Verified Solution Physics Units and Measurements

The dimensions of the quantity (electric charge)² / (permittivity * length * force) are equivalent to the dimensions of:

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Solution ✔ Verified
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Explanation

From Coulomb's Law, F = (1 / 4πε₀) * (q1 q2 / r²). So, 1/ε₀ = F r² / q². The given quantity is q² / (ε * L * F). Let's use ε₀ as an example. (q² / ε₀) / (L * F) = (F r²) / (L * F) = r² / L. The quantity is Q = q² / (ε L F). From F = (1/4πε) (q²/L²), we have (q²/ε) = F L². So, Q = (F L²) / (L F) = L. This is not dimensionless. Let's re-evaluate. Dimensions of charge (q) = [A T]. Dimensions of permittivity (ε) = [M^-1 L^-3 T⁴ A²]. (From F = (1/4πε) q1 q2 / r², ε = q² / (F r²) = [A T]² / ([M L T^-2] [L²]) = [A² T²] / [M L³ T^-2] = [M^-1 L^-3 T⁴ A²]) Dimensions of Length (L) = [L]. Dimensions of Force (F) = [M L T^-2]. So, q² / (ε L F) = ([A T])² / ([M^-1 L^-3 T⁴ A²] * [L] * [M L T^-2]) = [A² T²] / ([M^(-1+1) L^(-3+1+1) T^(4-2) A²]) = [A² T²] / ([M⁰ L^-1 T² A²]) = [A² T²] * [L¹ T^-2 A^-2] = [L]. The dimensions are [L], which is length. Still not dimensionless. Let's check the options again. If it is [L], then the options are incorrect. Is there an identity that simplifies to dimensionless? Maybe it's (charge)² / (permittivity * energy)? That would be F r² / E = (MLT^-2 L²)/(ML² T^-2) = LT⁰ = L. This is a standard competitive exam question, and it usually results in dimensionless or Energy. Let me re-verify my epsilon dimension. ε₀ = [M^-1 L^-3 T⁴ A²]. This is correct. So q² / (ε L F) = [L].

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