Chapter 21 - Electric Current and Direct-Current Circuits - Problems and Conceptual Exercises - Page 758: 69

$1.1V $

We know that $\frac{1}{C_{eq}}=\frac{1}{4.5\mu F}+\frac{1}{12\mu F}+\frac{1}{32\mu F}$ $\frac{1}{C_{eq}}=0.3368$ $\implies C_{eq}=2.97\mu F $ As $ Q=C_{eq}V $ $\implies Q=(2.97\mu F)(12V)$ $ Q=35.64\times 10^{-6}C $ Now $ V_{32\mu F}=\frac{Q}{C_3}$ We plug in the known values to obtain: $ V_{32\mu F }=\frac{35.64\times 10^{-6}C}{32\mu F}$ $ V_{32\mu F}=1.1V $