Answer:
a) [tex]C=4.92\times10^{-9}F [/tex]
b) [tex]Q=6.88\times10^{-8}C[/tex]
c) [tex]E=15555.6V/m[/tex]
Explanation:
a) The capacitance of a parallel-plate capacitor with plates of area A separated a distance d with a dielectric of dielectric constant k is given by the formula:
[tex]C=\frac{k\epsilon_0A}{d}[/tex]
where [tex]\epsilon_0=8.85\times10^-12F/m[/tex] is the permittivity.
We use our values in S.I.:
[tex]C=\frac{k\epsilon_0A}{d}=\frac{(5)(8.85\times10^-12F/m)(0.1m^2)}{(0.0009m)}=4.92\times10^{-9}F [/tex]
b) The charge stored Q on a capacitor of capacitance C connected to a voltage V is given by the formula Q=CV.
Using our values:
[tex]Q=CV=(4.92\times10^{-9}F)(14V)=6.88\times10^{-8}C[/tex]
c) The electric field E between plates separated a distance d where a voltage V is applied is obtained with the formula [tex]E=\frac{V}{d}[/tex].
We use our values:
[tex]E=\frac{V}{d}=\frac{14V}{0.0009m}=15555.6V/m[/tex]
