Internal discharge in disc with 0.1 mm void

Internal discharge occurs if the stress in the void exceeds the dielectric strength of the gas in the void. Since the void and the solid are in series and $D$ is continuous, the stress in the void is $\varepsilon_r$ times the stress in the solid.

Given

$d = 154$ mm $= 15.4$ cm, void $t = 0.1$ mm $= 0.01$ cm, $\varepsilon_r = 10$, $V = 40$ kV, void strength $= 46$ kV/cm.

Solution

$$\begin{aligned} E_a &= \varepsilon_r E_s, \qquad V = E_s(d - t) + E_a t = E_s\left[(d - t) + \varepsilon_r t\right] \\ E_s &= \frac{40}{(15.4 - 0.01) + 10\times0.01} = \frac{40}{15.49} = 2.582\ \text{kV/cm} \\ E_a &= 10\times2.582 = 25.82\ \text{kV/cm} \end{aligned}$$ $$E_a = 25.82\ \text{kV/cm} < 46\ \text{kV/cm}$$

Even taking 40 kV as rms, the peak stress $\sqrt2\times25.82 = 36.5$ kV/cm is still below 46 kV/cm.

Answer: Stress in the void is 25.82 kV/cm, less than 46 kV/cm, so internal discharge will not occur. Discharge would begin only at $V = 46\times15.49/10 = 71.3$ kV.