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 $\sigma/2\epsilon_0$?


1 Answer 1


By definition, capacitance measures the ability of a system to hold charge. Mathematically, it can be stated as the amount of charge the system can hold per unit potential difference across it.

$$CV = q \tag{1}$$

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.

An electric field is what causes a change in potential. The relationship between the change in electric potential and the electric field its simplest form can be expressed as follows:

$$V = E.d \tag{2}$$

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.