Answer
See answers.
Work Step by Step
a. Use the formula for the equivalent capacitance of capacitors in parallel.
$$C_{eq}= 0.20\mu F +0.45\mu F =0.65\mu F $$
Use the equivalent capacitance to calculate the stored energy.
$$PE=\frac{1}{2}C_{eq}V^2=\frac{1}{2}(0.65\times10^{-6} F)(24V)^2=1.9\times10^{-4}J$$
b. Use the formula for the equivalent capacitance of capacitors in series.
$$C_{eq}= (\frac{1}{0.20\mu F }+\frac{1}{0.45\mu F })^{-1} =0.1385\mu F $$
Use the equivalent capacitance to calculate the stored energy.
$$PE=\frac{1}{2}C_{eq}V^2=\frac{1}{2}(0.1385\times10^{-6} F)(24V)^2=4.0\times10^{-5}J$$
c. Use Q=CV to find the charge.
$$Q_{parallel}=CV=(0.65\times10^{-6}F)(24V)=1.6\times10^{-5}C$$
$$Q_{series}=CV=(0.1385\times10^{-6}F)(24V)=3.3\times10^{-6}C$$