Respuesta :
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 [tex]x^{4/3}[/tex]
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
Electrodes are used to establish contact with non-metallic parts of the circuit such as electrolytes. The value of constant for the given vacuum tube diode is [tex]4174 \times 10^3 \rm \ V / m^{3/4}[/tex].
Electrodes:
The value of the constant is directly proportional to the potential difference between electrodes.
The value of the constant C,
V = C
[tex]C = \dfrac {V }{ x^{\frac 43}}\\[/tex]
Where,
[tex]C[/tex]- constant
[tex]V[/tex]- potential difference between electrodes = 220 V
[tex]x[/tex] - distance between the cathode and anode = 11.0 mm = 0.11 m
Put the values
[tex]C = \dfrac {220 }{(11\times 10^{-2})^{\frac 43}}\\ C = 4174 \times 10^3 \rm \ V / m^{3/4}[/tex]
Therefore, the value of constant for the given vacuum tube diode is [tex]4174 \times 10^3 \rm \ V / m^{3/4}[/tex].
Learn more about electrodes:
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