Let [epsilon0] be the dimensional formula of the permittivity of free space. With M = mass, L = length, T = time and A = electric current, which expression is correct?

[epsilon0] = [M⁻¹ L⁻³ T⁴ A²]
[epsilon0] = [M L⁻³ T² A]
[epsilon0] = [M⁻¹ L² T⁻¹ A⁻²]
[epsilon0] = [M⁻¹ L² T⁻¹ A]

Correct answer: [epsilon0] = [M⁻¹ L⁻³ T⁴ A²]

Solution

Solving Coulomb's law for epsilon0 gives epsilon0 = q² / (F * r²). Substituting dimensions of charge, force and length yields the dimensional formula.

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