Prove the following

Question:

Let $\left[\epsilon_{0}\right]$ denote the dimensional formula of the permittivity of vacuum. If $M=$ mass, $L=$ Length, $\mathrm{T}=$ Time and $\mathrm{A}=$ electric current, then :

 

  1. $\left[\epsilon_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{2} \mathrm{~A}\right]$

  2. $\left[\epsilon_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]$

  3. $\left[\epsilon_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{2} \mathrm{~T}^{-1} \mathrm{~A}^{-2}\right]$

  4. $\left[\epsilon_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{2} \mathrm{~T}^{-1} \mathrm{~A}\right]$


Correct Option: , 2

Solution:

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