How to calculate capacitor value

Today, we will discuss the capacitance calculation law for capacitors in DC and AC circuits. We will also cover how to calculate the total equivalent capacitance of capacitors, whether connected in series or parallel, in addition to understanding the method of connecting or combining capacitors.

You can check out the online capacitor capacitance calculation program.

Capacitor Capacitance Law

The law for capacitor capacitance is given by the following relation:

C = q/V

Where C denotes the capacitor’s capacitance, measured in Farads (farad). q is the electrical charge stored on one of the plates, measured in Coulombs (coulomb), and V is the voltage (potential difference) applied across the capacitor, measured in Volts (Volt). Thus, we deduce that the capacitance of a capacitor is the ratio of the charge on one plate to the potential difference applied across it.

From the above, we conclude that 1 Farad = 1 Coulomb / 1 Volt.

However, the Farad is actually a very large unit. Therefore, parts of a Farad are used in measuring capacitor capacitance. The most common division is the microfarad, denoted by the symbol (μF).

1 microfarad = 10^-6 Farad

And sometimes we resort to the picofarad (pF) division, which equals (1/10)^-12 of a Farad.

A capacitor can be briefly defined as an electrical device or component for storing electrical charge. It is charged by applying a potential difference across its terminals. When the applied potential difference is removed, it retains the electrical charge for a period during which it discharges. Thus, according to this definition, we can write the capacitor capacitance law as a function of time according to the relation:

(t)C (t)= q/V

Factors Affecting Capacitor Capacitance

Since we know that a Farad is a large capacitance, we would need a very large capacitor to store 1 Farad of capacitance. The capacitor capacitance law according to its specifications is given by the relation:

C = (ε₀ x A) / d

ε₀ represents the permittivity of free space. Assuming the dielectric between the plates is vacuum (free space), then the permittivity varies from one dielectric to another. A is the common surface area of the plates, and d is the distance between the plates, measured in meters (SI unit).

Exercise: If you know that the permittivity ε_o = 8.85 x 10^-12 Nm² C^-2 for free space. And you have a capacitor with a distance of 1 meter between its plates. Calculate the side length of each plate of a square-shaped capacitor if its capacitance is 1 Farad.

Solution: Using the previous law:

C = (ε₀ x A) / d ⇒ 1 = (8.85 x 10^-12 x A)/1

⇒ A = 8.85 x 10¹²

Thus, if each plate is square-shaped, the side length of each is:

L = √(8.85 x 10¹²) ≈ 3000000 m = 300 Km

From this, we conclude that a Farad is an enormous capacitance, and we would need a capacitor with a side length of 300 km to store this capacitance. To reduce this enormous length, we can bring the distance closer. Therefore, the smaller the distance between the capacitor’s terminals, the greater its capacitance.

Exercises on Capacitors

1. First Exercise

A capacitor with two parallel square plates, each with a side length of 6 cm. The distance between them is 1 mm, and the dielectric between the plates is free space.

1. Calculate the capacitance of the capacitor.
2. If a potential difference of 12 volts is applied across the capacitor, calculate the charge stored on each plate (knowing that ε₀ = 8.85 x 10^-12).

Solution:

1. Applying the capacitor capacitance law:

C = (ε₀ x A) / d ⇒ C = (8.85 x 10^-12 x 36 x 10^-4)/1×10^-3

C = 318.6 x 10^-13 Farad = 31.86 pF

2. According to the law of charge stored on each side of the capacitor C = q/V, the charge on each plate after substituting the numbers is:

q = CV = 318.6 x 10^-13 x 12 = 38.232 x 10^-13 coloumb = 3.8232 pC

2. Second Exercise

A capacitor whose dielectric between its plates is Mica, with a permittivity ε_r = 5. The capacitor is supplied with a constant voltage of 12 volts. The surface area of each plate is 6 m², and the distance between them is 6 mm. Required:

1. Calculate the capacitance of the capacitor and the stored charge.

Solution

C = (ε_r ε₀ x A) / d ⇒ C = (5 x 8.85 x 10^-12 x 6)/6×10^-3 = 44.25×10^-9 F