Problem $6\mathrm{.}5\mathrm{.}2$: Show that the capacitance defined as $C=\frac{Q}{}V$ of a parallel plate capacitor $($neglecting fringing field effects$)$ is given by $C=\frac{A}{d},$ where $$ is the permittivity of the dielectric, $A$ is the area of the plates and $d$ is the plate separation, and calculate the capacitance of

a parallel plate capacitor with dielectric constant, ${}_r=100,$ plate area $A=1m^2$ and separation, $d={10}^{-3}m.$