Two identical thin metal plates has charge $$q_1$$ and $$q_2$$ respectively such that $$q_1 > q_2$$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is:

A

$$\frac{q_1+q_2}{C}$$

B

$$\frac{q_1-q_2}{C}$$

C

$$\frac{q_1-q_2}{2C}$$

D

$$\frac{2(q_1-q_2)}{C}$$