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(B) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{Q}{\varepsilon_0 A}$,where $Q$ is the charge on the

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$i/2$

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(B) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{Q}{\varepsilon_0 A}$,where $Q$ is the charge on the capacitor at time $t$.For a surface of area $A' = A/2$ placed parallel and symmetrically between the plates,the electric flux $\phi_E$ through this area is $\phi_E = E \cdot A' = \left( \frac{Q}{\varepsilon_0 A} \right) \left( \frac{A}{2} \right) = \frac{Q}{2\varepsilon_0}$.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 $i_d = \varepsilon_0 \frac{d}{dt} \left( \frac{Q}{2\varepsilon_0} \right) = \frac{1}{2} \frac{dQ}{dt}$.Since the charging current $i = \frac{dQ}{dt}$,the displacement current is $i_d = \frac{i}{2}$.

$A$ 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 symmetrically between the plates. The displacement current through this area will be:

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