Displacement current is set up between the plates of a capacitor when the potential difference across the plates is ..............

Match the following Maxwell’s equations: (The symbols used here have their usual meanings)

List-$I$	List-$II$
$(a)$ Gauss’ law for electrostatics	$(i)$ $\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$
$(b)$ Gauss’ law for magnetism	(ii) $\oint \vec{B} \cdot d\vec{l} = \mu_0 [i_c + \epsilon_0 \frac{d\phi_E}{dt}]$
$(c)$ Faraday’s law	(iii) $\oint \vec{B} \cdot d\vec{A} = 0$
$(d)$ Ampere-Maxwell’s law	(iv) $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$

$A$ 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

$A$ 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:

$A$ parallel plate capacitor of area $60\, cm^2$ and separation $3\, mm$ is charged initially to $90\, \mu C$. If the medium between the plates becomes slightly conducting and the plate loses the charge at the rate of $2.5\times10^{-8}\, C/s$,what is the magnetic field between the plates?

Choose the $CORRECT$ expression for the Ampere-Maxwell law.