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Question of 67

Q.The dimension of permittivity [epsilon_0] are ___. Take Q as the dimension of charge.

(a) M^1 L^-2 T^-2 Q^-2
(b) M^-1 L^2 T^-3 Q^-1
(c) M^-1 L^-3 T^2 Q^2
(d) M^-1 L^3 T^-2 Q^-2
Gujarat GsebGSEB Higher Secondary Certificate (HSC) Examination 2018MCQ· 1mImportance★★★★★
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Rearranging Coulomb's law gives epsilon_0 = q1 q2 / (4piF*r^2), and substituting dimensions yields M^-1 L^-3 T^2 Q^2.

Coulomb's law: F = (1/4piepsilon_0) * (q1 q2 / r^2), so epsilon_0 = q1 q2 / (4piF*r^2).

Dimensions:

  • Charge product q1 q2 -> Q^2
  • Force F -> M L T^-2 …

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