Give your numerical result in joules and in electron volts. 23.57 CALC A vacuum

Question

Give your numerical result in joules and in electron volts. 23.57 CALC A vacuum tube diode consists ice- of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode the electric potential between the electrodes is not a linear-function, of the position, even with planar geometry, but is given by where x is the distance from the cathode and C is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of C. (b) Obtain a formula for the electric field between the electrodes as a function of x. (c) Determine the force on an electron when the electron is halfway between the electrodes.

Explanation / Answer

Part 1:
14.0 mm = 0.014 m
(0.014 m)^(4/3) = .00337 m^(4/3)
290 V / .00337 m^(4/3) = ......................... << 85946.32 V/m^(4/3) >>

Part 2:
Take the derivative of V
dV/dx = (4/3)*x^(1/3)
Ex = ........................................? << -C(4/3)*x^(1/3) >>

Part 3:
F = E*q
q = -1.602*10^(-19)
E = -(85946.32) * (4/3) * (.014/2)^(1/3)
E = -21921.25
F = (-1.602*10^(-19)) * (-21921.25) = ................. << 3.51*10^(-15) >>

Part 4:
The electron has a negative charge;
it will be repelled by the negative
cathode and attracted to the
positive anode ........................................? << Towards the positive anode >>

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