Bias-T circuit design
C = 1 / (2 * 3.14 * 2000MHz * 2Ohms) = 80pF

This calculation gives you a capacitive reactance of 2 ohms at 2000 MHz, and proportionally increasing reactance at any frequency below 2000 MHz. That is, it will give you 4 ohms at 1000 MHz, 40 ohms at 100 MHz, and 400 ohms at 10 MHz. But what you want is to have very low capacitive reactance throughout the pass-band of the bias tee.
So you should choose the capacitor to have a low reactance at your lower band edge, and then it will have even lower reactance throughout the operating band. You might choose to make it 50 ohms (for 3 dB insertion loss in a 50 ohm system) at 10 MHz or some lower value (but that will move your 3-dB &quot;cut-off&quot; frequency below the 10 MHz you specified.
You will of course also have to make sure that the inductive parasitic behavior of your capacitor does not produce too much inductive reactance at the upper end of your frequency band.