Why does the electrostatic capacitance increase when the capacitor becomes thinner?

　　1. The reason why the capacitance increases even though the capacitor becomes thinner

　　According to the mathematical expression C = ε × S / d, there are three ways to increase the electrostatic capacitance of the capacitor:

　　①Increase ε (dielectric constant)

　　②Increase S (electrode area)

　　③Reduce d (dielectric thickness)

　　It is easy to imagine ①② here, but it is the opposite for ③. It is always thought that thick dielectrics can accumulate a lot of charge, but this is not the case. This is because the charge accumulates on the two electrodes, not in the dielectric. First, I will explain how to derive the calculation formula based on the above points. Below, I will list boring mathematical formulas, please forgive me.

　　2. Derivation of C = ε × S / d

　　Figure 1: Flat-plate capacitor

　　As shown in FIG1 , when a voltage is applied across the space between electrodes, the generated electric field strength is E[V/m], the voltage is V[V], the distance between electrodes is d[m], and equation (1) is obtained.

　　E=V/d [V/m] ……(1)

　　Although the electric field is generated by the charge from the power supply, if the electric field is described by electric lines, according to Gauss's theorem, Q/ε[roots] of electric lines start from the charge of +Q[C], so in Figure 1, Q/ε[roots] of electric lines start from electrode A and then reach electrode B.

　　Since the electric line density is the same as the electric field strength, if the area of ​​the electrode is set to S [m2], then the relationship of mathematical expression (2) holds.

　　V/d=(Q/ε)/S ……(2)

　　If the charge Q entering from the power supply is rectified, the mathematical expression (3) is obtained.

　　Q = ε × SV / d [C] ……(3)

　　From mathematical expression (3), it can be seen that since the charge Q is proportional to the applied voltage, the performance of the capacitor is better reflected by the amount of charge accumulated per unit applied voltage. If the electrostatic capacitance is set to C[F], then the following mathematical expression holds.

　　C=Q/V [C/V=F] ……(4)

　　Since it can be seen from this mathematical expression that the electrostatic capacitance C is proportional to the charge Q, the larger the charge Q accumulated by electrodes A and B in Figure 1 is, the better it is to increase the electrostatic capacitance. So, how can we increase the charge Q? From the mathematical expression (3), it can be seen that the charge Q is inversely proportional to the distance d between the electrodes. In other words, the smaller the distance between the electrodes, the larger the charge Q.

　　To simply summarize the above content, the smaller the distance d between the electrodes, the greater the charge Q accumulated by the electrodes A and B. Because the accumulated charge Q increases, the electrostatic capacitance C also increases. If you understand it this way, I think everyone will have a slightly more intuitive feeling.

　　From mathematical expressions (3) and (4), we can derive a similar expression (5). We can conclude from the mathematical expression that the smaller the distance d between electrodes, the larger the electrostatic capacitance C.

　　Then the following conclusion can be drawn.

　　C = ε × S / d [F] ……(5)