Capacitance of a single charged plate?
$$C = \frac Q V$$

which means \$Q = C × V \$.

For a parallel plate capacitor:

$$C = \frac {8.85 × 10^{-12} \frac F m × A × k} {d} $$

Dielectric constant k = 1, since we had better assume a vacuum.  Let \$A = 1 m^2\$.

$$C = \frac {8.85 × 10^{-12} Fm } {d\ m} $$

A really small number divided by a really large number.

$$\lim_{n\to \infty} \frac 1 n = 0$$

As d approaches \$\infty\$.  C = 0, which means Q = 0.

There is an inverse relationship with separation, so as d gets larger, capacitance gets smaller.

So the math and I agree with you.