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

1. [ε0] = [M⁻¹ L⁻³ T⁴ A⁻²]
2. [ε0] = [M⁻¹ L⁻² T² A²]
3. [ε0] = [M⁻¹ L⁻³ T³ A]
4. [ε0] = [M⁻¹ L⁻³ T⁴ A²]

Correct answer: [ε0] = [M⁻¹ L⁻³ T⁴ A²]

Solution

The correct option accurately represents the dimensional formula of permittivity of vacuum, which is derived from the relationship between electric field, charge, and force in electromagnetic theory, confirming that it has dimensions of mass to the power of -1, length to the power of -3, time to the power of 4, and electric current to the power of -2.

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