Urgent: A Level physics help!

When an uncharged capacitor is charged by a constant current of 4.5 μA for 60 s the pd
across it becomes 4.4 V.
(i) Calculate the capacitance of the capacitor.
capacitance ......................................... F (3)
(ii) The capacitor is charged using a circuit The battery emf is 6.0 V and its internal resistance is negligible. In order to keep the current constant at 4.5 μA, the resistance of the variable resistor R is decreased steadily as the charge on the capacitor increases.
Calculate the resistance of R when the uncharged capacitor has beencharging for 30 s.
resistance ........................................ Ω (3)
I got the first part, 6.14 x 10^-5 F, but for part two, I couldn't do it. Mark scheme says :
since V(of capacitor) was 4.4V after 60s, when t = 30s VC = 2.2 (V)
∴ pd across R is (6.0 – 2.2) = 3.8 (V)
Using V=IR, R=844kohms
My question is, why do u assume that during half the time, u get half the voltage? I swear its not linear its exponential right?

Its a variable resistor which is varied to keep the current constant. So after 1/2 the time, then 1/2 the voltage.

its talking abt the voltage of the capacitor. it says V subscript C.

Sure, but if you redid part a) for 30s instead of 60s (assuming you know the calculated capacitance), what would the answer (voltage across the capacitor) be?

well its just that the capacitance will half, irrelevant to this