polarised condensor microphone
On the input side of \$C_c\$, you have a voltage source (I assume of value E0) connected through \$R_c\$ to three capacitive loads \$C_s\$, \$C_t\$ and your signal \$\Delta\$C(t).  In the steady state, since there's no DC path to ground and \$\Delta\$C(0)=0, the junction of \$R_c\$ and the capacitors will be E0.

However, when 
\$\Delta\$C(t) changes, the coulomb count in your capacitor array cannot change instantly, so as a result the voltage changes.  The change in voltage is proportional to the change in capacitance, which is

\${\Delta C(t)\over C_s + C_t}\$

Multiply that by E0 and you've got the dynamic component of your waveform.