Respuesta :
Answer:
The value is [tex]I_D =8.501 * 10^{-8} \ A [/tex]
Explanation:
From the question we are told that
The length of one side of the plate is [tex]d = 8.6 \ cm = 0.086 \ m[/tex]
The rate at which the electric field is changing is [tex] \frac{dE}{dt} = 1.3*10^{6} \ V/m \cdot s[/tex]
Generally the area is mathematically represented as
[tex]A = d^2[/tex]
=> [tex]A = 0.086^2[/tex]
=> [tex]A = 0.007396 \ m^2 [/tex]
Generally the displacement current [tex]I_D[/tex] is mathematically represented as
[tex]I_D = \epsilon_o * A * \frac{dE}{dt}[/tex]
Here [tex]\epsilon_o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} C/(V⋅m [/tex]
=> [tex]I_D = 8.85*10^{-12}* 0.007396 * 1.3*10^{6} [/tex]
=> [tex]I_D =8.501 * 10^{-8} \ A [/tex]
The displacement between the plates is [tex]8.51*10^-^8m[/tex]
Data;
- Distance between the square plate = 8.6cm
- Rate of change of electric field = 1.3x10^6 V/m.s
Displacement of the Current
This can be calculated using a formula
[tex]d = e_o A \frac{\delta E}{\delta t} \\[/tex]
where
- d = displacement
- ε = epsilon
- A = area
Let's substitute the values and solve for the displacement between the plates.
[tex]d = e_o A \frac{\delta E}{\delta t} \\d = 8.85*10^-^1^2 * (0.086)^2 * 1.30*10^6\\d = 8.51*10^-^8m[/tex]
The displacement between the plates is [tex]8.51*10^-^8m[/tex]
Learn more on displacement on a charged plates here;
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