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Electric Potential Distribution in a Vacuum Diode
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[QUOTE="HotFurnace, post: 6073131, member: 652208"] Here's what I had done: - Apply Gauss's laws for a tiny box region, with surface S, thickness dx, we end up with this equation: $$ρ=ε*\frac {dEe} {dx} (1)$$ in which ρ is the charge density, Ee is the electric field created by the charge (the field created by the voltage U is Ep, the total field is E). - The current flowing inside this diode must be a constant everywhere inside the diode, regardless of position x. According to the definition of current: $$I=\frac {dq} {dt}=ρSv$$ in which v is the velocity of the electron at position x. Take the differentiation of I we get: $$0=v\frac {dρ} {dx}+ρ\frac {dv} {dx}$$ - The Second Newton laws of motion for one electron is $$e*E=m\frac {dv} {dt}=mv\frac {dv} {dx}$$ - Use substitution, we get the following equation: $$\frac {dρ} {ρ^3}=\frac {eS^2} {mI^2}Edx$$, integrate it, and we get $$-\frac {1} {ρ^2}=\frac {2eS^2} {mI^2}V+C$$ and now I'm unsure what to do, maybe we shall set C=0 and use equation (1) then we will end up with $$ω\sqrt \frac {U} {V} =ε\frac {\partial^2 V} {\partial x^2} $$ ω is the electron density at the anode. Solve this don't works as the solution is very complex. [/QUOTE]
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Electric Potential Distribution in a Vacuum Diode
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