Let ε _ 0 denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then dimensions of permittivity is given as [ M ^ p L ^ q T ^ r A ^ s ] . Find the

Correct answer: 3 From Coulomb's law, F = q _ 1 q _ 2 4π ε _ 0 r ^ 2 ∴ ε _ 0 = q _ 1 q _ 2 4 r ^ 2 = ( A ^ 1 T ^ 1 ) ( A ^ 1 T ^ 1 ) [ M ^ 1 L ^ 1 T ^ - 2 ] [ L ^ 2 ] = [ M ^ - 1 L ^ - 3 T ^ 4 A ^ 2 ] ∴ p = - 1, q = - 3, r = 4, s = 2 ∴ (p - q + r)/(s) = (- 1 - ( - 3 ) + 4)/(2) = (6)/(2) = 3

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Units and Dimensions

NTA Abhyas 2020 · Physics · Units and Dimensions

Let ε _ 0 denote the dimensional formula of the permittivity of...

Let

{ε}_{0}

denote the dimensional formula of the permittivity of vacuum. If

M =

mass,

\text{L} =

length,

\text{T} =

time and

\text{A} =

electric current, then dimensions of permittivity is given as \left[{\text{M}}^{\text{p}} {\text{L}}^{\text{q}} {\text{T}}^{\text{r}} {\text{A}}^{\text{s}}\right]

. Find the value of

\dfrac{\text{p} - \text{q} + \text{r}}{\text{s}}

Correct answer

Step-by-step solution

From Coulomb's law,

F = \dfrac{{q}_{1}{q}_{2}}{4\pi{ε}_{0}{r}^{2}}

∴ {ε}_{0} = \dfrac{{q}_{1}{q}_{2}}{4\piF{r}^{2}} = \dfrac{( {A}^{1}{T}^{1} ) ( {A}^{1}{T}^{1} )}{[ {M}^{1}{L}^{1}{T}^{- 2} ] [ {L}^{2} ]} = \left[{M}^{- 1}{L}^{- 3}{T}^{4}{A}^{2}\right]

∴ p = - 1, q = - 3, r = 4, s = 2

∴ \dfrac{p - q + r}{s} = \dfrac{- 1 - ( - 3 ) + 4}{2} = \dfrac{6}{2} = 3

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