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(A) The displacement current is given by the formula:$I_d = \varepsilon_0 \frac{d\phi_E}{dt} = \varepsilon_0 \frac{\Delta \phi_E}{\Delta t}$Since the

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$3.5$

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(A) The displacement current is given by the formula:$I_d = \varepsilon_0 \frac{d\phi_E}{dt} = \varepsilon_0 \frac{\Delta \phi_E}{\Delta t}$Since the electric flux $\phi_E = E \cdot A$ and the electric field $E = \frac{V}{d}$, we have:$I_d = \varepsilon_0 \frac{A}{d} \frac{\Delta V}{\Delta t}$Given values are:$\Delta V = 200 \, V$$\Delta t = 1 \, \mu s = 10^{-6} \, s$$d = 1 \, mm = 10^{-3} \, m$$A = 20 \, cm^2 = 20 	imes 10^{-4} \, m^2$$\varepsilon_0 = 8.85 	imes 10^{-12} \, F/m$Substituting these values:$I_d = \frac{8.85 	imes 10^{-12} 	imes 20 	imes 10^{-4} 	imes 200}{10^{-3} 	imes 10^{-6}}$$I_d = \frac{8.85 	imes 20 	imes 200 	imes 10^{-16}}{10^{-9}}$$I_d = 35400 	imes 10^{-7} \, A = 3.54 	imes 10^{-3} \, A \approx 3.5 \, mA$

Consider a parallel plate capacitor which is maintained at a potential of $200 \, V$. The separation distance between the plates of the capacitor and the area of the plates are $1 \, mm$ and $20 \, cm^2$, respectively. Calculate the displacement current in $1 \, \mu s$. (in $ \, mA$)

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