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Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousEasy2 minQB

PhysicsEasysingle choice

Using mass ( $M$ ), length ( $L$ ), time ( $T$ ) and current ( $A$ ) as fundamental quantities, the dimension of permittivity is

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Answer:

Solution:

From $F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}} : [ε_{0}]$
$= \frac{([ A T ] ) ^{2}}{[ M L T ^{- 2} ] [ L ^{2} ]}$
$= [M^{- 1} L^{- 3} T^{4} A^{2}]$ .

Stream:JEESubject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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