Dielectric Slab in a Capacitor: Safe Voltage Judgement
A teaching laboratory has a parallel-plate capacitor with circular plates of area 2.0 × 10^-2 m² and an adjustable gap. A 0.90 mm thick slab of dielectric (κ = 4.0, dielectric strength 20 kV/mm) is available, but it is slightly smaller than the plates. A student sets the plate separation to 1.00 mm, lays the slab on the lower plate, and plans to apply 12 kV across the capacitor for a demonstration. The laboratory air breaks down at about 3.0 × 10^6 V/m. The student argues that 12 kV across 1 mm is safe because the slab is rated at 20 kV/mm. A technician must approve or reject the run. The slab is flat, its faces are parallel, and it covers the whole plate area, but a 0.10 mm air layer remains between the slab and the upper plate. The supply can be set to any voltage up to 12 kV.
1. Calculate the electric field in the air layer for the proposed 12 kV run and decide whether the technician should approve it. State the assumption used about the field directions.
Read the worked answer
Assume the field is uniform and normal to the plates. The air layer and dielectric are in series, and the normal component of D is continuous: ε₀E_air = κε₀E_d, so E_d = E_air/κ. The applied voltage divides as
V = E_air t_air + E_d t_d = E_air(t_air + t_d/κ).
Here t_air = 0.10 mm and t_d = 0.90 mm, so t_air + t_d/κ = 0.10 + 0.90/4 = 0.325 mm = 3.25 × 10^-4 m.
Thus E_air = 12,000 / (3.25 × 10^-4) ≈ 3.7 × 10^7 V/m. This is more than ten times the air breakdown field of 3.0 × 10^6 V/m. The field in the slab is E_air/4 ≈ 9.2 × 10^6 V/m = 9.2 kV/mm, below its 20 kV/mm rating, but the air layer breaks down first. The technician should reject the run.
2. Find the maximum safe voltage with the 0.90 mm slab and the 0.10 mm air layer. Compare it with the maximum safe voltage of the empty 1.00 mm capacitor and explain the counter-intuitive result.
Read the worked answer
Set E_air equal to the breakdown value:
V_max = E_breakdown(t_air + t_d/κ) = (3.0 × 10^6)(3.25 × 10^-4) ≈ 975 V.
For the empty capacitor, the gap is 1.00 mm of air, so V_max = (3.0 × 10^6)(1.00 × 10^-3) = 3000 V.
The slab therefore reduces the safe voltage from about 3.0 kV to about 0.98 kV. The reason is that the dielectric has a higher permittivity, so for the same normal D the field inside it is lower. To keep the voltage fixed, most of the voltage appears across the low-permittivity air layer, and the air field becomes very large. A thin air gap is more dangerous than a thick air gap in this arrangement. The slab should not be used with a trapped air layer at these voltages.
3. The student suggests pressing the slab tightly so that it exactly fills the 1.00 mm gap with no air layer. Under that ideal condition, is 12 kV acceptable? What practical precautions remain necessary?
Read the worked answer
With no air layer, the field is uniform in the slab: E = V/d = 12 kV / 1.00 mm = 12 kV/mm. This is below the slab's dielectric strength of 20 kV/mm, so 12 kV is acceptable in the ideal case. The maximum ideal voltage would be 20 kV/mm × 1.00 mm = 20 kV.
However, the approval should still be cautious. Perfect contact is difficult to maintain; even a very thin air layer can concentrate field and cause partial discharge. Sharp edges of the plates also enhance the field. Practical precautions include rounding the electrodes, using a dielectric fluid or compliant layer to exclude air, and starting at a lower voltage while checking for corona. The safe conclusion is that the original 0.10 mm air gap makes the run unsafe, while a genuinely void-free slab could support 12 kV.