Question

A cylindrical capacitor is filled with two cylindrical layers of dielectric with permittivities $\varepsilon_1$ and $\varepsilon_2$. The inside radii of the layers are equal to $R_1$ and $R_2 > R_1$. The maximum permissible values of electric field strength are equal to $E_{1m}$ and $E_{2m}$ for these dielectrics. At what relationship between $\varepsilon, R$, and $E_m$ will the voltage increase result in the field strength reaching the breakdown value for both dielectrics simultaneously?

Answer 1

The relationship between $\varepsilon_1, \varepsilon_2, R_1, R_2, E_{1m}$, and $E_{2m}$ for simultaneous breakdown is $\varepsilon_1 R_1 E_{1m} = \varepsilon_2 R_2 E_{2m}$.

Answer 2

The electric field in a cylindrical capacitor decreases with radius ($E \propto 1/r$). The maximum field in each dielectric layer occurs at its inner boundary ($R_1$ for the inner layer, $R_2$ for the outer layer). Breakdown occurs when the maximum field reaches the dielectric strength ($E_{1m}$ or $E_{2m}$). For simultaneous breakdown, the maximum field in both layers must reach their limits at the same charge $Q$. The electric field is proportional to $Q$ and inversely proportional to $\varepsilon$ and $r$. Setting the maximum fields equal to the breakdown strengths ($E_{1m} = \frac{Q}{2\pi \varepsilon_1 R_1}$ and $E_{2m} = \frac{Q}{2\pi \varepsilon_2 R_2}$) and equating the expressions for $Q$ gives the relationship $\varepsilon_1 R_1 E_{1m} = \varepsilon_2 R_2 E_{2m}$.