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Electric Potential Distribution in a Vacuum Diode
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[QUOTE="TSny, post: 6073293, member: 229090"] I think you are essentially OK up to here, except I don't believe you should have a negative sign on the left. The signs are a headache due to the fact that the electron has a negative charge, the E-field points in the negative x direction, the potential difference equals the negative of the integral of E, etc. Yes, you can take the constant C to be zero. It is assumed that the speed ##v## of the electrons can be taken to be essentially zero at the cathode. So the charge density ##\rho## must be very large at the cathode because the current ##\rho vS## has a finite value that is independent of ##x##. So, ##\frac{1}{\rho^2}## can be taken to be negligible at the cathode. The equation ##ω\sqrt \frac {U} {V} =ε\frac {d^2 V} {d x^2} ## can be solved easily by first multiplying both sides by ##\frac {dV}{dx}## and then integrating. Another way to derive the differential equation for ##V## is to set up conservation of energy as ##\frac{1}{2}mv^2 = eV## and solve for ##v##. Sub this into ##\rho = -\frac{I}{vS}## where ##I## is the magnitude of the current (so that ##I## is a positive number). Then use this in your equation (1) to get the desired second-order differential equation for ##V##. This way, you avoid having to set up and solve a differential equation for ##\rho##. [/QUOTE]
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Electric Potential Distribution in a Vacuum Diode
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