You may have better luck simulating your voltage source as an equivalent resistor with that amount of thermal noise, as a resistor with no parasitic inductance or capacitance has a flat noise amplitude if you keep the bandwidth the same for each measurement.
Equally you really need to specify the specific bandwidth of your current measuring device, as infinite bandwidth = infinite white noise power = infinite white noise current, how I have approached this in the past has been to break it up into many little steps to brute force it.
So you have a noisy 10000 ohm thermal resistor, lets say you look at it in slices 1Hz apart e.g. 1-2Hz, that is a bandwidth of 1Hz (same value for 100-100 or 25000-25001, if it is different than 1, then it becomes the square root of the bandwidth), at a temperature of 20C it has a thermal noise of 12.72nV RMS,
your resistor has some resistance so it can equally be modeled as a noise current, in this case 1.272pA RMS (12.72E-9 / 1E4), this is with all of the current flowing through that resistor, so anything in parrellel will divert some of this, but never all of it
As your capacitor has some reactance for the frequency you are looking at for that given step (technically impedance, as all capacitors have some equivalent series resistance) you can treat it like finding the current in 1 leg of 2 parallel resistors,
E.g. the step of 1000-1001Hz, the capacitor has an impedance of roughly 159 Ohm, so the current through the capacitor would be Noise Current * (Rz / (Cz + Rz)) note that this value changes with each frequency step, so you would be left to root square sum all of these currents over the bandwidth your interested in,