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The dimensions of permittivity [ε₀] are ________. Take Q as the dimension of charge.

Asked in GSEB Board July 2015 · Units and dimensions

• (1) M¹L⁻²T⁻²Q⁻²
• (2) M⁻¹L²T⁻³Q⁻¹
• (3) M⁻¹L⁻³T²Q²
• (4) M⁻¹L³T⁻²Q⁻²

Step-by-step solution

Why the other options are wrong

• (1) No single slip produces M¹L⁻²T⁻²Q⁻² -- the signs on M, T and Q are all reversed and the length exponent is wrong as well. Sanity check: ε₀=(q₁q₂)/(4π Fr²) fixes it at M⁻¹L⁻³T²Q².
• (2) M⁻¹L²T⁻³Q⁻¹ carries only one power of the charge, and that in the denominator. Coulomb's law has the product q₁q₂, so the charge appears squared -- and in the numerator of ε₀, giving Q².
• (4) M⁻¹L³T⁻²Q⁻² comes from writing the ratio upside down: (Fr²)/(q₁q₂) is 1/(4πε₀), which works out to M¹L³T⁻²Q⁻² -- this option is that, with the mass exponent left negative.

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