Electrostatics
Intermediate
Capacitor
Source: High school physics (Chinese)
Problem
In the circuit of Figure, $C_1 = 0.25\ \mu\mathrm{F}$, $C_2 = 0.15\ \mu\mathrm{F}$, $C_3 = 0.20\ \mu\mathrm{F}$. $C_2$ and $C_3$ are connected in parallel with each other, and that parallel combination is in series with $C_1$ between terminals $A$ and $B$. The voltage across $C_1$ is $50$ V.
Find the voltage $U_{AB}$ between $A$ and $B$.
$U_{AB} \approx 85.7$ V.
Charge on $C_1$ (the series element):
$$Q = C_1 U_{C_1} = 0.25 \times 10^{-6} \times 50 = 1.25 \times 10^{-5}\ \mathrm{C}.$$The same charge $Q$ flows onto the parallel combination $C_2 \parallel C_3$ of capacitance $C_2 + C_3 = 0.35\ \mu\mathrm{F}$, so the voltage across that parallel block is:
$$U_{23} = \dfrac{Q}{C_2 + C_3} = \dfrac{1.25 \times 10^{-5}}{0.35 \times 10^{-6}} \approx 35.7\ \mathrm{V}.$$Total voltage across the series chain:
$$U_{AB} = U_{C_1} + U_{23} = 50 + 35.7 \approx 85.7\ \mathrm{V}.$$