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

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