8\,\mathrm{nC}.$$

Physical check

The conserved quantity is signed charge on each connected node, not the sum of initial charge magnitudes.