= Solution
Take the dot product of <Ampère-Maxwell equation> with $E$ and of <Faraday's law> with $B/\mu_0$. The <divergence of a cross product> identity
$$
\nabla\mathbin{\cdot}(E\times B)=B\mathbin{\cdot}(\nabla\times E)-E\mathbin{\cdot}(\nabla\times B)
$$
then gives the local <Poynting theorem>
$$
\frac{\partial}{\partial t}\left(\frac{\varepsilon_0E^2}{2}+\frac{B^2}{2\mu_0}\right)+J\mathbin{\cdot}E+\nabla\mathbin{\cdot}\left(\frac1{\mu_0}E\times B\right)=0.
$$
Integration and the <divergence theorem> produce the stated equation. Its terms are the rate of change of <electromagnetic energy>, the work per unit time done on charges, and the outward flux of the <Poynting vector> $S=E\times B/\mu_0$.
Let the plate separation be $d$ and their radius be $R$, so $A=\pi R^2$. Between the plates,
$$
E=\frac{q}{\varepsilon_0A}\,\hat z.
$$
For a circular <Amperian loop> of radius $r<R$, the <displacement current> in the <Ampère-Maxwell equation> gives
$$
B_\phi(r)=\frac{\mu_0\varepsilon_0r}{2}\dot E
=-\frac{\mu_0\lambda q r}{2A}.
$$
Therefore
$$
\boxed{S=\frac{\lambda q^2r}{2\varepsilon_0A^2}\,\hat r,}
$$
which points radially outward. Its flux through the cylindrical side is $\lambda q^2d/(\varepsilon_0A)$, exactly $-dU/dt$ for the stored <capacitor energy> $U=q^2d/(2\varepsilon_0A)$.
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