Voltage sharing between air gap and press board

The air gap and the press board form two capacitors in series. The normal flux density $D$ is the same in both, so the field in each is inversely proportional to its permittivity: $\varepsilon_0 E_1 = \varepsilon_0\varepsilon_r E_2$.

Given

• Total gap $d = 2.0$ cm, press board $d_2 = 0.2$ cm, so air gap $d_1 = 2.0 - 0.2 = 1.8$ cm
• $\varepsilon_{r2} = 4$ (press board), $\varepsilon_{r1} = 1$ (air), $V = 60$ kV

  ======================== upper plate (V)
        air   d1 = 1.8 cm      V1
  ------------------------
   press board d2 = 0.2 cm     V2
  ======================== lower plate (0)

Solution

$$\begin{aligned} E_1 &= \varepsilon_r E_2 = 4E_2 \\ V &= E_1 d_1 + E_2 d_2 = E_1\left(d_1 + \frac{d_2}{\varepsilon_r}\right) \\ E_1 &= \frac{60}{1.8 + 0.2/4} = \frac{60}{1.85} = 32.43\ \text{kV/cm} \\ E_2 &= \frac{E_1}{4} = 8.11\ \text{kV/cm} \end{aligned}$$

Voltages:

$$\begin{aligned} V_1 &= E_1 d_1 = 32.43\times1.8 = 58.38\ \text{kV (air gap)} \\ V_2 &= E_2 d_2 = 8.11\times0.2 = 1.62\ \text{kV (press board)} \end{aligned}$$

Check: $58.38 + 1.62 = 60$ kV.

Answer: Voltage across the air gap $V_1 \approx$ 58.38 kV; voltage across the press board $V_2 \approx$ 1.62 kV.

Comment

Without the press board the air stress would be $60/2 = 30$ kV/cm; with it, the air stress rises to 32.43 kV/cm. Inserting a solid of higher permittivity does not relieve the air; it increases the stress in the air. If 60 kV is the rms value, the peak air stress is $\sqrt2\times32.43 = 45.9$ kV/cm, above the 30 kV/cm (peak) strength of air, so the air gap would break down (or discharge) first, and then the full voltage would appear across the thin press board. This is why composite insulation must be designed with the stress in the weaker, lower-permittivity material in mind.