Charge Amplifier

For dc component, $C_1$ and $C_T$ can be considered as open circuit. Then,

$$\frac{V_o}{V_i} = \frac{10^8 + 10k}{10^8} = 1.0001 $$

$$gain = 20log1.0001 \approx 0dB$$

For ac component at specific frequency, $C_1$ and $C_T$ can be considered as short circuit. Then,

$$\frac{V_o}{V_i} = \frac{100 + 10k}{100} = 101 $$

$$gain = 20log101 \approx 40dB$$