## College Physics (4th Edition)

We can find the equivalent capacitance when we put two capacitors in parallel: $9.0~pF+9.0~pF = 18.0~pF$ We can find the equivalent capacitance when we put another capacitor in series with these two capacitors: $\frac{1}{C_{eq}} = \frac{1}{18.0~pF}+\frac{1}{9.0~pF}$ $\frac{1}{C_{eq}} = \frac{3}{18.0~pF}$ $C_{eq} = 6.0~pF$ We can put two capacitors in parallel, and then we can put another capacitor in series with these two capacitors.