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Question

Physics Question on Dimensional Analysis

The dimensional formula for permittivity of free space ((ϵ)0)\left(\left(\epsilon \right)_{0}\right) in the equation F=14πϵ0q1q2r2F=\frac{1}{4 \pi \epsilon _{0}}\frac{q_{1} q_{2}}{r^{2}} , where symbols have their usual meaning, is

A

[M1L3A2T4]\left[M^{1} \, L^{3} \, A^{- 2} \, T^{- 4}\right]

B

[M1L3T4A2]\left[M^{- 1} \, L^{- 3} \, T^{4} \, A^{2}\right]

C

[M1L3A2T4]\left[M^{- 1} \, L^{- 3} \, A^{- 2} \, T^{- 4}\right]

D

[M1L3T2A4]\left[M^{1} \, L^{3} \, T^{2} \, A^{- 4}\right]

Answer

[M1L3T4A2]\left[M^{- 1} \, L^{- 3} \, T^{4} \, A^{2}\right]

Explanation

Solution

As, F=14πϵ0q1q2r2F = \frac{1}{4 \pi \epsilon _{0}} \frac{q_{1} q_{2}}{r^{2}} ϵ0=q1q24πFr2\Rightarrow \epsilon _{0} = \frac{q_{1} q_{2}}{4 \pi F r^{2}} =[TA]2[L2][MLT2]=\frac{\left[T A\right]^{2}}{\left[L^{2}\right] \left[M L T^{- 2}\right]} =[M1L3T4A2]=\left[M^{- 1} L^{- 3} \, T^{4} A^{2}\right]