In the previous analogy, we formed two chambers in a rigid vesssel by using a flexible membrane to divide them. This enabled us to utilise the pressure in each chamber to assist the transfer of air between the chambers and therefore to transfer more air before the limiting pressures where reached. You will now see, that in a similar way, by bringing the plates of a capacitor into close proximity, we can use the build up of charge on one plate, to increase the accumulation of charge on the other. This will enable us to transfer more charge before the voltage between the capacitor plates is equal to the battery voltage.
Assume that the plates are initially isolated and have been fully charged as described earlier.
Considering first the positive plate. It has lost electrons and has therefore built up a large positive charge. This resists the removal of further electrons, despite the “pull”, from the positive terminal of the battery. Imagine now, that we bring the strongly charged negative plate in to close proximity. The remaining electrons on the positive plate, will now experience repulsion,from the concentration of electrons on the negative plate. This effectively creates a “push” on the electrons on the positive plate, which assists the pull of the positive battery terminal. This causes more electrons to be removed from the positive plate, building up more positive charge with the same battery voltage.
Similarly, we can consider the effect of the positive plate, on the charge accumulated on the negative plate. Here, the strong attraction from the positive plate, will help pull more electrons onto the negative plate.
The net effect, is that bringing the plates into close proximity, has increased the amount of charged stored using the same battery voltage. i.e. It has increased the capacitance of the capacitor.
In fact C is proportional to 1/d.
i.e. If distance halves, capacitance doubles.
