Answer:
[tex]1.88\cdot 10^{-5} A[/tex]
Explanation:
The capacitance of a parallel plate capacitor is given by:
[tex]C=\frac{\epsilon_0 A}{d}[/tex] (1)
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
The charge stored on the capacitor is given by
[tex]Q=CV[/tex] (2)
where C is the capacitance and V is the voltage across the capacitor.
The displacement current in the capacitor is given by
[tex]J=\frac{Q}{t}[/tex] (3)
where t is the time elapsed
Substituting (1) and (2) into (3), we find an expression for the displacement current:
[tex]J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}[/tex]
where we have
[tex]A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2[/tex]
[tex]d = 0.58 mm = 5.8\cdot 10^{-4} m[/tex]
[tex]\frac{V}{t}=500,000 V/s[/tex]
Substituting into the equation, we find
[tex]J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A[/tex]
