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

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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Single Correct Medium Published on 18th 08, 2020
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