Capacitor with a Dielectric — Question 5

Question 5

A partial-thickness insertion can identify an unknown permittivity

A dielectric slab of thickness t=.600d fills the plate area but leaves vacuum in the rest of the gap. The measured capacitance doubles relative to the empty capacitor C0. Neglect fringing.

1. Derive the effective-gap expression for capacitance.

2. Infer the dielectric relative permittivity and check that it is physically consistent with the measured ratio.

Question 5 - Solution

A partial-thickness insertion can identify an unknown permittivity

(a) Method and calculation

$$V=\frac{Q}{\epsilon_0A}\left(d-t+\frac t\kappa\right).$$$$\frac C{C_0}=\frac d{d-t+t/\kappa}=\frac1{1-f+f/\kappa},\quad f=t/d.$$

(b) Interpretation and further calculation

$$\frac12=.400+\frac{.600}{\kappa}\Rightarrow\kappa=6.00.$$$$1<C/C_0<1/(1-f)=2.50,$$so the observed ratio 2 lies in the finite-positive-permittivity range for $\kappa>1$.

Physical check

A slab with arbitrarily high permittivity cannot remove the voltage drop across the remaining vacuum.