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Let \left [ \epsilon _{0} \right ] denote the dimensional formula of the permittivity of vacuum. If M= mass,L=length,T=time and A=electric current,then:   

 

 

  • Option 1)

    \left [ \epsilon _{0} \right ]=\left [ M^{-1}L^{2}T^{-1}A \right ]

  • Option 2)

    \left [ \epsilon _{0} \right ]=\left [ M^{-1}L^{-3}T^{2}A \right ]

  • Option 3)

    \left [ \epsilon _{0} \right ]=\left [ M^{-1}L^{-3}T^{4}A^{2} \right ]

  • Option 4)

    \left [ \epsilon _{0} \right ]=\left [ M^{-1}L^{2}T^{-1}A^{-2} \right ]

 

Answers (1)

best_answer

As we have learned

Permittivity of free space -

\epsilon_{o}=\ M^{-1}L^{-3}T^{4}A^{2}^{}

 

- wherein

CN^{-1}m^{-2}

 

 For Coulumbs law 

F = \frac{1}{4 \pi \epsilon _0 }\cdot \frac{q_1q_2}{r^2 }

or \\\epsilon _0 = \frac{q_1 q_2 }{Fr^2}

[\epsilon _0] = \frac{[AT][AT] }{[MLT^{-2}][L^2]}=[ M^{-1}L^{-3}T^4A^2]

 

 

 

 

 

 


Option 1)

\left [ \epsilon _{0} \right ]=\left [ M^{-1}L^{2}T^{-1}A \right ]

Option 2)

\left [ \epsilon _{0} \right ]=\left [ M^{-1}L^{-3}T^{2}A \right ]

Option 3)

\left [ \epsilon _{0} \right ]=\left [ M^{-1}L^{-3}T^{4}A^{2} \right ]

Option 4)

\left [ \epsilon _{0} \right ]=\left [ M^{-1}L^{2}T^{-1}A^{-2} \right ]

Posted by

SudhirSol

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