electrostatics - Why does the area of the plates affect the capacitance?

Why does the area of the plates affect the capacitance? Lets say I have a parallel plate capacitor with a charge of 10C and a potential difference of 5V. By the definition $C=Q/V$, the capacitance is 2 farads. Now if I increase the area of the plates, the charge definitely doesn't change. By the definition $V=kq/r$, the voltage doesn't change either. So why does the capacitance increase if we increase the surface area?

Answer

The capacitance is the ratio of charge on the plates over the voltage applied. $$C = \frac{Q}{V} \Leftrightarrow Q = C \cdot V$$ The calculation you show determines the capacitance from measured voltage and charge on the plates. You basically know the result you want and determine the size of the capacitor you need.

A larger capacitor, with a larger capacity, will hold a bigger charge at the same voltage. Doubling the area will double the capacitance (in case of a plate capacitor), so for 4 farads of capacity you get $$Q = C \cdot V = 4 F \cdot 5 V = 20 C$$

The pysics works as follows: The voltage is a driving force, pushing electrons through the wires an onto the plates of the capacitor (or sucking them off on the positive pole), until the mutual repulsion of the electrons leads to a balance of foces. If you have a larger plate, the charge can distribute over a larger area, there is less "pileup" and therefore a smaller "pushback force". This is why, with larger plates, you get a bigger charge into your capacitor with the same voltage.