= Solution
\b[Distributed line parameters.] The geometry is the wide <parallel-plate transmission line> described in the question, with absolute <permittivity> $\varepsilon=\epsilon\varepsilon_0$. A length $dx$ has charge $C V\,dx$ and field $E=V/d$, giving
$$
\boxed{C=\frac{\varepsilon a}{d}.}
$$
The opposed plate currents have sheet density $I/a$ and confine the leading magnetic field $H=I/a$ between the plates. The magnetic <energy> per unit length is $\mu_0H^2ad/2=LI^2/2$, so
$$
\boxed{L=\frac{\mu_0d}{a}.}
$$
Here $C$ and $L$ are per unit length, in $\mathrm{F\,m^{-1}}$ and $\mathrm{H\,m^{-1}}$, not lumped component values. Internal conductor fields and edge corrections are neglected.
\b[Voltage and current evolution.] The current entering a short length minus that leaving it is the rate of increase of its plate charge:
$$
I(x)-I(x+dx)=C\,dx\,\partial_tV.
$$
The voltage drop across that length is its inductive drop, $V(x)-V(x+dx)=L\,dx\,\partial_tI$. Dividing by $dx$ gives the lossless <Telegrapher's equations>,
$$
\boxed{\partial_xI+C\partial_tV=0,\qquad
\partial_xV+L\partial_tI=0.}
$$
\b[Propagation speed.] Differentiate these equations once more and eliminate the other variable:
$$
\partial_x^2V=LC\,\partial_t^2V,\qquad
\partial_x^2I=LC\,\partial_t^2I.
$$
The <wave equation> speed is
$$
\boxed{v=\frac1{\sqrt{LC}}
=\frac1{\sqrt{\mu_0\varepsilon_0\epsilon}}
=\frac c{\sqrt\epsilon}.}
$$
\b[Travelling-wave impedance.] For a forward wave $V=F(t-x/v)$, take $I=V/Z_0$. Substitution into $\partial_xI+C\partial_tV=0$ gives $1/(vZ_0)=C$, hence the <characteristic impedance>
$$
\boxed{Z_0=\sqrt{\frac LC}
=\frac da\sqrt{\frac{\mu_0}{\epsilon\varepsilon_0}}.}
$$
The backward-wave current has the opposite sign relative to its voltage; the fixed ratio above applies to one travelling direction.
\b[Matching the antenna.] The voltage <reflection coefficient> is $\Gamma=(Z_{\rm load}-Z_0)/(Z_{\rm load}+Z_0)$. No reflection requires $Z_0=5\ \Omega$. With $a=0.020\ \mathrm m$, $\epsilon=1.5$, and $\sqrt{\mu_0/\varepsilon_0}\simeq376.7\ \Omega$,
$$
\boxed{d=a\frac{Z_0\sqrt\epsilon}{376.7\ \Omega}
\simeq3.25\times10^{-4}\ \mathrm m=0.325\ \mathrm{mm}.}
$$
The resulting $d/a\simeq0.016$ is consistent with the wide-plate approximation.
\b[Both transverse forces.] The matched <transverse electromagnetic mode> has electric attraction and magnetic repulsion. For time-averaged power $P=100\ \mathrm{kW}$,
$$
V_{\rm rms}=\sqrt{PZ_0}\simeq707\ \mathrm V,\qquad
I_{\rm rms}=\sqrt{P/Z_0}\simeq141\ \mathrm A.
$$
The force on one plate excludes its own field. Thus the average electric attraction and magnetic repulsion per unit length are respectively
$$
f_E=\frac{\varepsilon aV_{\rm rms}^2}{2d^2},\qquad
f_B=\frac{\mu_0I_{\rm rms}^2}{2a}.
$$
The <electric and magnetic pressure cancellation in a matched TEM line> follows directly from $V_{\rm rms}=Z_0I_{\rm rms}$:
$$
\boxed{f_E=f_B\simeq0.628\ \mathrm{N\,m^{-1}},\qquad
f_{\rm net}=0.}
$$
The equality also holds instantaneously for a single travelling wave. Quoting either $0.628\ \mathrm{N\,m^{-1}}$ alone would omit the opposing contribution. This is the ideal electromagnetic force in the stated lossless uniform-dielectric model; a standing wave would not generally have the same cancellation.
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