Capacitance Formula

A capacitor is an electronic device about which quite a few people know. Also, after going through this topic you will be able to define capacitance, capacitance formula, and will be able to solve question-related to capacitance.

Capacitance

It is an electric device which is practically present in almost every electronic device. Moreover, every electronic device has a capacitor of some kind in it. Furthermore, the capacitor store charge (electrical charge) in them until they get full and release it in the form of burst.

Also, the capacitor helps the device to store charge so that it does not die instantly in the absence of external power and allow the recovery process to complete. Besides, you need a circuit to get regular ‘pulse’ of energy every x amount of time. Furthermore, capacitors come in a wide variety of sizes that can hold from a tiny amount of energy to a large amount of energy.

For example, we use capacitors on television, high-band pass filters, computer, car starters, and many more. In addition, nearly every electronic device we use includes a capacitor. Besides, the capacitance is the measure of a capacitor’s capability to store a charge that we measure in farads; also, a capacitor with a larger capacitance will store more charge.

Capacitance Formula

The capacitance formula is as follows:

C = \(\frac {Q}{V}\)

Derivation of the Formula

C = refers to the capacitance that we measure in farads
Q = refers to the equal charge that we measure in coulombs
V = refers to the voltage that we measure in volts

Besides, there is another formula which appears like this:

C = \(\frac{k \varepsilon_{0}A}{d}\)

Derivation 

C = refers to the capacitance
K = refers to the relative permittivity
\(\varepsilon_{0}\) = refers to the permittivity of free space
A = refers to the surface area of the plates
d = refers to the distance between places measured

Solved Example on Capacitance Formula

Example 1

Assume that you have a capacitor of area 0.1 meters squared, that has plates 0.01 meters away from each other. Also, there is air between the plates. So, what will be the charge that the plates can store if it is connected to a 9V battery?

Solution:

For solving this firstly, write down what we know. So, the area is 0.1 meters squared, then A = 0.1. Moreover, the plates are 0.01 meters distant from each other, so d = 0.01. Furthermore, there is air between the plates thus K is approximately 1. Also, the voltage is also given which is V =9 volts. Besides, we have to find the charge that is Q.

In addition, we can’t solve for Q since we have V, but we do not have C. Thus, we will need to find C first.

Let’s plug numbers into that equations, we get the capacitance is equal to 1 multiplied by 8.854 × \(10^{-12}\) multiplied by 0.1 and divided by 0.01; that gives us a capacitance of 8.854 × \(10^{-11}\) farads. So, we can now plug that capacitance into the first equation, and rearranging algebraically to make Q the subject, we find that the charge Q is equal to 8.854 × \(10^{-11}\) multiplied by 9 that is 8 × \(10^{-10}\) coulombs.

Or

C = \(\frac{k \varepsilon_{0}A}{d}\)
C = 8.854 × \(10^{-12}\) × 0.1 ÷ 0.01 = 8.854 * \(10^-{11}\)
Q = 8.854 × \(10^{-11}\) × 9
Q = 8 × \(10^{-10}\) coulombs.

Example 2

Suppose a parallel plate capacitor contains two plates with a total surface area of 100 \(cm^{2}\). Then what will be the capacitance in pico-farads, (pF) of the capacitor if the plate separation is 0.2 cm? and the dielectric medium is air.

Solution:

C = \(\varepsilon \frac{A}{d}\) , \(\varepsilon\) = 8.85 pF/m
A = 100\(cm^{2}\) = 0.01 \(m^{2}\)
d = 0.2 cm 0.002 m
So, C = 8.85 × \(10^{-12}\) × \(\frac{0.01 m^{2}}{0.002 m}\) = 44 pF
So, the answer is 44 pF.