Answer
a) $C=9.08\times10^{-12}F$
b) $Q=8.17\times10^{-11}C$
c) $E=225\frac{V}{m}$
d) $E_P=3.68\times10^{-10}J$
3) The answer to a, b, and d will change.
Work Step by Step
a) $C=\epsilon_o\frac{A}{d}=(8.85\times10^{-12}\frac{C^2}{N\cdot m^2})\frac{0.0410m^2}{0.04m}$
$=9.08\times10^{-12}F$
b) $Q=CV=(9.08\times10^{-11}F)(9V)=8.17\times10^{-11}C$
c) $E=\frac{V}{d}=\frac{9V}{0.04m}=225\frac{V}{m}$
d) $E_P=\frac{QV}{2}=\frac{(8.17\times10^{-11}C)(9V)}{2}=3.68\times10^{-10}J$
3) If a dielectric is inserted, you increase the dielectric constant, increasing the capacitance. Changing the capacitance will change the charge and changing the charge will change the electric potential. The answer to a, b, and d will change.