What is Permittivity?
Permittivity is the ability of a material to oppose the passing of electric field lines through it. It can be better understood by Coulomb’s law.
Let’s understand permittivity from Coulomb’s law.
Coulomb’s law states that when two charges are kept at a distance from each other. The distance between their centers is ‘r’. Then, the electrostatic force between them is given by F = k\[\frac{q1q2}{r^{2}}\].
Where k depends upon the medium, and its value is \[\frac{1}{4πε}\].
So, the above formula becomes:
F = \[\frac{1}{4πε}\] . \[\frac{q1q2}{r^{2}}\]…(1)
Here, ε is the permittivity, and it is inversely proportional to the force. This means more is the permittivity, lesser is the electrostatic force between the two charges.
So, permittivity is a property of a medium that permits its field to decrease the interaction field between the two charges.
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In this article, we are going to derive the dimensional formula of permittivity. Before, starting with our derivation, let’s understand why do we study the dimension of a physical quantity.
Dimensions of a Physical Quantity
The dimensions of a physical object are the powers to which the base (or the fundamental) units are based to represent the physical quantities.
As we know, we can derive the units from the fundamental units, i.e., mass, length, and time. So, in dimension, we represent mass by [M], length by [L], and time by [T].
Suppose we have to find the dimension of force, then we know that force = mass x acceleration.
So, mass is [M], acceleration = velocity /time = [LT⁻²]
We get for force as [MLT⁻²]
Now, let’s find out the unit and dimension of permittivity.
Unit and Dimension of Permittivity
If we have to find the unit and dimension of permittivity, we can use the above formula.
F = \[\frac{1}{4πε}\] . \[\frac{q1q2}{r^{2}}\] or ε = \[\frac{q1q2}{F4πr^{2}}\]...(1)
We know that unit of force is N or Kgms⁻²
Distance = meter (m)
Charges q1 and q2 = C² = A²s²
4, π is a constant value.
Then, the unit of permittivity is:
\[\frac{A^{2}s^{2}}{Kgms^{-2}m^{2}}\] = A\[^{2}\]s\[^{4}\]Kg\[^{-1}\]m\[^{-3}\]
We know that 1 Farad or F = s\[^{4}\]A\[^{2}\]Kg\[^{-1}\]m\[^{-2}\]
⇒ A\[^{2}\]s\[^{4}\]Kg\[^{-1}\]m\[^{-3}\] = Fm\[^{-1}\]
Here, we converted one system of the unit, i.e., s\[^{4}\]A\[^{2}\]Kg\[^{-1}\]m\[^{-2}\] to F.
So, the unit of permittivity is F/m.
Now, let’s find out the dimension of permittivity.
We know that the dimensional formula of F = [MLT⁻²]
q1 and q2 = Current x time
So, the dimension of q * q = [A²T²]
∵ 4, and π are constant values, so their dimension will be [M⁰L⁰T⁰].
Here, we know one thing that 4, and π are dimensionless constants because these constants don’t possess dimensions.
The dimension of r² = [L²]
Now, putting these values in equation (1), we get
= \[\frac{[A^{2}T^{2}]}{[MLT^{-2}] \ast [M^{0}L^{0}T^{0}] \ast [L^{2}]}\] = [M\[^{-1}\]A\[^{2}\]T\[^{4}\]L\[^{-3}\]]
This is the dimension formula for permittivity.
Unit and Dimension of Permittivity of Free Space
What is the Permittivity of Free Space?
We understood that permittivity of a medium restricts or opposes the electrostatic field between the two charges kept in itself, and it is inversely proportional to this field.
If we talk about vacuum, at vacuum the value of this permittivity is minimum. This means there is a maximum of electrostatic field lines between the two charges kept in a vacuum.
Therefore, the force between these charges is also greater.
At a minimum (in a vacuum), ε → εo
This means ε (Epsilon) becomes εo (Epsilon-naught).
The minimum value for the permittivity of free space is 8.85 x 10⁻¹²C²N⁻¹m⁻².
We know that permittivity is a physical quantity that has a constant value.
We know that the quantities that have a constant value and possess dimensions are known as dimensional constants. So, permittivity is also a dimensional constant.
Now, let’s find out the unit and dimension of permittivity of free space.
1. Unit of Permittivity of Free Space
When ε → εo, then the formula becomes εo = \[\frac{q1q2}{F4 \pi r^{2} }\]
You can see that the formula for permittivity and permittivity of free space is the same.
So, the unit of permittivity of free space = A\[^{2}\]s\[^{4}\]Kg\[^{-1}\]m\[^{-3}\] = Fm\[^{-1}\].
2. The Dimension of Permittivity of Free Space
Having the same formula as permittivity, the dimension of permittivity of vacuum is also [M\[^{-1}\]A\[^{2}\]T\[^{4}\]L\[^{-3}\]].
Q1: What is the Formula for Relative Permittivity?
Ans: We know that εr is the ratio of the capacitance of a capacitor filled with the dielectric in between the plates to the capacitance of the same capacitor with vacuum or air between the plates, i.e., without the dielectric material.
So, ε_{r} = C_{m}/C_{o}
Where C_{m} = KC_{o} = ε_{o}A/d = εA/d
Q2: What is the Difference Between Permeability and Permittivity?
Ans: The table below describes the difference between permeability and permittivity:
Permeability | Permittivity |
Permeability measures the degree of magnetization, a material gains in response to the applied magnetic field. | Permittivity is an ability of a material to oppose the electrostatic force between the two charges kept inside it. |
Q3: The Capacity of a Capacitor Becomes 20 μF When the Air Between the Two Plates is Removed by a Dielectric slab of K = 4. What is the Capacity of a Condenser with Air in Between the Plates?
Ans: Here, C_{m} = 20 μF, K = 4, C_{o} = ?
As K= C_{m}/C_{o} ∴ C_{m} = C_{o}/K = 20/4 = 5 μF
Q4: What is the Value of Absolute Permittivity of Water?
Ans: Absolute permittivity is the value of permittivity of free space. Its value is 8.85 x 10⁻¹²C²N⁻¹m⁻².