Cylinder Capacitor – How to Determine Resultant Permittivity of Two Dielectrics

capacitorcharge

Cylinder capacitor

I am trying to calculate the density charge of the inner cylinder in the above picture.
If there was only one dielectric I would use the following equation:

$$
\frac{q_{enc}^f}{L} = \frac{2\pi \varepsilon_0 \varepsilon_r V (r_a)}{\ln (\frac{r_b}{r_a})}
$$

But how do I calculate with \$\varepsilon_{r_{1}}\$ and \$\varepsilon_{r_{2}}\$?

Material with \$\varepsilon_{r_{1}}\$ permittivity is in the area where \$r_1 < r < r_2\$ and \$\varepsilon_{r_{2}}\$ is where \$r_2 < r < r_3\$.

Best Answer

You can reduce this problem to a series combination of two separate cylinder capacitors.

  • One with inner radius \$r_0\$, outer radius \$r_1\$ and permitivity \$\epsilon_{r1}\$ and
  • the other with inner radius \$r_1\$, outer radius \$r_2\$ and permittivity \$\epsilon_{r2}\$.

Since they are in series, charge (per length) will be equal for both capacitors.