Verified Solution • 2025–26 Session moderate

QPages 121-124: Q1: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q1 Solution: Find the dimensional formula of .

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat electric charge as a primitive dimension instead of expressing it as current times time ($[q]=[I][T]=[A][T]$), which causes the $A^2$ factor to be omitted or sign‑flipped in the final result. Always write the Coulomb‑law expression with force dimension $[F]=[M][L][T^{-2}]$ and substitute $[q]=[A][T]$ before simplifying the exponents.

Find the dimensional formula of ε0\varepsilon_0.

Step-by-Step Verified Solution

**Answer:** The dimensional formula of __ is [M__ L__ T_ A_]. **Step-by-Step Explanation:** From Coulomb's law: F=14πε0q1q2r2F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}.
ε0=q1q24πFr2\varepsilon_0 = \frac{q_1q_2}{4\pi F r^2}
Dimensions: [q] = [A T], [F] = [M L T__], [r] = [L].
[ε0]=[AT]2[MLT−2][L]2=[A2T2][ML3T−2]=[M−1L−3T4A2][\varepsilon_0] = \frac{[A T]^2}{[M L T^{-2}][L]^2} = \frac{[A^2 T^2]}{[M L^3 T^{-2}]} = [M^{-1} L^{-3} T^4 A^2]

All Questions in this Chapter

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