Statement
A parallel plate capacitor with a separation d has a voltage V applied when free space is the only dielectric. Assume a dielectric strength for air of 30 kV/cm. Show why the air breaks down when a thin piece of glass of thickness d2 is inserted into the gap.
System Parameters
Arbitrarily set area:
Permittivity of free space:
Solution
Before the glass is inserted, the voltage drop is the same as the applied voltage,
and the electric field in the space between the plates is
just below the breakdown field for the air between the plates.
The problem becomes one of two capacitors in series:
As shown in the previous problem, the voltage drop across the air after the glass is inserted is given by
Notice that the area drops out of the problem here.
which exceeds the dielectric strength of air. The dielectric strength is really the ionization potential of a material. Above this field strength, the material -- in this case, air -- ionizes, and becomes conductive. As emphasized previously, conductors cannot support electric field within them, except transiently. The capacitor will cease to be a capacitor and become a single conductor.
P.20 |
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P.21 Breakdown Voltage |
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