When a pin is set to analogue or digital input it will have a very high impedance and the typical current that will flow into/out of the pin will be less than 1uA. In your example the only current that will flow will be through the 10k pot and will follow ohms law. In this case 5/10,000 A (500uA) will flow through the resistor and less than 1uA will flow through the 1k resistor into the pic pin. The 1k resistor and capacitor are not really needed unless you are doing very fast acquisitions and even then it would be rare they would be required.
A typical pullup resistor would be 20k and this is all that is required. The 1-4.7k resistor in Burt's diagram is not needed. And has has already been mentioned it is more normal to have pullups rather than pulldowns
Edit, BTW, the above assumes you are not going to do something silly like making the pin an output.
Mike.