Answer to Question #84115 in Electricity and Magnetism for Nads

Question

The relative permittivity of a dielectric is 2.25. (a) What is its permittivity? (b) What will be the capacitance if this dielectric is inserted into a 7.8 uF air capacitor?

Answer 1

a) The relative permittivity

ε_r=ε/ε_0 ,

where ε is the absolute permittivity; ε_0 is the vacuum permittivity, ε_0=8.85∙〖10〗^(-12) F/m. We have

ε=ε_r∙ε_0=2.25∙8.85∙〖10〗^(-12)=19.91∙〖10〗^(-12) F/m.

b) The capacitance of a capacitor with this dielectric

ε_r=C/C_0 → C=ε_r∙C_0=2.25∙7.8=17.55 μF.