# 1.05 An analogy to explain how the area of the plates affects the capacitance of a capacitor.

Consider the analogy with the air pump described previously. However, in this case rather than just blocking the inlet and outlet, consider connecting them to two rigid sealed vessels. Vessel A is connected to the inlet and vessel B is connected to the outlet. If we turn the pump handle with the same constant force as before, we will find that we can now transfer more air before the handle stops turning.
Considering first the outlet:- We have now increased the volume at the outlet, so more air can be transferred, before the air in chamber B, gets compressed to the level at which the handle stops turning.
Considering the inlet:- there is now more air at the inlet (from chamber A), that has to be removed, to create the level of suction at which the handle will stop turning.
The net result, is that by increasing the volume at the input and output, we can now displace more air, before we create the same limiting pressure levels as before.

In the next section, you will see that in a similar way for an electrical circuit, we can transfer more charge, by increasing the area available for storing charge. (Note. In our “air-pump analogy”, we would tend to think of an accumulation of air in chamber B and loss of air from chamber A. However when we think of an electrical circuit, we consider the accumulation of charge at both plates, ( i.e. the removal of electrons from the positive plate, results in the accumulation of positive charge!).