Capacitors in Parallel and in Series — Question 2

Question 2

Series equality of charge comes from a neutral floating junction

C1=3.00 nF and C2=6.00 nF are in series across VA-VB=18.0 V. The internal junction M was neutral and remains isolated.

1. Use junction charge conservation to derive the equivalent capacitance.

2. Find the charge magnitude, both voltage drops and VM if VB=0.

Question 2 - Solution

Series equality of charge comes from a neutral floating junction

(a) Method and calculation

$$-Q_1+Q_2=0\Rightarrow Q_1=Q_2=Q.$$$$V=Q/C_1+Q/C_2\Rightarrow C_{\mathrm{eq}}=\frac{C_1C_2}{C_1+C_2}=2.00\,\mathrm{nF}.$$

(b) Interpretation and further calculation

$$Q=C_{\mathrm{eq}}V=36.0\,\mathrm{nC}.$$$$V_A-V_M=12.0\,\mathrm V,\quad V_M-V_B=6.00\,\mathrm V,\quad V_M=6.00\,\mathrm V.$$

Physical check

The smaller capacitance carries the larger voltage when the series charge magnitudes are equal.