Tlhaloso ea Bokhoni ba li-capacitor
Khalase e nyane e ka ba le metsi a manyane, ha khalase e kgolo e ka ba le metsi a mangata. Ha khalase e le kgolo, metsi a mangata a ka ba le metsi a mangata. Kahoo, khalase ka 'ngoe e na le bokgoni kapa boholo ba bokgoni ba ho ba le metsi. Joalo ka khalase, di-capacitor di ka boela tsa boloka ditefiso tsa motlakase le matla a motlakase. Moqapi bokgoni ba ho boloka tjhaja ya motlakase le matla a motlakase bo bitswa bokhoni.
Mabaka a amang bokgoni ba ho sebetsa
Boholo ba bokgoni ba khalase ba ho boloka metsi bo laolwa ke bophahamo ba khalase. Ho thoe'ng ka di-capacitor, ke eng e etsang qeto ya boholo ba bokgoni ba capacitor ba ho boloka tjhaja ya motlakase?
The figure on the side shows a simple capacitor consisting of two-conductor plates separated by a certain distance. Before connecting to a voltage source, such as a battery, the two plates are not electrically charged. Then one of the plates is connected to the positive pole of the battery and the other plate is connected to the negative pole of the battery using a cable.
After connected to the battery’s positive pole, the positive charge on the battery draws negatively charged electrons on the plate so that the electron moves to the positive pole of the battery. This causes the plates to lack electrons (negative charge) and excess protons (positive charge) so that the plate becomes positively charged.
After being connected to the battery’s negative pole, the positive charge on the plate attracts negatively charged electrons on the negative pole of the battery so that the electrons move to the plate. This causes the plate to overload the electron so that the plate becomes negatively charged. The process of moving electrons between plates and batteries stops after the potential difference between the two plates is equal to the potential difference between the two battery poles.
How do you increase the electrical charge on both conductor plates? In other words, what should be done to transfer electrons to and from the positive pole of the battery? Electron displacement occurs only when the electric potential difference between the two battery poles is greater than the electric potential difference between the two conductors. In order for the electron to move again so that the electrical charge in each conductor plate increases, the battery used is replaced by another voltage source that has a greater electrical potential difference. The displacement of electrons stops when the potential difference in the voltage source is equal to the potential difference of the capacitor. Therefore, if the potential difference in the voltage source is greater than, the capacitor potential difference is also greater.
Based on the above review, it can be concluded that the greater the electrical charge stored in each conductor plate, the greater the electrical potential difference between the two conductor plates. So, the electric charge (Q) is proportional to the electric potential difference (V). The relationship between an electric charge and an electric potential difference is stated in the following comparison:
Q α V
The above comparisons are converted into equations by adding the constant C:
Q = C V or C = Q / V
Q = electric charge (Coulomb), V = electric potential difference or voltage (Volt), C = constant which is called the capacitor capacitance.
The capacitance value does not depend on the electrical charge and electrical voltage, but depends on the shape and size of the conductor plate. Mathematical proof that capacitance depends on the shape and size of the conductor plate is explained in the topic of the parallel plate capacitor. In the topic, it is assumed that between the two conductors there is a vacuum.
The capacitance of a capacitor also depends on the nature of the material that is between the two conductor plates. The material that is between the two conductor plates is called a dielectric. The capacitance of capacitors that have dielectrics is discussed in depth in the topic of dielectric constants.
Unit of capacitance
The unit of electric charge is Coulomb and the unit of the electric potential difference is Volt, so that based on the capacitance equation above. The capacitance unit is Coulomb per Volt (C/V), also called Farad (F) which comes from the name of the British scientist Michael Faraday (1791-1867). So, 1 Farad = 1 Coulomb/Volt.
Suppose that a capacitor having a value of 2 Farad means that the capacitor stores an electrical charge of +2 Coulomb on one of the conductor plates and -2 Coulomb on the other conductor plate. Where the two conductor plates have a potential difference of 1 Volt. If the 12 Volt battery is connected to the capacitor, one of the electrically charged conductor plates is Q = C V = (2) (12 Volts) = +24 Coulomb while the other conductor plate is -24 Coulomb.
Kindly note that Farad is a huge capacitance unit so that it is usually used as a smaller unit, namely microfarad abbreviated μF (10-6 Farad) or picofarad abbreviated as pF (10-12 Farad). Mathematical calculations to show that Farad is a giant unit, discussed in the topic of the parallel plate capacitor.
