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(B) The displacement current $I_d$ is defined as $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,where $\Phi_E$ is the electric flux through the surface.For a p

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$I/2$

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(B) The displacement current $I_d$ is defined as $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,where $\Phi_E$ is the electric flux through the surface.For a parallel plate capacitor,the electric field $E$ between the plates is uniform and given by $E = \frac{q}{\epsilon_0 A}$,where $q$ is the charge on the plates.The electric flux $\Phi_E$ through an area $A'$ parallel to the plates is $\Phi_E = E \cdot A' = \frac{q}{\epsilon_0 A} \cdot A'$.Substituting $A' = A/2$,we get $\Phi_E = \frac{q}{\epsilon_0 A} \cdot \frac{A}{2} = \frac{q}{2\epsilon_0}$.The displacement current is $I_d = \epsilon_0 \frac{d}{dt} \left( \frac{q}{2\epsilon_0} \right) = \frac{1}{2} \frac{dq}{dt}$.Since the charging current $I = \frac{dq}{dt}$,we have $I_d = \frac{I}{2}$.

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

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