Question

There is a double-layer cylindrical capacitor whose parameters are shown in Fig. 3.16. The breakdown field strength values for these dielectrics are equal to $E_1$ and $E_2$ respectively. What is the breakdown voltage of this capacitor if $\epsilon_1 R_1 E_1 < \epsilon_2 R_2 E_2$?

Answer 1

$R_1 E_1 \ln\frac{R_2}{R_1} + \frac{\epsilon_1 R_1 E_1}{\epsilon_2} \ln\frac{R_3}{R_2}$

Answer 2

The electric field at a distance $r$ from the axis in the region $R_1 < r < R_2$ is given by $E_1(r) = \frac{\lambda}{2\pi \epsilon_1 r}$, where $\lambda$ is the charge per unit length on the inner conductor. The maximum electric field in this region is at $r=R_1$, $E_{1,max} = \frac{\lambda}{2\pi \epsilon_1 R_1}$. Breakdown occurs in this layer when $E_{1,max} = E_1$, which corresponds to a charge per unit length $\lambda_{c1} = 2\pi \epsilon_1 R_1 E_1$.

The electric field at a distance $r$ from the axis in the region $R_2 < r < R_3$ is given by $E_2(r) = \frac{\lambda}{2\pi \epsilon_2 r}$. The maximum electric field in this region is at $r=R_2$, $E_{2,max} = \frac{\lambda}{2\pi \epsilon_2 R_2}$. Breakdown occurs in this layer when $E_{2,max} = E_2$, which corresponds to a charge per unit length $\lambda_{c2} = 2\pi \epsilon_2 R_2 E_2$.

The breakdown of the capacitor occurs when the applied voltage results in a charge per unit length $\lambda$ that reaches the minimum of $\lambda_{c1}$ and $\lambda_{c2}$. Given the condition $\epsilon_1 R_1 E_1 < \epsilon_2 R_2 E_2$, we have $2\pi \epsilon_1 R_1 E_1 < 2\pi \epsilon_2 R_2 E_2$, so $\lambda_{c1} < \lambda_{c2}$. Therefore, the breakdown occurs when $\lambda = \lambda_{max} = \lambda_{c1} = 2\pi \epsilon_1 R_1 E_1$.

The potential difference between the inner and outer conductors is given by:

$V = \int_{R_1}^{R_3} E(r) dr = \int_{R_1}^{R_2} E_1(r) dr + \int_{R_2}^{R_3} E_2(r) dr$

$V = \int_{R_1}^{R_2} \frac{\lambda}{2\pi \epsilon_1 r} dr + \int_{R_2}^{R_3} \frac{\lambda}{2\pi \epsilon_2 r} dr$

$V = \frac{\lambda}{2\pi \epsilon_1} [\ln r]_{R_1}^{R_2} + \frac{\lambda}{2\pi \epsilon_2} [\ln r]_{R_2}^{R_3}$

$V = \frac{\lambda}{2\pi} \left( \frac{1}{\epsilon_1} \ln(\frac{R_2}{R_1}) + \frac{1}{\epsilon_2} \ln(\frac{R_3}{R_2}) \right)$

The breakdown voltage is the voltage when $\lambda = \lambda_{max} = 2\pi \epsilon_1 R_1 E_1$.

$V_{breakdown} = \frac{2\pi \epsilon_1 R_1 E_1}{2\pi} \left( \frac{1}{\epsilon_1} \ln(\frac{R_2}{R_1}) + \frac{1}{\epsilon_2} \ln(\frac{R_3}{R_2}) \right)$

$V_{breakdown} = \epsilon_1 R_1 E_1 \left( \frac{1}{\epsilon_1} \ln(\frac{R_2}{R_1}) + \frac{1}{\epsilon_2} \ln(\frac{R_3}{R_2}) \right)$

$V_{breakdown} = R_1 E_1 \ln(\frac{R_2}{R_1}) + \frac{\epsilon_1 R_1 E_1}{\epsilon_2} \ln(\frac{R_3}{R_2})$