Maxwell fourth equation incorrect?

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mdn
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Hi all
this question seems to be silly but i want to where i am wrong. Maxwell was made modification in Amperes law curl H=J and he added new quantity displacement current Jd, reason was that Div.J=0 (true in DC current but not in AC charging capacitor).
but if i say Div J=0 in capacitor because the amount of current entering and leaving is exactly equal in Gaussian loop (loop has half part in space between plate) on the same back direction as it is AC currant and can not flow through capacitor plate, where i am wrong?
and second doubt is that current can not added in series because it is same therefore fourth equation should be
Curl H=2J.
 
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mdn said:
Hi all
...reason was that Div.J=0 (true in DC current but not in AC charging capacitor).
but if i say Div J=0 in capacitor because the amount of current entering and leaving is exactly equal in Gaussian loop (loop has half part in space between plate) on the same back direction as it is AC currant and can not flow through capacitor plate, where i am wrong?

The reason for adding displacement current term into the right-hand side of the equation

$$
\nabla\times \mathbf B = \mu_0 \mathbf j
$$

was that divergence of the left-hand side is zero everywhere for any field ##\mathbf B##, while divergence of ##\mathbf j## is not always zero.
 
The current density between the plates of a capacitor is 0.If you igore dE/dt you get that the curl of B is 0 in this area, and this conflicts with reality.
It's easier to see in integral form, where the curl of B integrated around the boundary of a surface is the same for any surface with the same boundary.
If you take a capacitor you can have one surface going through one of the leads, and another surface going through the gap between the plates. If you integrate only the real current and not the displacement current you won't get the same outcome.
 
div of J is zero or not in case of capacitor? if yes then problem in fourth equation
and if no can you prove?