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(B) The displacement current $I_D$ is given by the formula $I_D = \varepsilon_0 \frac{d\phi_E}{dt}$.Since the electric field $E = \frac{V}{d}$ and the

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$1.062 	imes 10^{-2} \, A$

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(B) The displacement current $I_D$ is given by the formula $I_D = \varepsilon_0 \frac{d\phi_E}{dt}$.Since the electric field $E = \frac{V}{d}$ and the electric flux $\phi_E = E \cdot A$,we have $I_D = \varepsilon_0 \frac{A}{d} \frac{dV}{dt}$.Given: $\varepsilon_0 = 8.854 	imes 10^{-12} \, F/m$,$V = 400 \, V$,$A = 60 \, cm^2 = 60 	imes 10^{-4} \, m^2$,$d = 2 \, mm = 2 	imes 10^{-3} \, m$,and $\Delta t = 10^{-6} \, s$.Substituting these values:$I_D = \frac{8.854 	imes 10^{-12} 	imes 60 	imes 10^{-4} 	imes 400}{2 	imes 10^{-3} 	imes 10^{-6}}$$I_D = \frac{8.854 	imes 60 	imes 4 	imes 10^{-12} 	imes 10^{-4} 	imes 10^2}{2 	imes 10^{-9}}$$I_D = \frac{2124.96 	imes 10^{-14}}{2 	imes 10^{-9}} = 1062.48 	imes 10^{-5} = 1.06248 	imes 10^{-2} \, A$.Thus,the correct option is $B$.

$A$ parallel plate capacitor with a plate separation of $2 \, mm$ is connected to an electric circuit with a source voltage of $400 \, V$. If the plate area is $60 \, cm^2$,what is the value of the displacement current for a time interval of $10^{-6} \, s$?

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