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medium
Let $[ {\varepsilon _0} ]$ denote the dimensional formula of the permittivity of vacuum. If $M =$ mass, $L=$ length, $T =$ time and $A=$ electric current, then:
A$[ {\varepsilon _0} ]=[M^{-1}L^{-3}T^2A]$
B$[ {\varepsilon _0} ]=[M^{-1}L^{-3}T^4A^2]$
C$[ {\varepsilon _0} ]=[M^{-1}L^2T^{-1}A^{-2}]$
D$[ {\varepsilon _0} ]=[M^{-1}L^2T^{-1}A]$
(JEE MAIN-2013)
Solution
As we know $F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{R^2}}} \Rightarrow {\varepsilon _0} = \frac{{{q_1}{q_2}}}{{4\pi F{R^2}}}$
Hence ${\varepsilon _0} = \frac{{{{\left[ {AT} \right]}^2}}}{{ML{T^{ – 2}}{L^2}}} = \left[ {{M^{ -1}}{L^{ – 3}}{T^{ 4}}{A^2}} \right]$
Hence ${\varepsilon _0} = \frac{{{{\left[ {AT} \right]}^2}}}{{ML{T^{ – 2}}{L^2}}} = \left[ {{M^{ -1}}{L^{ – 3}}{T^{ 4}}{A^2}} \right]$
Standard 11
Physics
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