# Question about the permitivity of free space (from OCR paper)

• bonbon22
In summary, the answer is 2.16*10^-11. The conversation discussed the use of the constant Eo 8.85418782 × 10-12 in a problem involving an insulator between capacitor plates. The variable A represents the area of one plate, and the entire region between the plates is filled with a dielectric material. The relative permittivity, ## \epsilon_r ##, is multiplied by the absolute permittivity, ## \epsilon_o ##, to get the value for the dielectric material. This value is reasonable, around 2.4.
bonbon22
Homework Statement
A capacitor of capacitance 7.2pF consists of two parallel metal plates separated by an
insulator of thickness 1.2mm. The area of overlap between the plates is 4.0 × 10−4m2.
Calculate the permittivity of the insulator between the capacitor plates.
Relevant Equations
capacitance = Eo Er A /d
what i don' t understand about the question is why did they not use the constant Eo 8.85418782 × 10-12
in this case as there is a insulator occupying the area inbetween does that mean then their is no free space ??
Also in this equation the variable A is the area of each capacitor plate or both together ?
They mentioned "area of overall" but I am not sure what they mean by this.

## \epsilon_r ## is the relative permittivity, but they are looking for the absolute permittivity which is the product ## \epsilon_r \epsilon_o ##. ## \\ ## For the area, it is simply the area of one face plate. ## \\ ## And the entire region between the plates is filled with a dielectric material=(like a plastic, etc). If you compute ## \epsilon_r ##, I think you get something like 2.4, which is reasonable.

Last edited:
collinsmark

## 1. What is the permitivity of free space?

The permitivity of free space, also known as the electric constant, is a physical constant that represents the ability of a vacuum to permit the flow of electric charge. Its value is approximately 8.85 x 10^-12 F/m.

## 2. How is the permitivity of free space measured?

The permitivity of free space is typically measured using a device called a capacitor, which consists of two conductive plates separated by a vacuum. The capacitance of the capacitor is then calculated using the formula C = Q/V, where Q is the charge on the plates and V is the voltage across the plates. The permitivity of free space can then be calculated using the formula ε0 = C/d, where d is the distance between the plates.

## 3. What is the significance of the permitivity of free space?

The permitivity of free space is an important constant in electromagnetism, as it is used to calculate the strength of electric fields in a vacuum. It is also used to define the unit of electric charge, the coulomb (C).

## 4. Can the permitivity of free space vary?

No, the permitivity of free space is a fundamental constant of nature and is considered to be a universal constant. It is not affected by external factors and has the same value everywhere in the universe.

## 5. How was the value of the permitivity of free space determined?

The value of the permitivity of free space was first determined by scientist Henry Cavendish in the late 18th century through experiments with electric charges and distances. It was later refined by other scientists, including James Clerk Maxwell and Albert Einstein, through various experiments and calculations.

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