Why has the capacitor become thinner, but the electrostatic capacity has increased?
At first glance, it seems difficult, but in fact it is super simple, just press the formula of the capacitor again and it's right!
1. The reason why the capacitor becomes thinner but the electrostatic capacity increases
According to the mathematical expression C = ε × S / d, there are three methods to increase the capacitance of the capacitor:
①Increase ε (dielectric constant)
② Increase S (electrode area)
③ Reduce d (dielectric thickness)
About ①② here, it is easy to imagine visually, but about ③, on the contrary, I always feel that a thick dielectric can accumulate a lot of charge, but this is not the case. This is because the charge is accumulated on the two electrodes, not the dielectric. First, I will explain how to derive the calculation formula on the basis of understanding the above points. Below, I will list the boring mathematical formulas, please understand.
2. Derivation C = ε × S / d
Figure 1 Flat Capacitor
As shown in Figure 1, when voltage is applied across the space between the electrodes, the resulting electric field strength is E [V / m], the voltage is V [V], and the distance between the electrodes is d [m], And get formula (1).
E = V / d [V / m] …… (1)
Although the electric field is generated by the charge from the power source, if the electric field is used to describe the electric field, according to Gauss's theorem, the power line of Q / ε [root] starts from the charge of + Q [C], then in Figure 1 , The power line of Q / ε [root] starts from electrode A and then reaches electrode B.
Because the power line density and the electric field strength are the same, 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 entered from the power supply is sorted out, then the mathematical expression (3) is obtained.
Q = ε × SV / d [C] …… (3)
It can be seen from the mathematical expression (3) that because 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 capacity is set to C [F] , Then the following mathematical expression holds.
C = Q / V [C / V = F] …… (4)
Because it can be seen from this mathematical expression that the electrostatic capacity C and the charge Q are proportional, for increasing the electrostatic capacity, the larger the charge Q accumulated in the electrodes A and B of FIG. So, how to increase the charge Q? Through 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.
Simply summarize the above, that is, the smaller the distance d between the electrodes, the larger the charge Q accumulated by the electrodes A and B. Because the accumulated charge Q increases, the electrostatic capacity C also becomes larger. If you understand this way, I think everyone has a little intuitive feeling.
Through mathematical expressions (3) and (4), a similar expression (5) can be derived. We can draw conclusions through mathematical expressions: the smaller the distance d between the electrodes, the larger the electrostatic capacity C.
Then the following conclusions can be drawn.
C = ε × S / d [F] …… (5)