Answer
a) $6.4\times10^{-11}C$
b) $6.4\times10^{-11}C$
c) $18V$
d) $1.9\times10^{-10}J$
Work Step by Step
a) $Q=CV=(8.85\times10^{12}C^2/Nm^2)(\frac{3.0\times10^{-4}m^2}{0.5\times10^{-3}m})(12V)=6.4\times10^{-11}C$
b) After the capacitor is charged, moving the plates will not have an impact on its charge. $6.4\times10^{-11}C$
c) $C=(8.85\times10^{12}C^2/Nm^2)(\frac{3.0\times10^{-4}m^2}{0.75\times10^{-3}m})$$=3.54\times10^{-12}$
$V=\frac{Q}{C}=\frac{6.4\times10^{-11}C}{3.54\times10^{-12}}=18V$
d) $W=\frac{QV_{ba}}{2}=\frac{(6.4\times10^{-11}C)(18V-12V)}{2}=1.9\times10^{-10}J$