The air between two parallel plates separated by a distance `d = 20 mm` is inoized by X-ray radisation. Each plate has an area `S = 500 cm^(2)`. Find the concentration of positive ions if at a voltage `V = 100 V` a current `I = 30 muA` flows between the paltes, which is well below the saturation current. The air ion mobilities are `u_(0)^(+) = 1.37 cm^(2)//(V .s)` and `u_(0) = 1.91 xm^(2)//(V . s)`

Explanation

`E = (V)/(d)`
So by the definition of the mobility
`v^(2) = u_(0)^(+) (V)/(d), v^(-) = u_(0)^(-) - (V)/(d)`
and `j = (n_(+) u_(0)^(+) + n_(-) u_(0)^(-)) (eV)/(d)`
(The negative ions move towards the anode and the positive ion towards the cathode and the total current is the sum of the currents due to them.)
On the other hand, in equilibrium `n_(+) = n_(-)`
So, `n_(+) = n_(-) = (I)/(S)//(u_(0)^(+) + u_(0)^(-)) (eV)/(d)`
`= (Id)/(e V S (u_(0)^(+) + u_(0)^(-))) = 2.3xx10^(8) cm^(-3)`