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Questions  

Using mass (M), length (L), time (T), and electric current (A) as fundamental quantities, the dimensions of permittivity will be

a
MLT−1A−1
b
MLT−2A−2
c
M−1L−3T4A2
d
M2L−2T−2A

detailed solution

Correct option is C

By Coulomb's laws, F=14πε0×q1q2r2          ε0=q1q24π×F×r2Taking dimensions, ε0=(AT)(AT)ML1T−2×L2=M−1L−3T4A2

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