Calculate the relative permittivity of the dialectric?

Question

I'm given a question: The charge taken by a capacitor is 0.2 (micro)C when terminal voltage is 62 V. The plate area of the capacitor is 33cm^2 and its dielectric thickness is 0.6 mm.

Permittivity of free space = 8.85 x 10^-12 F/m

I have never learned this nor was I ever taught by my professor, so how do I answer this?

Answer 1

Capacitors with air between the plates have a dielectric constant, ε0, (the permittivity of free space). Once the space between the plates if filled with a dielectric the permittivity changes and the dielectric has a relative permittivity or dielectric constant, ε, = Κε₀.

Using the standard equations for parallel plate capacitors we have:

C=Q/V=ε(A/d)

solving for ε

ε=(Q/V)*(d/A)

(Q/V)=[(0.2x10^-6C)/(62V)] = 3.23x10^-9 F

(d/A)=[(0.6x10^-3m)/(33x(10^-2)²m²] = .182 m^-1

ε=(Q/V)*(d/A)= 3.23x10^-9 F*1.82 m^-1 = 5.9x10^-10 F/m (to two significant figures as given in problem)