A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a

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A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a

ccumulation 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 V(x)=Cx4/3V(x)=Cx4/3 where xx is the distance from the cathode and CC is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 11.0 mmmm and the potential difference between electrodes is 220 VV

Physics

1 answer:

serg [3.5K] · Answer 1 · 2026-03-07T11:52:22+03:00

Answer:

C = 4,174 10³ V / m^{3/4}, E = 7.19 10² / ∛x, E = 1.5 10³ N/C

Explanation:

In this problem, we are tasked with determining the constant value and the generated electric field.

We will begin with computing the constant C:

V = C $x^{4/3}$

C = V / x^{4/3}

C = 220 / (11 10⁻²)^{4/3}

C = 4,174 10³ V / m^{3/4}

Next, we will find the electric field by utilizing the formula:

V = E dx

E = dx / V

E = ∫ dx / C x^{4/3}

E = 1 / C x^{-1/3} / (- 1/3)

E = 1 / C (-3 / x^{1/3})

We consider the evaluation from the lower limit x = 0 where E = E₀ = 0 to the upper limit x = x, resulting in E = E:

E = 3 / C (0- (-1 / x^{1/3}))

E = 3 / 4,174 10³ (1 / x^{1/3})

E = 7.19 10² / ∛x

Substituting x = 0.110 cm:

E = 7.19 10² /∛0.11

E = 1.5 10³ N/C