Physics

The quantity $$X$$ is given by $${\varepsilon _o}L\frac{{\Delta V}}{{\Delta T}}$$, where $${\varepsilon _o}$$ is the permittivity of free space, $$L$$ is a length, $${\Delta V}$$ is a potential difference and $${\Delta T}$$ is a time interval. The dimensional formula for $$X$$ is the same as that of 


ANSWER

current


SOLUTION
We know than, from coulomb's law
$$E_0 =\dfrac{1}{4\pi}\,\,\,\,\,\, \dfrac {Q_1 Q_2}{F R^2}$$
$$  \Rightarrow [E_0]= \dfrac {[Q_1][Q_2]}{[F][R^2]}$$
$$\Rightarrow [E_0]=\dfrac {A^{2}T^{2}}{mLT^{-2}L^2}=\left [ \dfrac {A^2T^2}{mLT^{3}L^{-2}} \right ]$$
$$ V = - \int \vec {E}. \vec{d\pi}$$
$$[V]= [E][L]$$
$$[V]= mLT^{-2}A^{-1}T^{-1}L= [mL^2A^{-1}T^{-3}]$$
$$ [T]= T$$
$$ [L]= L$$
$$X = \epsilon _0L\dfrac {\Delta V}{\Delta  T}$$
$$[X]= \left [ \dfrac {A^2T^2}{mLT^{3}L^{-2}} \right ] [L]\left [ \dfrac {mL^2A^{-1}T^{-3}}{T} \right ]$$
$$[X]= [A]$$
$$[A]$$ is the dimension of current.

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