Parallel plate capacitor

Definition of the parallel plate capacitor

Parallel plate capacitor 1The parallel plate capacitor is a capacitor that consists of two parallel conductor plates, each plate having an equal cross-sectional area (A) and two plates separated by a certain distance (d), as shown in the figure left. One of the conductor plates is positively charged (+Q) while the other conductor plate is negatively charged (-Q), where the amount of electric charge on each plate is equal. So that the charge does not move to the air molecule, the capacitor is isolated from the environment, and between the two plates, there is a vacuum.

The difference in the types of electrical charges in both conductor plates creates an electric field and an electric potential difference between the two plates. The positively charged plate has a higher electric potential while the negatively charged plate has a lower electric potential. As explained in the topic of the electric potential, if there is an electric potential difference between the two plates of the capacitor, there is an electric potential energy in the capacitor. The electric potential energy stored in capacitors has many uses.

Factors affect the capacitance of the parallel plate capacitor

Parallel plate capacitor 2The capacitor at first is not electrically charged or electrically neutral. In order to be charged, charging the capacitor is done by connecting the capacitor with a voltage source such as a battery using a cable. One of the conductor plates is connected to the positive pole of the battery while the other conductor plate is connected to the negative pole of the battery.

Remember that like charges repel each other while unlike charges attract each other. The electrons are negatively charged and easy to move because they are on the surface of the atom, whereas protons are positively charged and cannot move because they are at the atomic nucleus. If the conductor plate has more protons than electrons, the plate is positively charged, on the contrary, if the number of electrons is more than the proton, the plate is negatively charged.

After the capacitor is connected to the battery, the positive pole of the battery is positively charged so that it pulls the electron from the conductor plate while the negative pole of the battery is negatively charged so that it rejects the electron to the conductor plate. The electric displacement between the battery and the conductor plate causes the conductor plate to lose electrons to be positively charged and the conductor plate receives electrons to be negatively charged. The movement of electrons stops after the electric potential difference between the two conductor plates is equal to the electric potential difference between the two battery poles.

Capacitor function to store the electric charge and the electric potential energy. The size of the capacitor’s ability to store the electric charge and the electric potential energy is called capacitance. The more electricity is stored, the electric potential energy stored is also greater, the greater the capacitance of the capacitor. What factors influence the capacitance of a parallel plate capacitor?

The surface area of the conductor plate

The parallel plate capacitor store the electric charge on the conductor plate. If the plate surface area is small then little electric charge is stored, otherwise if the plate surface area is large then a lot of the electric charge is stored. The more electric charges stored on the conductor plate, the greater the electric potential energy possessed by the capacitor. So the greater the electric potential energy of the capacitor, the greater the capacitance of the capacitor. Based on this review, it can be concluded that the capacitance of the capacitor (C) is proportional to the surface area of the conductor plate (A).

The distance between the two conductor plates

The transfer of electrons stops after the potential difference between the two conductor plates is equal to the potential difference between the two battery poles. How do you increase the electric charge on both conductor plates? One way is to reduce the distance between the two conductors (minimizing d). When the two conductors are brought closer, the amount of electric charge remains so that the electric field produced by the electric charge is fixed. Based on the equation V = E d, when the electric field (E) is constant, the electric potential difference (V) decreases if the distance between the two conductor plates decreases (d).

The potential difference between the two conductor plates is reduced so that it is smaller than the potential difference between the two battery poles. This causes the displacement of electrons again until the electrical charge of each conductor plate increases. The transfer of electrons stops after the potential difference between the two plates is equal to the potential difference between the two battery poles.

When the distance between the two conductor plates is reduced, the electric charge on each plate increases so that the electric potential energy of the capacitor also increases. If the electric potential energy of the capacitor increases, the capacitance of the capacitor also increases. If the distance is reduced, the charge increases so that the capacitance increases, when the distance is increased, the charge decreases so that the capacitance decreases. It can be concluded that the capacitance of the capacitor (C) is inversely proportional to the distance between the two conductor plates (d).

The equation of the capacitance of the parallel plate capacitor

Previously, things that affect the capacitance of parallel plate capacitors have been explained. The capacitance value can be clearly known through calculations using the equation. In a topic about determining the electric field using Gauss’s law, the formula for calculating the electric field near the electrically charged conductor plate is E = σ / εo, where σ = Q / A so that the electric field equation changes to E = Q/A: εo = Q/A x 1/εo = Q / A εo. The equation of the electric potential is V = E d, where E = Q / A εo so that the equation changes to V = Q d / A εo. The equation of capacitance is C = Q/V, where V = Q d / A εo so that the equation of capacitance changes to C = Q: Q d / A εo = Q x A εo / Q d = A εo / d.

Based on the equation of capacitance C = A εo / d it can be concluded that capacitance (C) is proportional to the surface area (A) and inversely proportional to the distance (d) between the two conductor plates.

E = electric field, σ = charge density, εo = permittivity vacuum = 8.85 x 10-12 F/m, Q = electric charge, A = surface area of the conductor plate, V = electric potential difference.

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