Okey, think of it another away.
When the electron is being deaccelerated, he's doing work on the system. That energy goes to the capacitor formed by grid and plate right? Because the capacitance is the same, that means that the voltage across the capacitor must change. Is my reasoning correct?
Yeah, the explanation is not really the best. My original question is why and how could I calculate this variation in the voltage. Is there any resource I could use to read about electrostatic induction in situations like this? Wikipedia wasn't really helpful
I'm afraid I did not understand this last thing. Is the voltage created by electrostatic induction or by the flow of the current (caused by the electrons hitting the grid and plate) through the inductances?
I attach the circuit I found in Terman, which I suppose is what you a referring to.
Two questions:
1.- Because the capacitance is fixed (i.e the electrodes are fixed) this causes a voltage to be created?
2.- Isn't the voltage fixed at the grid and plate due to the external power supplies?
Thanks for all the help, I think I'm getting there
Why does this traveling charge alter the potential in the grid and plate? Is there a name for this phenomenon? How could I calculate this variation (obviously assuming a simple geometry like parallel planes)?
Hi:
Due to the quarantine I have more free time, I've decided to learn more about electron tubes. Currently, I'm trying to understand how the Barkhausen–Kurz electron oscillator works.
Now, I've been able to understand the calculation of electron motion inside the device and why they...
Hi:
I myself have some experience building Ion Chambers (and more specifically the models presented in that website). Some clarifications should be in order:
1.- The Darlington Pair must be soldered in air: the leakage current in the protoboard is simply too great. After it has been...
Oh. True. Same mistake twice. So how could the conversion occur? Multiplying the result by the sinus and cosinus (y and x) of the angle formed by the tangent line of the shape to the horizontal?
When the angle is 45 degrees, isn't the effective force sin(45)mg, from which we find the acceleration to be (√2/2)*g, and subsequently v= (gt√2)/2 and then x = (gt^2√2)/4?
Shouldn't both expressions be the same?
How does the result emerge? When I take x=phi, so that y=x=phi:
t*√2*√g = 2*√2*√phi
√phi = (t*√g)/2
Since phi=x
x = (t^2*g)/4
Which is not the expected result. Where have I done wrong?
Ok, so assuming the straight line y=x previously discussed and plugin the equation in Wolfram:
Solution (the energy E is N, because otherwise it interpreted it has the number e).
It seems as the constant c1 should play a role, since by setting c1 = 0 (E=0, thus assuming we start at the origin...
This is where my knowledge of calculus is just unable to keep up. Just to check if I have structured the equation correctly:
f(x) = x^2 (for example)
E = mgx^2 + 1/2m((dx/dt)^2+(x^2dx/dt)^2)
And I would put this expression in some online solver.
Thanks for all the help. The conclusion is that I...
The result is zero, and that is basically the principle of conservation of energy i.e energy is invariant and thus it's derivative must be zero (a constant). Correct?