Question 19.4: Dry air will support a maximum electric field strength of ab...

Dry air will support a maximum electric field strength of about 3.0 \times 10^{6} V / m . Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?

Strategy
We are given the maximum electric field E between the plates and the distance d between them. The equation V_{ AB }=E d can thus be used to calculate the maximum voltage.

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The potential difference or voltage between the plates is

V_{ AB }=E d.                      (19.28)

Entering the given values for E and d gives

V_{ AB }=\left(3.0 \times 10^{6} V / m \right)(0.025 m )=7.5 \times 10^{4} V                  (19.29)

or

V_{ AB }=75 kV.                 (19.30)

(The answer is quoted to only two digits, since the maximum field strength is approximate.)
Discussion
One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5 cm (1 in.) gap, or 150 kV for a 5 cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage will cause a spark if there are points on the surface, since points create greater fields than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry days.

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