Dimensions of varepsilon_0 are | vedclass

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(A) We know that the electrostatic force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's Law:$F = \frac{1}{4

Vedclass · Accepted Answer

$[M^{-1} L^{-3} T^4 A^2]$

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(A) We know that the electrostatic force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's Law:$F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}$Rearranging the formula to solve for permittivity $\varepsilon_0$:$\varepsilon_0 = \frac{q_1 q_2}{4 \pi F r^2}$Now,substituting the dimensional formulas for each physical quantity:$[q] = [AT]$$[F] = [MLT^{-2}]$$[r] = [L]$Substituting these into the expression for $\varepsilon_0$:$[\varepsilon_0] = \frac{[AT][AT]}{[MLT^{-2}][L^2]} = \frac{[A^2 T^2]}{[ML^3 T^{-2}]}$$[\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]$

Dimensions of $\varepsilon_0$ are

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