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Question:

Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M=mass, L=length, T=time and A=electric current, then what is the dimensional formula for [ε0]?

[ε0]=[M⁻¹L⁻³T⁴A²]

[ε0]=[M⁻¹L⁻²T⁴A²]

[ε0]=[M⁻¹L⁻³T²A]

[ε0]=[M⁻¹L⁻²TA]

Solution:

14πε0q2r2=F
ε0=[A2T2][MLT⁻²L²]=[M⁻¹L⁻³A²T⁴]