The wires have a relaitvely small effective area, and are much farther apart than the capacitor plates, so the capacitance between the wires will normally be much less than that of the capacitor. 1) If the wires are right beside each other (like in a circuit board), the distance is around the same as a capacitor.
Since the whole thing acts as one big capacitor, the charge wouldn't just gather at the capacitor, it would spread out over the whole wire and the capacitor, meaning there would be less charge in the capacitor. And if this is true why doesn't the equation for capacitance take the position of the wires into account?
A capacitor is characterised by its capacitance (C) typically given in units Farad. It is the ratio of the charge (Q) to the potential difference (V), where C = Q/V The larger the capacitance, the more charge a capacitor can hold.
The capacitance C of a capacitor is defined as the ratio of the maximum charge Q that can be stored in a capacitor to the applied voltage V across its plates. In other words, capacitance is the largest amount of charge per volt that can be stored on the device: C = Q V
A capacitor will have a large plate area, with very closely placed plates, to give a large capacitance relative to its size. The wires have a relaitvely small effective area, and are much farther apart than the capacitor plates, so the capacitance between the wires will normally be much less than that of the capacitor.
If your parallel-plate cap is much bigger than the capacitance you may be able to ignore the wire capacitance. A home-made adjustable capacitor made with twisted wires is often called a "gimmick capacitor". With a voltage source, there is not any less charge on the capacitor. There's just a tiny additional charge on the wires, too.