Question

In M.K.S system of units permitivity of free space equals

Answer 1

The MKS units of permittivity of free space is C²N⁻¹m⁻².

We have to find the MKS system of units of permittivity of free space.

MKS system of units :

• MKS is just the short form of Meter - Kilogram - second.
• SI unit is nothing but MKS system.

Permittivity :

• Permittivity is an intrinsic property of a material.
• It is the ability of a material to store the energy within the material.
• It measures the resistance offered against the formation of electric field.

In Coulomb's law of electric force between two charges,

$F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Here ∈₀ is the permittivity of free space.

Now you can easily find the permittivity of free space using units' concept.

∴ units of F = (units of q₁q₂)/(units of ∈₀ × units of r²)

Here,

• MKS units of charge, q = C
• MKS units of Force, F = N
• MKS units of distance, r = m

⇒ N = (C²)/(units of ∈₀ × m²)

⇒ units of ∈₀ = C²/(Nm²)

Therefore the MKS units of permittivity of free space is C²N⁻¹m⁻².

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