Generals 2005 II 6

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[edit] Part a

To add nonzero SEE δ:

n_{e}\left(1-\delta\right)\sqrt{\frac{kT_{e}}{2\pi m_{e}}}\exp\left(\frac{-e\left(V_{sp}-V_{f}\right)}{kT_{e}}\right)=0.6n_{e}\sqrt{\frac{kT_{e}}{m_{i}}}


[edit] Part b

Taking the log of both sides:

V_{f}-V_{sp}=\frac{kT_{e}}{e}\log\left[0.6\left(1+\delta\right)\sqrt{\frac{2\pi m_{e}}{m_{i}}}\right]


So if δ were to increase, the floating potential will increase.


[edit] Part c

For Vf − Vsp = 0:

\left(1-\delta\right)=0.6\sqrt{\frac{2\pi m_{e}}{m_{i}}}


So, for hydrogen, δ = .96 will cause the probe to float at the space charge potential. If no material has this potential, the probe could be heated, so that it emits electrons on its own. This would add a term to the right of the current balance, increasing the allowed incident electron current.


[edit] Part d

Baising the limiter to the space potential would mean increasing the voltage (since it is usually at the floating potential), and so would increase the power flow to the limiter (since the electrons would no longer be repelled).


[edit] Part e

There would be no difference between an insulating limiter and a conducting limiter, since the charge for the limiter comes from the plasma and so does not need to be brought in from ground. This means that there will be less power flow than part d.

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