NCERT · Class 12 · Physics · Electric Charges and FieldsWhat is the dimensional formula of the permittivity of free space ($\ mepsilon0$)?
Step-by-Step Solution
From Coulomb's Law, $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \implies \epsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$.\nDimensional formula: $$[\epsilon_0] = \frac{[A T]^2}{[M L T^{-2}][L^2]} = [M^{-1} L^{-3} T^4 A^2]$$
Detailed Options Breakdown
Option : $[M^{-1} L^{-3} T^4 A^2]$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 1: $[M L^3 T^{-4} A^{-2}]$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electric Charges and Fields.
Option 2: $[M^{-1} L^3 T^{-4} A^2]$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electric Charges and Fields.
Option 3: $[M L^{-3} T^4 A^{-2}]$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electric Charges and Fields.
💡 Study Guide: This question tests core syllabus concepts from Electric Charges and Fields. For formulas, key summaries, and mock exam reference guides, read the full Electric Charges and Fields Revision Notes.