Answer:
2.121876 mC
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
A = Area = [tex]0.037\times 12\ m^2[/tex]
d = Thickness = 0.0225 mm
E = Dielectric strength = [tex]1\times 10^8\ V/m[/tex]
k = Dielectric constant = 5.4
Capacitance is given by
[tex]C=\dfrac{k\epsilon_0A}{d}\\\Rightarrow C=\dfrac{5.4\times 8.85\times 10^{-12}\times 0.037\times 12}{0.0225\times 10^{-3}}\\\Rightarrow C=9.43056\times 10^{-7}\ F[/tex]
Maximum voltage is given by
[tex]V_m=E_md\\\Rightarrow V_m=1\times 10^8\times 0.0225\times 10^{-3}\\\Rightarrow V_m=2250\ V[/tex]
Maximum charge is given by
[tex]Q_m=CV_m\\\Rightarrow Q_m=9.43056\times 10^{-7}\times 2250\\\Rightarrow Q_m=0.002121876\ C=2.121876\ mC[/tex]
The maximum charge that can be stored in this capacitor is 2.121876 mC
