From the following combinations of physical c | 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

From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is

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