(a) 7.18
The electric field within a parallel plate capacitor with dielectric is given by:
[tex]E=\frac{\sigma}{k \epsilon_0}[/tex] (1)
where
[tex]\sigma[/tex] is the surface charge density
k is the dielectric constant
[tex]\epsilon_0[/tex] is the vacuum permittivity
The area of the plates in this capacitor is
[tex]A=100 cm^2 = 100\cdot 10^{-4} m^2[/tex]
while the charge is
[tex]Q=8.9\cdot 10^{-7}C[/tex]
So the surface charge density is
[tex]\sigma = \frac{Q}{A}=\frac{8.9\cdot 10^{-7} C}{100\cdot 10^{-4} m^2}=8.9\cdot 10^{-5} C/m^2[/tex]
The electric field is
[tex]E=1.4\cdot 10^6 V/m[/tex]
So we can re-arrange eq.(1) to find k:
[tex]k=\frac{\sigma}{E \epsilon_0}=\frac{8.9\cdot 10^{-5} C/m^2}{(1.4\cdot 10^6 V/m)(8.85\cdot 10^{-12} F/m)}=7.18[/tex]
(b) [tex]7.66\cdot 10^{-7}C[/tex]
The surface charge density induced on each dielectric surface is given by
[tex]\sigma' = \sigma (1-\frac{1}{k})[/tex]
where
[tex]\sigma=8.9\cdot 10^{-5} C/m^2[/tex] is the initial charge density
k = 7.18 is the dielectric constant
Substituting,
[tex]\sigma' = (8.9\cdot 10^{-5} C/m^2) (1-\frac{1}{7.18})=7.66\cdot 10^{5} C/m^2[/tex]
And by multiplying by the area, we find the charge induced on each surface:
[tex]Q' = \sigma' A = (7.66\cdot 10^{-5} C/m^2)(100 \cdot 10^{-4}m^2)=7.66\cdot 10^{-7}C[/tex]
