Question

A power source of 6.0 V is attached to the ends of a capacitor. The charge is 12 C.

Answer 1

From your given, A power source of 6.0 V is attached to the ends of a capacitor. The charge is 12 C., the answer would be 2.0 μF.

 The formula would be:
C=q/Vab
SOLUTON:Vab=6VQ=2uAC

C=12uAC/6=2uAF

Answer 2

The electric field at any point in the region between the conductors is proportional to the magnitude Q of charge on each conductor. It follows that:

"The potential difference Vab between the conductors is also proportional to Q"

If we double the magnitude of charge on each conductor, the charge density at each point doubles, the electric field at each point doubles, and the potential difference between conductors doubles; however, the ratio of charge to potential difference does not change. This ratio is called the capacitance C of the capacitor:

$C= \frac{Q}{V_{ab}}$

Given that:

$V_{ab}=6V$ and $Q=12\mu C$ 

Lastly, the capacitance is given by:

$C= \frac{12\mu C}{6V}=2\mu F$