A vacuum tube diode consists of concentric cylindrical electrodes, the negative

Question

A vacuum tube diode consists 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 V(x)=Cx^(4/3) 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 220 V .

a)Determine the value of C.
b)Obtain a formula for the electric field between the electrodes as a function of x. Express your answer in terms of the variables C and x.
c)Determine the magnitude of the force on an electron when the electron is halfway between the electrodes.

Explanation / Answer

Given,

V(x) = C x^(4/3) ; d = 13 mm ; V = 220 V

a)from the given eqn

C = V/x^(4/3)

C = 220/(0.013)^4/3 = 220/0.0031 = 70968 = 7.1 x 10^4

Hence, C = 7.1 x 10^4 V/m^4/3

b)We know that

E = dV/dx

E = d(C x^(4/3))/dx = C x 4/3 x^1/3

Hence, E = 4/3 C x^(1/3)

c)We know that

F = qE

E = 1.6 x 10^-19 x 7.1 x 10^4 x 4/3 x (0.013/2)^(1/3) = 2.83 x 10^-15 N

Hence, F = 2.83 x 10^-15 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.