A constant 1ma is applied to a capacitor of 1 uf. The voltage required to maintain this current increases over time. This is because the more +q which accumulates on one plate, the more the additional charge is repelled.
For more mathematically minded individuals, see the equation for capacitance. C ( farads) = q (coulombs) / V (volts)
Rearrange this equation to see that
V = q / C.
Meaning the voltage is proportional to the amount of charge stored and inversely proportional to the capacitance. This current is 0.001 C /s, so the voltage goes up to 1000 V after 1s. (0.001 C/0.000001 F) = 1000 V
