Answer
a) $E_{Pi}=2.66\times10^{-4}J$
b) $E_P=1.13\times10^{-4}J$
c) $\Delta E=-1.53\times10^{-4}J$
Work Step by Step
a) $E_{P1}=\frac{CV^2}{2}=\frac{(3.70\mu F)(12.0V)^2}{2}=2.66\times10^{-4}J$
$E_{P2}=0J$
$E_{Pi}=2.66\times10^{-4}J$
b) After they are connected, they will have the same voltage.
$Q_{1f}+Q_{2f}=Q_1$
$C_1V_f+C_2V_f=C_1V_1$
$V_f=\frac{C_1V_1}{C_1+C_2}=\frac{(3.70\mu F)(12.0V)}{3.70\mu F+5.00\mu F}=5.10V$
$E_P=\frac{C_1V_f^2}{2}+\frac{C_2V_f^2}{2}=\frac{(3.70\mu F)(5.10V)^2}{2}+\frac{(5.00\mu F)(5.10V)^2}{2}=1.13\times10^{-4}J$
c) $\Delta E=1.13\times10^{-4}J-2.66\times10^{-4}J=-1.53\times10^{-4}J$
