Internal discharge in disc with 0.01 mm void

An internal discharge occurs when the electric stress in a gas-filled void inside a solid exceeds the breakdown strength of the gas in the void. Because the void has a lower permittivity than the solid, the stress in it is higher than in the solid.

Given

• Disc thickness $d = 154$ mm $= 15.4$ cm
• Void thickness $t = 0.01$ mm $= 0.001$ cm
• $\varepsilon_r = 10$, applied voltage $V = 40$ kV
• Strength of air for this thin void: $46$ kV/cm

  +V ===========================
       solid  (er = 10)
       ----- void t = 0.01 mm --   (air)
       solid
   0 ===========================
          total d = 154 mm

Field in the void

The void and the solid are in series. The normal component of flux density is continuous:

$$\varepsilon_0\,E_a = \varepsilon_0\varepsilon_r E_s \;\Rightarrow\; E_a = \varepsilon_r E_s$$

Applied voltage:

$$\begin{aligned} V &= E_s(d - t) + E_a t = E_s\left[(d - t) + \varepsilon_r t\right] \\ E_s &= \frac{40}{(15.4 - 0.001) + 10\times0.001} = \frac{40}{15.409} \\ &= 2.596\ \text{kV/cm} \\ E_a &= \varepsilon_r E_s = 10\times2.596 = 25.96\ \text{kV/cm} \end{aligned}$$

Check

$$E_a = 25.96\ \text{kV/cm} < 46\ \text{kV/cm}$$

Even if 40 kV is taken as rms and the peak stress is used, $E_{a,peak} = \sqrt2\times25.96 = 36.7$ kV/cm, still below 46 kV/cm.

Voltage across the void $= E_a t = 25.96\times0.001 = 0.026$ kV $= 26$ V (rms), far below the minimum Paschen sparking voltage of air (about 327 V), which confirms the result.

Answer: Stress in the void $= 25.96$ kV/cm, which is less than 46 kV/cm, so internal discharge will not occur during the 40 kV withstand test.

The applied voltage needed to start discharge would be $46\times15.409/10 = 70.9$ kV.