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(D) The displacement current $I$ in a parallel plate capacitor is given by the formula $I = \varepsilon_0 \frac{d\phi_E}{dt}$,where $\phi_E$ is the el

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$\frac{I}{\varepsilon_0 E}$

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(D) The displacement current $I$ in a parallel plate capacitor is given by the formula $I = \varepsilon_0 \frac{d\phi_E}{dt}$,where $\phi_E$ is the electric flux.Since the electric flux $\phi_E$ through the area $A$ of the plates is given by $\phi_E = E_{field} \cdot A$,where $E_{field}$ is the electric field between the plates.Substituting this into the displacement current formula,we get $I = \varepsilon_0 \frac{d}{dt}(E_{field} \cdot A)$.Since the area $A$ is constant,we have $I = \varepsilon_0 A \frac{dE_{field}}{dt}$.Given that the rate of change of the electric field is $E$,we substitute $\frac{dE_{field}}{dt} = E$.Thus,$I = \varepsilon_0 A E$.Solving for the area $A$,we get $A = \frac{I}{\varepsilon_0 E}$.

If the rate of change of electric field across the plates of a parallel plate capacitor is $E$ and the displacement current is $I$,then the area of one plate of the capacitor is ($\varepsilon_{0}$ is permittivity of free space).

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