How to select value of resistor to dicharge a capacitor?
The relation between momentary voltage across the cap \$V(t)\$, the resistance used for discharge \$R\$, the initial voltage \$V_0\$ and the capacity \$C\$ is as follows:

$$V(t) = V_0e^{-\frac{t}{RC}}$$

Lower \$R\$ will give you a faster time to reach a safe voltage, say below 100V. The problem with this is, that the bleeder resistor will constantly drain current from your power source and turn it into heat. Assuming we want to reach 100V within the first 10 seconds of power-off. Solving the above equation for \$R\$ yields:

$$R=\frac{t}{\ln(V_0/V)C}$$

Then we'll need a resistor equal to or higher less than 440kOhm.
It will dissipate a power equal to $$(10\,\text{kV})^2/440\,\text{k}\Omega \approx 230\,\text{W} $$
Oops, that's alot. Can your power supply give 230W just for the bleeder? From this simple calculation you can see that making this PS safe after power-off won't be easy (let alone during power-on, but that's off-topic).

Thus my suggestions:


Use a lower-value switched resistor (~500kOhm or less), which, under normal power-down conditions, discharges the caps quickly. You might get away with peak power ratings for the resistor instead of continuous power ratings here. (see HV switching)
Use something on the order of 13MOhm as permanent bleeder, that'll give you around 5 Minutes discharge if anything goes wrong with the switched bleeder.
If any high voltage is accessible from outside, add a signal (e.g. an LED) to make sure the user is aware of the fact that there's still a dangerous voltage. This LED should also be on in normal power-on operation.
Add a similar signal to the board itself, incase someone is working on it with opened case (most probably you).


Using yourself as a bleeder for 10kV in 5µF will kill you.