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What is the SI unit of the permittivity of free space, ε₀?

1. coulomb²/(newton-metre)²
2. coulomb/newton-metre
3. newton-metre²/coulomb²
4. coulomb²/newton-metre²

Correct answer: coulomb²/newton-metre²

Solution

Coulomb's law F = q^2/(4*pi*eps0*r^2) gives eps0 = q^2/(F r^2). The units are coulomb^2/(newton x metre^2) = C^2 N^-1 m^-2, i.e. coulomb^2/newton-metre^2.

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