Units & Dimensions

JEE Main level, single correct, medium

139 of 149

Question 139 of 149

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

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Answer: (D) [∈0]=[M−1L−3T4A2]\left[{\in }_{0}\right]=\left[{\text{M}}^{-1}{\text{L}}^{-3}{\text{T}}^{4}{\text{A}}^{2}\right]

F=14πϵ0q1q2R2\text{F}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{{\text{q}}_{1}{\text{q}}_{2}}{{\text{R}}^{2}}

ϵ0=q1q24πFR2{\epsilon }_{0}=\dfrac{{\text{q}}_{1}{\text{q}}_{2}}{4\pi {\text{FR}}^{2}}

Hence, ϵ0=C2N . m2=[AT]2MLT−2. L2=[M−1L−3T4A2]{\epsilon }_{0}=\dfrac{{\text{C}}^{2}}{{\text{N . m}}^{2}}=\dfrac{{\left[\text{AT}\right]}^{2}}{\text{ML}{\text{T}}^{-2}{\text{. L}}^{2}}=\left[{\text{M}}^{-1}{\text{L}}^{-3}{\text{T}}^{4}{\text{A}}^{2}\right]

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