Permittivity of Free Space
Last Updated : 01 May, 2024
Permittivity of Free Space or Permittivity of Vacuum is the property of material that measures the material’s ability to resist the formation of an electric field within it or to permit the passage of electric flux in a vacuum, as the resistivity offered by the free space is the lowest.
Permittivity of Free Space is introduced to students in class 12. In this article, we will explore the concept of Permittivity of Free Space in detail, including its definition, value, and formula.
What is Permittivity in Physics?
Permittivity as the name suggests, must be something related to giving permission. Permittivity is the property of a matter or a medium to permit its own field to decrease the electrostatic force/electronic interaction between two charges.
Permittivity can also be defined as the measure of electric polarizability of a dielectric, polarization of material causes induction of charges which in turn creates electric fields within the material, thus opposing the electricity between the original charged particles and also storing energy at the same time.
What is Permittivity of Free Space?
The resistivity offered by the free space is the lowest and is the fundamental constant, it is referred as permittivity of free space and is denoted by ε₀. It has a value of 8.85×10-12 Farad/meter.
The permittivity of free space, often denoted by the symbol ε₀ (epsilon naught), is a physical constant that represents the ability of a vacuum or free space to permit the transmission of electric field lines. Its value is approximately 8.85×10-12 farads per meter (F/m) in the International System of Units (SI).
Vacuum Permittivity Definition
Vacuum permittivity is a fundamental physical constant that characterizes the ability of a vacuum (empty space) to permit the electric field lines to propagate through it.
Vacuum permittivity or permittivity of free space is denoted by the symbol ε₀ (epsilon naught), and it plays a crucial role in the formulation of Coulomb's law and Maxwell's equations, which describe the behavior of electric fields.
What is the Value of Permittivity of Free Space?
The value of Permittivity of free space is approximately 8.85×10-12 farads per meter (F/m).
Permittivity of Free Space is also sometimes expressed in terms of its inverse, known as the vacuum permeability (μ₀), which is related to the speed of light (c) in a vacuum:
c = 1/ε₀μ₀
Where,
- c is the speed of light in a vacuum.
- ε₀ is the vacuum permittivity.
- μ₀ is the vacuum permeability.
SI Unit of Permittivity of Free Space
The SI unit of Permittivity of free space farads per meter (F/m) in the International System of Units (SI).
Other than this, C2N-1m-2 is also unit of Permittivity of Free Space.
Dimension of Permittivity of Free Space
The Dimension of Permittivity of free space is M-1L-3T4I2.
Dielectric material also shows permittivity, it is represented as the ratio of absolute permittivity of the medium or material and the permittivity of free space, known as relative permittivity. Thus relative permittivity is a dimensionless quantity, represented as εr.
Formula for Relative Permittivity (εr) is given as:
εr = ε/ε₀
Where,
- εr is the relative permittivity of the material,
- ε is the absolute permittivity of the material, and
- ε₀ is the permittivity of the free space.
Read More,
Sample Problems on Permittivity of Free Space
Some sample problems based on the formula discussed in the article are:
Problem 1: The relative permittivity εr of a material between the plates of a capacitor is 5. If the permittivity of free space ε₀ is 8.854 ×10-12 F/m, find the permittivity i.e., ε of the material.
Solution:
Using the formula εr = ε/ε₀, we can rearrange it to find ε:
⇒ ε = εr × ε₀
Substitute the given values:
⇒ ε = 5 × 8.854 × 10-12 F/m
⇒ ε = 4.427 × 10-11 F/m
Thus, the required value of permittivity is 4.427 × 10-11 F/m
Problem 2: A capacitor with a relative permittivity of 3 is placed in a vacuum. Calculate the permittivity of the material between the plates.
Solution:
Using the formula εr = ε/ε₀, we can rearrange it to find ε:
⇒ ε = εr × ε₀
Substitute the given values:
⇒ ε = 3 × 8.854 × 10-12 F/m
⇒ ε = 2.656 × 10-11 F/m
Thus, the required value of permittivity is 2.656 × 10-11 F/m.
Practice Problems on Permittivity of Free Space
Problem 1: Calculate the capacitance of a parallel-plate capacitor with a plate area of 200cm2 and a separation distance of 2 mm filled with air. The permittivity of free space is 8.85 ×10-12 F/m.
Problem 2: The electric field between the plates of a parallel-plate capacitor is 2 × 104 N/C. If the separation distance between the plates is 0.01 m, determine the charge density on the plates. Assume the medium between the plates is vacuum.
Problem 3: A capacitor with a capacitance of 10 nF has plates initially separated by 1mm. If the separation distance is doubled, calculate the new capacitance.
Problem 4: A parallel plate capacitor with air between the plates has a capacitance of 20pF. If a dielectric material is inserted that increases the capacitance to 80pF, calculate the relative permittivity of the dielectric material.