भौतिक स्थिरांकों के निम्नलिखित संयोजन से (अपन | vedclass

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Question

$\begin{array}{l}
The\,{m{Dimensional}}\,{m{formulae}}\,{m{of}}\
{m{e}}\,{m{ = }}\left[ {{M^0}{L^0}{T^1}{A^1}} ight]\
{\varepsilon _0

Vedclass · Accepted Answer

$\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}$

Vedclass · Answer

$\begin{array}{l}
The\,{m{Dimensional}}\,{m{formulae}}\,{m{of}}\
{m{e}}\,{m{ = }}\left[ {{M^0}{L^0}{T^1}{A^1}} ight]\
{\varepsilon _0} = \left[ {{M^{ - 1}}{L^{-3}}{T^4}{A^2}} ight]\
G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} ight]\
and\,{m_e} = \left[ {{M^1}{L^0}{T^0}} ight]\
Now,\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}
\end{array}$

$\begin{array}{l}
 = \frac{{{{\left[ {{M^0}{L^0}{T^1}{A^1}} ight]}^2}}}{{2\pi \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} ight]\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} ight]{{\left[ {{M^1}{L^0}{T^o}} ight]}^2}}}\
 = \frac{{\left[ {{T^2}{A^2}} ight]}}{{2\pi \left[ {{M^{ - 1 - 1 + 2}}{L^{ - 3 + 3}}{T^{4 - 2}}{A^2}} ight]}}\
 = \frac{{\left[ {{T^2}{A^2}} ight]}}{{2\pi \left[ {{M^0}{L^0}{T^2}{A^2}} ight]}} = \frac{1}{{2\pi }}
\end{array}$

$\begin{array}{l}
\frac{1}{{2\pi }}\,is\,{m{Dimensionl}}ess\,thus\,the\,combination\
\frac{{{e^2}}}{{2\pi {\varepsilon _0}Gm_e^2}}\end{array}$

would d have the same value in diffierent systems of units

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