c , G तथा frac{ e ^{2}}{4 pi varepsilon_{0}} | vedclass

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Question

Dimensions of 
$\frac{{{e^2}}}{{4\pi {\varepsilon _0}}} = \left[ {F 	imes {d^2}} ight] = \left[ {M{L^3}{T^{ - 2}}} ight]$
Dimensions of $G

Vedclass · Accepted Answer

$\frac{1}{{{c^2}}}$$\sqrt {\frac{{G{e^2}}}{{4\pi \varepsilon_0}}} $

Vedclass · Answer

Dimensions of 
$\frac{{{e^2}}}{{4\pi {\varepsilon _0}}} = \left[ {F 	imes {d^2}} ight] = \left[ {M{L^3}{T^{ - 2}}} ight]$
Dimensions of $G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} ight]$
Dimensions of $c = \left[ {L{T^{ - 1}}} ight]$
$ l\, \propto \,{\left( {\frac{{{e^2}}}{{4\pi {\varepsilon _o}}}} ight)^p}{G^q}{c^r}$
$	herefore \left[ {{L^1}} ight] = {\left[ {M{L^3}{T^{ - 2}}} ight]^p}{\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} ight]^q}{\left[ {L{T^{ - 1}}} ight]^r}$

On comparing both sides and solving we get

$P = \frac{1}{2},\,q = \frac{1}{2}\,and\,r =  - 2$
$	herefore \,\,l = \frac{1}{{_c2}}{\left[ {\frac{{G{e^2}}}{{4\pi {\varepsilon _0}}}} ight]^{1/2}}$

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