Internal discharge in square disc with void

Internal discharge starts when the stress in the gas-filled void reaches the breakdown strength of that gas. The void (permittivity $\varepsilon_0$) and the solid ($\varepsilon_0\varepsilon_r$) are in series, so the stress in the void is $\varepsilon_r$ times that in the solid.

Given

• Disc thickness $d = 154$ mm $= 15.4$ cm, $\varepsilon_r = 10$
• Void thickness $t = 0.1$ mm $= 0.01$ cm, strength of void $= 46$ kV/cm
• Applied voltage $V = 40$ kV (taken as rms)

The disc being square does not matter: for a thin flat void parallel to the electrodes, the field depends only on thicknesses.

Stress in the void

$$\begin{aligned} V &= E_s(d - t) + E_a t, \qquad E_a = \varepsilon_r E_s \\ E_s &= \frac{V}{(d - t) + \varepsilon_r t} = \frac{40}{15.39 + 0.10} = 2.582\ \text{kV/cm} \\ E_a &= 10\times2.582 = 25.82\ \text{kV/cm (rms)} \\ E_{a,peak} &= \sqrt2\times25.82 = 36.52\ \text{kV/cm (peak)} \end{aligned}$$

Check

The void is only 0.1 mm thick. By Paschen's law a very thin air gap has a higher strength than the 30 kV/cm (peak) of air in centimetre gaps, which is why the void strength is given as 46 kV/cm. The correct comparison is therefore with 46 kV/cm.

Voltage needed for discharge: $V_i = 46\times\dfrac{15.49}{10} = 71.3$ kV.

Answer: Stress in the void $= 25.82$ kV/cm (rms) $< 46$ kV/cm, so internal discharge will not occur at 40 kV. (If the 30 kV/cm peak value for large air gaps were wrongly applied, the 36.5 kV/cm peak stress would suggest discharge.)