Let ∈ 0 denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T =…

Let ∈ 0 denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then :

  1. $[\isin_{0}] = [M^{-1}L^{2}T^{-1}A^{-2}]$
  2. $[\isin_0] = [M^{-1}L^2T^{-1}A]$
  3. $[\isin_0] = [M^{-1}L^{-3}T^{2}A]$
  4. $[\isin_0] = [M^{-1}L^{-3}T^{4}A^{2}]$

Solution

Here's the corrected LaTeX formatted text: $F = \frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{R^{2}}$ $\epsilon_{0} = \frac{q_{1} q_{2}}{4 \pi F R^{2}}$ Hence, $\epsilon_{0} = \frac{C^{2}}{N \cdot m^{2}} = \frac{[AT]^{2}}{MLT^{-2} \cdot L^{2}} = [M^{-1} L^{-3} T^{4} A^{2}]$

Asked in: JEE Main 2013 (07 Apr)

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