Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M= mass, L= length, T= time and A= electric current, then:
[ε0]=[M−1L−3T4A2]
[ε0]=[M−1L2T−1A−2]
[ε0]=[M−1L2T−1A]
[ε0]=[M−1L−3T2A]
A
[ε0]=[M−1L−3T2A]
B
[ε0]=[M−1L2T−1A]
C
[ε0]=[M−1L2T−1A−2]
D
[ε0]=[M−1L−3T4A2]
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Solution
Verified by Toppr
14πε0q2r2=F ε0=[A2T2][MLT−2L2]=[M−1L−3A2T4]
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Let [ε0] denote the dimensional formula of the permittivity of the vaccum and [μ0] denote the permeability of the vacuum. If M= mass, L= length, T= time and I= electric current, then