S parameters of a capacitor
ThePhoton has given a very good answer. However, below is an expanded version with complete derivation of the formula and check in LTSpice, as well as power transfer analysis. 

I. The S-matrix for the 2-port system is 

$$\begin{pmatrix}
{\frac{Z}{{2{Z_0} + Z}}}&amp;{\frac{{2{Z_0}}}{{2{Z_0} + Z}}}\\
{\frac{{2{Z_0}}}{{2{Z_0} + Z}}}&amp;{\frac{Z}{{2{Z_0} + Z}}}
\end{pmatrix},$$  

where \$Z \equiv {1 \over {j\omega C}}\$, \${Z_0}\$ is the real characteristic impedance of the system. The matrix is true for any Z, both with resistive and reactive components. 



II. Derivation 

For real   \${Z_0}\$
$${a_i} = {{{V_i} + {I_i}{Z_0}} \over {2\sqrt {{Z_0}} }},\:{b_i} = {{{V_i} - {I_i}{Z_0}} \over {2\sqrt {{Z_0}} }}$$
Source: "Power Waves and the Scattering Matrix", K. Kurokawa, IEEE, 1965, URL.  

From voltage divider rule
$${V_1} = {V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}},\:{V_2} = {V_s}{{{Z_0}} \over {2{Z_0} + Z}}$$

From Ohm's law
$${I_1} = {{{{V_s}} \over {2{Z_0} + Z}}}$$  

Then the value for the input reflection coefficient is 

$${S_{11}} = {\left. {{{{b_1}} \over {{a_1}}}} \right|_{{a_2} = 0}} = {{{V_1} - {I_1}{Z_0}} \over {{V_1} + {I_1}{Z_0}}} = {{{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} - {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \over {{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}}} = {Z \over {2{Z_0} + Z}}$$  

Applying \${I_2} =  - {I_1}\$, the value for the forward gain is
$${S_{21}} = {\left. {{{{b_2}} \over {{a_1}}}} \right|_{{a_2} = 0}} = {{{V_2} - {I_2}{Z_0}} \over {{V_1} + {I_1}{Z_0}}} = {{{V_2} + {I_1}{Z_0}} \over {{V_1} + {I_1}{Z_0}}} = {{{V_s}{{{Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \over {{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}}} = {{2{Z_0}} \over {2{Z_0} + Z}}$$ 

Note that \${S_{11}} + {S_{21}} = 1\$, what was expected if no incoming wave at the second port   (\${a_{2}}=0\$). 

III. Conservation of energy  

As for the law of conservation of energy
$${\left| {{a_1}} \right|^2} - {\left| {{b_1}} \right|^2} = {\left| {{b_2}} \right|^2} + \Delta, $$
where the \${\left| {{a_1}} \right|^2} - {\left| {{b_1}} \right|^2}\$ is the power supplied by the generator, \${\left| {{b_2}} \right|^2}\$ is the power consumed by the load with the delta being the power consumed by the DUT. 

Pls, refer to the same paper by  Kurokawa if not clear why these particular expressions are used to calculate power transfers.

Let's show that, in case of a capacitor, \$\Delta \$ is zero (no power is consumed by the DUT). Remembering \$Z \equiv {1 \over {j\omega C}}\$ 

$${\left| {{a_1}} \right|^2} - {\left| {{b_1}} \right|^2} = {\left| {{{{V_1} + {I_1}{Z_0}} \over {2\sqrt {{Z_0}} }}} \right|^2} - {\left| {{{{V_1} - {I_1}{Z_0}} \over {2\sqrt {{Z_0}} }}} \right|^2} = {1 \over {4{Z_0}}}\left( {{{\left| {{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \right|}^2} - {{\left| {{V_s}{{Z + {Z_0}} \over {2{Z_0} + Z}} - {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \right|}^2}} \right) = {{{V_s}^2} \over {4{Z_0}}}\left( {{{\left| {{{2{Z_0} + Z} \over {2{Z_0} + Z}}} \right|}^2} - {{\left| {{Z \over {2{Z_0} + Z}}} \right|}^2}} \right) = {{{V_s}^2} \over {4{Z_0}}}\left( {1 - {{{{\left| Z \right|}^2}} \over {{{\left| {2{Z_0} + Z} \right|}^2}}}} \right) = {{{V_s}^2} \over {4{Z_0}}}\left( {1 - {{{1 \over {{C^2}{\omega ^2}}}} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}}} \right) = {{{V_s}^2} \over {4{Z_0}}}{{4Z_0^2} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}} = {V_s}^2{{{Z_0}} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}}$$

Now, the same for the transmitted wave:
$${\left| {{b_2}} \right|^2} = {\left| {{{{V_2} - {I_2}{Z_0}} \over {2\sqrt {{Z_0}} }}} \right|^2} = {\left| {{{{V_2} + {I_1}{Z_0}} \over {2\sqrt {{Z_0}} }}} \right|^2} = {1 \over {4{Z_0}}}\left( {{{\left| {{V_s}{{{Z_0}} \over {2{Z_0} + Z}} + {{{V_s}} \over {2{Z_0} + Z}}{Z_0}} \right|}^2}} \right) = {{{V_s}^2} \over {4{Z_0}}}{\left| {{{2{Z_0}} \over {2{Z_0} + Z}}} \right|^2} = {{{V_s}^2} \over {4{Z_0}}}{{4Z_0^2} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}} = {V_s}^2{{{Z_0}} \over {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}}} = {\left| {{a_1}} \right|^2} - {\left| {{b_1}} \right|^2},\, \Rightarrow \Delta  = 0.$$

IV. Extraction of s-parameters in LTSpice and check of the analytical formula  

Finally, let's extract an s-parameter with LTSpice and compare it with our analytical formula. Let's do it for, say, \${S_{21}}\$:

$$\left| {{S_{21}}} \right| = \left| {{{2{Z_0}} \over {2{Z_0} + Z}}} \right| = {{2{Z_0}} \over {\sqrt {4Z_0^2 + {1 \over {{C^2}{\omega ^2}}}} }} = 0.30,\,$$
at \${Z_{0}}\$=50 Ohm, \$\omega\$=10 MHz and C=50 pF.

Extraction in LTSpice:  

  

What this chart says is that
1) at lower frequencies forward gain is low (nothing breaks through to the load), as the capacitor behaves as open,
2) at higher frequencies forward gain is close to one: incoming power wave is fully transmitted to the load as the capacitor behaves as short.