Two parallel plate capacitors,C₁ and C₂ are connected in series to a 9.0-V battery.If capacitor C₁ has a capacitance of 10 nF and capacitor C₂ has a capacitance of 40 nF,what is the magnitude of the charge on each plate of capacitor C₂?

Question

Quizplus · Accepted Answer

In series, capacitors have the same charge. The equivalent capacitance $C_{eq}$ of capacitors in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$. For $C_1 = 10$ nF and $C_2 = 40$ nF, $C_{eq} = \frac{1}{\frac{1}{10} + \frac{1}{40}} = \frac{1}{\frac{1}{10} + \frac{1}{40}} = \frac{1}{\frac{5}{40}} = 8$ nF. The charge $Q$ is given by $Q = C_{eq}V = 8 \times 10^{-9} F \times 9 V = 72 \times 10^{-9} C = 72$ nC.

Two Parallel Plate Capacitors,C1 and C2 Are Connected in Series

Two Parallel Plate Capacitors,C₁ and C₂ Are Connected in Series