Question

How much potential is required to place a 7×10-3 C charge on a 3000 μF capacitor? 

Answer 1

$Q = 3000 \times 10^{-6} = 3 \times 10^{-3}\;F \\ C = 7 \times 10^{-3} \;C \\ V = \frac{Q}{C} = \frac{3 \times 10^{-3}}{7 \times 10^{-3}} = 0.428 \;V$