= Solution
Insert $\nabla\times\mathbf E=-\partial_t\mathbf B$ and $\nabla\times\mathbf H=\mathbf J+\partial_t\mathbf D$ into the given identity. For a linear, time-independent, nondispersive medium, $\mathbf D=\epsilon\mathbf E$ and $\mathbf B=\mu\mathbf H$, giving
$$
\nabla\cdot(\mathbf E\times\mathbf H)=-\mathbf J\cdot\mathbf E-\partial_tu,
\qquad u=\frac12\mathbf E\cdot\mathbf D+\frac12\mathbf H\cdot\mathbf B.
$$
The <divergence theorem> then yields
$$
\boxed{\oint_S(\mathbf E\times\mathbf H)\cdot d\mathbf S=-\int_V[\mathbf J\cdot\mathbf E+\partial_tu]\,dV.}
$$
The left side is outward electromagnetic <power> flow. The current term is <power> delivered to matter, becoming <Joule heating> for an <Ohmic conductor>, and $\partial_tu$ is the change in stored electric plus <magnetic energy>. The minus sign says outward flux reduces stored <energy> unless supplied locally.
Follow the drawing's coordinates: lower strip at $z=0$, upper strip at $z=d$, current $I$ along $+x$ in the upper strip and back along the lower. The wording “both strips lie in the $xy$-plane” must mean parallel to that plane, since the drawing separates them in $z$. Neglect the side and end fringes using $l\gg a\gg d$. Then $I=V_0/(R+R_L)$ and the top-strip potential is $V(x)=V_0-IRx/l$, while the lower strip is equipotential at zero.
The harmonic function matching those two surfaces is
$$
\boxed{\phi=\frac zd\left(V_0-\frac{IRx}{l}\right).}
$$
Indeed $\nabla^2\phi=0$, and it has the required potentials at $z=0,d$. Thus
$$
\mathbf E=\left(\frac{IRz}{ld},0,-\frac{V(x)}d\right),\qquad
\mathbf H=\left(0,\frac Ia,0\right).
$$
The two current sheets reinforce their <magnetic field> between the strips; its direction follows the right-hand rule. The <Poynting flow in resistive parallel strips> follows from their cross product:
$$
\boxed{\mathbf S=\left(\frac{IV(x)}{ad},\ 0,\ \frac{I^2Rz}{ald}\right).}
$$
<Power> flows along the gap towards the load and upwards into the resistive strip. Across a transverse area $ad$, $P_x(x)=IV(x)$ falls from $IV_0$ to $I^2R_L$. The upper surface receives $\int_0^l aS_z(z=d)\,dx=I^2R$, exactly the strip's heating; the ideal lower strip receives none. In the charge-free gap, $\partial_xS_x+\partial_zS_z=0$, consistent with a static field and no local dissipation. The small longitudinal <electric field> cannot be discarded if this upward <energy> transfer is to be represented.
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