JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2013
- Medium
- MCQ
Let \([ {\varepsilon _0} ]\) denote the dimensional formula of the permittivity of vacuum. If \(M =\) mass, \(L=\) length, \(T =\) time and \(A=\) electric current, then:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\;{\varepsilon _0}=M^{-1}L^{-3}T^4A^2\)
Step-by-step Solution
Detailed explanation
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 \({\varepsilon _0} = \frac{{{{\left[ {AT} \right]}^2}}}{{ML{T^{ - 2}}{L^2}}} = \left[ {{M^{ -1}}{L^{ - 3}}{T^{ 4}}{A^2}} \right]\)
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