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)=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

Answer 1

Answer:

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

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of 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}

To calculate the electric field we use the expression

            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 evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

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

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

for x = 0.110 cm

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C