Question

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

• A. \(\varepsilon_0 = [M^{-1}L^{-3}T^{2}A]\)
• B. \(\varepsilon_0 = [M^{-1}L^{-3}T^{4}A^{2}]\)
• C. \(\varepsilon_0 = [M^{1}L^{2}T^{1}A^{2}]\)
• D. \(\varepsilon_0 = [M^{1}L^{2}T^{1}A]\)

Hint:

Use Coulomb's law \(F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}\) and \(q = AT\).

Answer 1

The Correct Option is B

Coulomb's law: \(F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r^2}\), so \(\varepsilon_0 = \frac{q^2}{4\pi F r^2}\).
\([q] = [AT]\), \([F] = [MLT^{-2}]\), \([r^2] = [L^2]\).
\([\varepsilon_0] = \frac{A^2T^2}{MLT^{-2}\cdot L^2} = M^{-1}L^{-3}T^{4}A^{2}\).
Answer: (2) \(\varepsilon_0 = [M^{-1}L^{-3}T^{4}A^{2}]\)