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Q.The unit of permittivity of vacuum is: (A) Newton m2^2/coulomb2^2
(B) coulomb2^2/Newton m2^2
(C) Newton/coulomb
(D) Newton volt/m2^2

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2025MCQ· 1mImportance★★★★★
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Rearranging Coulomb's law gives ε0\varepsilon_0 the units C2 N−1 m−2C^2\,N^{-1}\,m^{-2} (coulomb2^2/newton·m2^2) — option (B).

Concept. Coulomb's law is

F=14πε0q1q2r2.F=\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^{2}}.

Solving for the permittivity of free space,

ε0=q1q24πFr2.\varepsilon_0=\frac{q_1 q_2}{4\pi F r^{2}}.

Units.

[ε0]=[coulomb]2[newton] [metre]2=C2 N−1 m−2.[\varepsilon_0]=\frac{[\text{coulomb}]^{2}}{[\text{newton}]\,[\text{metre}]^{2}}=C^{2}\,N^{-1}\,m^{-2}. …

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