According to coulomb’s law, the force of interaction between two charges q1 and q2 separated by a distance r in vacuum is
f0 = 1 /4π∈0 q1 q2 / r2
Where ∈0 is absolute electrical permittivity of free space.
If the same two charges are held the same distance apart in a material medium then the force of interaction between them is
fm = 1 / 4π∈ q1 q2 / r2
Dividing (6) by (7) we get
f0 / fm = ∈/∈0 = ∈r or k
Where ∈r is relative electrical permittivity or dielectric constant (k) of the material medium.
From we may also define dielectric constant of a medium as the ratio of the electrostatic force of interaction between two given point charges held certain distance apart in vacuum/air tot eh force of interaction between the same two charges held the same distance a part in the material medium.
Faraday observed that when an insulating material is fully occupying the space between the charged plates of a capacitor, its capacitance increases. The factor by which capacitance is multiplies depends on the nature of the dielectric introduced, and is called relative permittivity (∈) or dielectric constant (k) of the material, we may write.
∈r or k = capacitance of a capacitor with dielectric in-between the plates / capacitance of the same capacitor with vacuum / air in between the plates
∈r or K = Cm / C0
Thus Cm = KC0 = ∈r ∈0A / d = ∈A / d
The values of ∈r or K for common dielectric materials are given in table
Table (d) dielectric constant for some common dielectrics
||ε r or K
||ε r or K
|Air (I atm)
||6 – 7
|Air (100 atm)
||3 – 4
||6 – 7
The capacity of a capacitor becomes 10 u F when air between the plates is replaced by a dielectric slab of K = 2 what is the capacity of the condenser with air in between the plates?
Sol: here Cm = 10 μF,
C0 =? K = 2
As k = Cm / C0 ∴ C0 = Cm / K = 10 / 2 = 5 u F
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