# Suppose you have a 9.00-V battery, a 2.00 - \mu F capacitor, and a 7.40

Suppose you have a 9.00-V battery, a 2.00 - $$\displaystyle\mu{F}$$ capacitor, and a $$\displaystyle{7.40}-\mu{F}$$ capacitor. (a) Find the charge and energy stored if the capacitors are connected to the battery in series

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Coldst
Identify the unknown:
The charge and energy stored if the capacitors are connected in series
Set Up the Problem:
Capacitance of a serie combination:
$$\displaystyle{\frac{{{1}}}{{{C}_{{S}}}}}={\frac{{{1}}}{{{C}_{{1}}}}}+{\frac{{{1}}}{{{C}_{{2}}}}}+{\frac{{{1}}}{{{C}_{{3}}}}}+\ldots$$
$$\displaystyle{\frac{{{1}}}{{{C}_{{S}}}}}={\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{7.4}}}}$$
$$\displaystyle{C}_{{S}}={\frac{{{2}\times{7.4}}}{{{2}+{7.4}}}}={1.575}\mu{F}={1.575}\times{10}^{{-{6}}}{F}$$
Capacitance of a series combitation is given by:
$$\displaystyle{C}_{{S}}={\frac{{{Q}}}{{{V}}}}$$
Then the charges stored in the series combination is:
$$\displaystyle{Q}={C}_{{S}}{V}$$
Energy stored in the series combination is:
$$\displaystyle{U}_{{C}}={\frac{{{1}}}{{{2}}}}{V}^{{2}}{C}_{{S}}$$
Solve the problem:
$$\displaystyle{Q}={1.575}\times{10}^{{-{6}}}\times{9}={1.42}\times{10}^{{-{5}}}{C}$$
$$\displaystyle{U}_{{C}}={\frac{{{1}}}{{{2}}}}\times{\left({9}\right)}^{{2}}\times{1.575}\times{1}{)}^{{-{6}}}={6.38}\times{10}^{{-{5}}}$$ J