Units and Measurements

Concept:The quantity $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is the characteristic impedance of free space. Using $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, we get $\sqrt{\frac{\mu_0}{\epsilon_0}} = \frac{1}{c \epsilon_0}$.Explanation:First, find the dimensions of $\epsilon_0$ from Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$. Charge $q$ has dimensions $[A\,T]$, force $F$ has $[M\,L\,T^{-2}]$, and distance $r$ has $[L]$. Thus, $[\epsilon_0] = \frac{[A\,T][A\,T]}{[M\,L\,T^{-2}][L^2]} = [M^{-1} L^{-3} T^4 A^2]$. The speed of light $c$ has dimensions $[L\,T^{-1}]$.Now, $[\sqrt{\mu_0/\epsilon_0}] = \frac{1}{[c][\epsilon_0]} = \frac{1}{[L\,T^{-1}][M^{-1}L^{-3}T^4 A^2]} = [M^1 L^2 T^{-3} A^{-2}]$.Let the numerical value in SI be $N_1$ and in the new system be $N_2$. Since $N_1 U_1 = N_2 U_2$, we have $N_2 = N_1 \left(\frac{U_1}{U_2}\right)$. Using the dimensional formula $[M^1 L^2 T^{-3} A^{-2}]$,$N_2 = N_1 \left[\frac{M_1}{M_2}\right]^1 \left[\frac{L_1}{L_2}\right]^2 \left[\frac{T_1}{T_2}\right]^{-3} \left[\frac{A_1}{A_2}\right]^{-2}$.Here $M_1=1\;\text{kg}$, $L_1=1\;\text{m}$, $T_1=1\;\text{s}$, $A_1=1\;\text{A}$; and $M_2=5\;\text{kg}$, $L_2=5\;\text{m}$, $T_2=5\;\text{s}$, $A_2=5\;\text{A}$.For one SI unit, $N_1 = 1$. Then$N_2 = 1 \times \left(\frac{1}{5}\right)^1 \left(\frac{1}{5}\right)^2 \left(\frac{1}{5}\right)^{-3} \left(\frac{1}{5}\right)^{-2}$.Summing exponents: $1+2-3-2 = -2$, so $N_2 = \left(\frac{1}{5}\right)^{-2} = 5^2 = 25$.Answer:25