Determine the dimensional formula for the permittivity of a vacuum (epsilon_0). Given the fundamental quantities: M = mass, L = length, T = time, and A = electric current.

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EEJ MAIN Physics QUESTION #7033

Question 1

Determine the dimensional formula for the permittivity of a vacuum ($\epsilon_0$). Given the fundamental quantities: $M$ = mass, $L$ = length, $T$ = time, and $A$ = electric current.

$[\epsilon_0] = [M^{-1} L^{-3} T^2 A]$
$[\epsilon_0] = [M^{-1} L^{-3} T^4 A^2]$✔️
$[\epsilon_0] = [M^{-1} L^2 T^4 A^{-2}]$
$[\epsilon_0] = [M^{-1} L^2 T^{-1} A]$

Correct Answer Explanation

According to Coulomb's Law, the force $F$ between two charges $q_1$ and $q_2$ separated by distance $r$ is:

$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$

Rearranging for permittivity:

$\epsilon_0 = \frac{q^2}{4\pi F r^2}$

Dimensional analysis:

Charge $[q] = [AT]$
Force $[F] = [MLT^{-2}]$
Distance $[r] = [L]$

$[\epsilon_0] = \frac{[A^2 T^2]}{[M L T^{-2}] [L^2]} = [M^{-1} L^{-3} T^4 A^2]$

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