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A physical quantity of the dimensions of length that can be formed out of $c, G$ and $\frac{e^2}{4\pi \varepsilon _0}$ is $[c$ is velocity of light, $G$ is the universal constant of gravitation and $e$ is charge $] $
$\frac{1}{{{c^2}}}$$\sqrt {\frac{{{e^2}}}{{G4\pi \varepsilon_0}}} $
$\frac{1}{{{c^{}}}}\frac{{G{e^2}}}{{4\pi \varepsilon_0}}$
$\frac{1}{{{c^2}}}$$\sqrt {\frac{{G{e^2}}}{{4\pi \varepsilon_0}}} $
${c^2}\;\sqrt {\frac{{G{e^2}}}{{4\pi \varepsilon_0}}} $
Solution
Dimensions of
$\frac{{{e^2}}}{{4\pi {\varepsilon _0}}} = \left[ {F \times {d^2}} \right] = \left[ {M{L^3}{T^{ – 2}}} \right]$
Dimensions of $G = \left[ {{M^{ – 1}}{L^3}{T^{ – 2}}} \right]$
Dimensions of $c = \left[ {L{T^{ – 1}}} \right]$
$ l\, \propto \,{\left( {\frac{{{e^2}}}{{4\pi {\varepsilon _o}}}} \right)^p}{G^q}{c^r}$
$\therefore \left[ {{L^1}} \right] = {\left[ {M{L^3}{T^{ – 2}}} \right]^p}{\left[ {{M^{ – 1}}{L^3}{T^{ – 2}}} \right]^q}{\left[ {L{T^{ – 1}}} \right]^r}$
On comparing both sides and solving we get
$P = \frac{1}{2},\,q = \frac{1}{2}\,and\,r = – 2$
$\therefore \,\,l = \frac{1}{{_c2}}{\left[ {\frac{{G{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{1/2}}$
