Displacement current and conduction current

The Maxwell’s fourth equation is

$$
vecnabla times vec H = vec J_c + vec J_d spacespace$$ $$ text{where} spacespace vec J_d = frac{partial vec D}{partial t}
$$

Taking the divergence on both side, because divergence of curl is zero, we get

$$
vec nabla cdot vec J_c + vec nabla cdot vec J_d spacespace = 0
$$

And from the continuity of current equation and Gauss’s law, we get

$$
vec nabla cdot vec J_d = – vec nabla cdot vec J_c =frac{partial rho_v}{partial t} = frac{partial}{partial t} (vec nabla cdot vec D) = vec nabla cdot frac{partial vec D}{partial t} $$

Therefore,
$$
vec J_d = frac{partial vec D}{partial t}
$$

So my question is that, following this logic, I think the below equation must also be true

$$
vec J_c = – frac{partial vec D}{partial t}
$$

which says that when there is a change in electric flux density, there must be a conduction current with the same magnitude as the displacement current but with the opposite sign. But we know that, in the case of a parallel capacitor, the conduction current is zero in between the plates and the displacement current is a nonzero value. Where am I wrong on this?

Please help me clarify this. Thanks!