Two conductors, which are isolated from each other and placed near, form a condenser.
The conductors forming the capacitor are charged with the same size and different charges.
A wide application in practice finds a flata capacitor consisting of two flat parallel metal plates separated by a dielectric layer. The distance between the plates is small in comparison with their size. Condenser plates are called capacitor plates.
To charge the plates equal to the oppositecharges, you can attach them to the poles of the electrical machine. Then a negative charge will pass to one plate, while the other will receive a positive charge.
You can connect one of the plates to the pole of the machine,and the other is grounded; Then on the other plate, by induction, there will be a charge of the same charge and equal in charge to the charge of the first plate. If the plate A is charged positively, then the plate B is negatively charged by induction; The positive charge of plate B is neutralized by electrons that have flowed to the plate from the ground, which is practically an inexhaustible source of them. When the positive charge B is drawn to the positively charged plate A, it will be located along the inner surface of the plate facing A.
If the plate A is charged negatively, then the free electrons of the plate B repel the plate A and go to the ground, the plate B is charged positively.
In both cases, the charges are concentrated only on the surfaces A and B facing each other.
The absence of charges on the external surfaces givesThe possibility of completely transferring charges to the capacitor by means of the outer sides of the plates. The charge of the capacitor is determined by the charge of one of its plates, since on the other by induction there is an equal in magnitude charge.
We connect one plate of the condenser with a rodelectrometer, and the other plate and the body of the electrometer are grounded. Using a test ball, we will transfer the capacitor in series in equal portions. We note that when the charge is increased by 2, 3, 4 or more times, respectively, in 2, 3, 4 or more times, the potential difference of the capacitor increases.
This value, which is measured by the ratio of the charge of the capacitor to the potential difference of its plates (or plates), is the capacity of the capacitor.
Denoting it with the letter C, we can write:
C = q / (φ1 - φ2).
The electric field of the capacitor is practically concentrated between the plates inside it, so the surrounding bodies do not affect the capacity of the capacitor.
We will carry out the experiment. Take a flat capacitor consisting of two metal plates A and B, fixed on insulators.
We connect the plate A to the electrometer, and the plateIn the ground. We charge the plate A, the electrometer will note a certain difference in potentials of the capacitor. If the plate B is approached to A, then it can be seen that the potential difference of the plates decreases.
Reducing the potential difference of the plates of the capacitor with an unchanged charge on it indicates an increase in its capacitance.
Thus, the capacitance of a flat capacitor will be the greater, the smaller the distance between the plates or the smaller the thickness of the dielectric enclosed between the plates.
By shifting the plate B relative to the plate A upwardsand down, we will change the area of the plates overlapping each other. While observing the readings of the electrometer, it can be established that the larger the area of mutually overlapping plates of the capacitor, the greater its capacitance. The larger the area of the capacitor plates, the greater the charge can be concentrated on them for a given potential equality.
We will make one more experience. To do this, we arrange the plates of capacitor A and B at some distance from each other, and charge plate A.
Note the magnitude of the potential difference, when inThe role of the dielectric is played by air. Now place between the plates a glass sheet or any other dielectric; we see that the potential difference decreases with them. To raise it to its previous level, it is necessary to add a charge to the plate A. It follows that the replacement of the air layer between the plates of the capacitor by any other dielectric increases its capacitance.
Let C₀ be the capacity of the capacitor, when between its plates there is a void or air, and C is its capacitance with the use of a dielectric.
Dividing C and C₀, we find the dielectric permittivity of the dielectric ԑ:
ԑ = С / С₀.
Thus, the greater the permittivity of the dielectric, the greater the capacitor's electrical capacitance.