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JEE Main 2013 · Units & Dimensions

Units & Dimensions question from JEE Main 2013

Q86JEE Main 2013Easy

Units & Dimensions

Let [ $\varepsilon_0$ ] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:

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Answer: (b)

Coulomb's law: $F = \frac{q^2}{4\pi\varepsilon_0 r^2}$ , so $\varepsilon_0 = \frac{q^2}{Fr^2}$ .
$[\varepsilon_0] = \frac{\mathrm{A^2T^2}}{\mathrm{MLT^{-2}}\cdot\mathrm{L^2}} = \mathrm{M^{-1}L^{-3}T^4A^2}$
So the answer is (b).

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