Internal discharge voltage in 1 cm specimen

An internal discharge begins when the voltage across the gas void reaches the breakdown voltage of the gas. The void and the rest of the dielectric behave as two capacitors in series.

Given

• Specimen thickness $d = 1$ cm, void thickness $t = 1$ mm $= 0.1$ cm
• $\varepsilon_r = 4.0$, applied voltage $80$ kV (rms)
• Breakdown strength of air $E_b = 30$ kV (peak)/cm

  +V (80 kV rms) =================
        solid   er = 4
       [ air void  t = 1 mm  ]
        solid
   0  ==========================
           d = 10 mm

Series capacitances (per unit area)

$$C_1 = \frac{\varepsilon_0 A}{t}\ (\text{void}), \qquad C_2 = \frac{\varepsilon_0\varepsilon_r A}{d - t}\ (\text{solid})$$

Voltage across the void:

$$V_1 = V\,\frac{C_2}{C_1 + C_2} = \frac{V\,t}{t + \dfrac{d - t}{\varepsilon_r}}$$

Voltage across the void at its breakdown

$$V_{1b} = E_b\,t = 30\times0.1 = 3.0\ \text{kV (peak)}$$

Applied voltage for internal discharge

$$\begin{aligned} V_i &= E_b\left[t + \frac{d - t}{\varepsilon_r}\right] \\ &= 30\left[0.1 + \frac{0.9}{4}\right] = 30\times0.325 \\ &= 9.75\ \text{kV (peak)} \\ V_{i,rms} &= \frac{9.75}{\sqrt2} = 6.89\ \text{kV (rms)} \end{aligned}$$

Answer: Internal discharge starts when the applied voltage reaches 9.75 kV (peak) = 6.89 kV (rms).

Since the applied voltage (80 kV rms) is much higher than 6.89 kV rms, the void will discharge many times in each half cycle and the specimen will deteriorate quickly.