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(B) The charge on the capacitor plates at time $t$ is $q = It$.The electric field $E$ between the plates at this instant is given by $E = \frac{q}{A \

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$I/2$

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(B) The charge on the capacitor plates at time $t$ is $q = It$.The electric field $E$ between the plates at this instant is given by $E = \frac{q}{A \varepsilon_0} = \frac{It}{A \varepsilon_0}$.The electric flux $\phi_E$ through the given area $A/2$ is $\phi_E = E \cdot \left(\frac{A}{2}\right) = \left(\frac{It}{A \varepsilon_0}\right) \cdot \left(\frac{A}{2}\right) = \frac{It}{2 \varepsilon_0}$.The displacement current $I_D$ is defined as $I_D = \varepsilon_0 \frac{d\phi_E}{dt}$.Substituting the expression for $\phi_E$,we get $I_D = \varepsilon_0 \frac{d}{dt} \left(\frac{It}{2 \varepsilon_0}\right) = \varepsilon_0 \cdot \frac{I}{2 \varepsilon_0} = \frac{I}{2}$.

$A$ parallel plate capacitor with plate area $A$ and separation between the plates $d$ is charged by a constant current $I$. Consider a plane surface of area $A/2$ parallel to the plates and drawn between the plates. The displacement current through the area is

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