# Why as distance decreases capacitance increases?

I understand mathematically that as the distance between the capacitor plates decreases, capacitance increases, but I find there is a small contradiction that I want to understand. If the capacitance equals the charge on the capacitor divided by the voltage, isn't the charge that is accumulated on the capacitor triggered by the force that results from the charge on the other plate, which is constant and equals to $\sigma/2\epsilon_0$?

$$CV = q \tag{1}$$
where $V$ is the potential difference (or potential) of the system, $C$ is the capacitance of the system and $q$ is the charge stored in the system.
$$V = E.d \tag{2}$$
Consider a parallel plate capacitor. If you keep the amount of charge on the system constant and then reduce the distance between the plates, the potential across the capacitor decreases. As the electric field between the plates depends solely on the surface charge density (assuming that the plates are very large), from equation $(2)$, you can infer that $V$ decreases as $d$ decreases. Therefore, you are holding the same amount of charge for a smaller potential difference. Aha! The capacitor is now able to store more charge per unit potential, therefore its ability to hold charge, that is capacitance, has increased.