There is another useful system of units,besides the $SI/MKS$. $A$ system,called the $CGS$ (centimeter-gram-second) system. In this system,Coulomb's law is given by $\vec F = \frac{{Qq}}{{{r^2}}} \cdot \hat r$ where the distance $r$ is measured in $cm$ $(= 10^{-2} \ m)$,$F$ in dynes $(= 10^{-5} \ N)$ and the charges in electrostatic units $(esu)$,where $1 \ esu$ of charge $= \frac{1}{[3]} \times 10^{-9} \ C$. The number $[3]$ actually arises from the speed of light in vacuum which is now taken to be exactly given by $c = 2.99792458 \times 10^8 \ m/s$. An approximate value of $c$ then is $c = 3 \times 10^8 \ m/s$.
$(i)$ Show that the Coulomb law in $CGS$ units yields $1 \ esu$ of charge $= 1 \ (dyne)^{1/2} \ cm$. Obtain the dimensions of units of charge in terms of mass $M$,length $L$ and time $T$. Show that it is given in terms of fractional powers of $M$ and $L$.
$(ii)$ Write $1 \ esu$ of charge $= xC$,where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi \epsilon_0}} = \frac{{10^{-9}}}{{{x^2}}} \frac{N \ m^2}{C^2}$. With $x = \frac{1}{[3]} \times 10^{-9}$,we have $\frac{1}{{4\pi \epsilon_0}} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$ or $\frac{1}{{4\pi \epsilon_0}} = (2.99792458)^2 \times 10^9 \frac{N \ m^2}{C^2}$ (exactly).
$(i)$ Show that the Coulomb law in $CGS$ units yields $1 \ esu$ of charge $= 1 \ (dyne)^{1/2} \ cm$. Obtain the dimensions of units of charge in terms of mass $M$,length $L$ and time $T$. Show that it is given in terms of fractional powers of $M$ and $L$.
$(ii)$ Write $1 \ esu$ of charge $= xC$,where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi \epsilon_0}} = \frac{{10^{-9}}}{{{x^2}}} \frac{N \ m^2}{C^2}$. With $x = \frac{1}{[3]} \times 10^{-9}$,we have $\frac{1}{{4\pi \epsilon_0}} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$ or $\frac{1}{{4\pi \epsilon_0}} = (2.99792458)^2 \times 10^9 \frac{N \ m^2}{C^2}$ (exactly).
(A) $(i)$ In $CGS$ units,Coulomb's law is $F = \frac{Qq}{r^2}$.
For $1 \ esu$ of charge,$F = 1 \ dyne$ and $r = 1 \ cm$.
$1 \ dyne = \frac{(1 \ esu)^2}{(1 \ cm)^2} \implies 1 \ esu = (1 \ dyne)^{1/2} \ cm$.
Since $[F] = [M^1 L^1 T^{-2}]$ and $[L] = [L^1]$,the dimensions of $1 \ esu$ are $[M^1 L^1 T^{-2}]^{1/2} \times [L^1] = [M^{1/2} L^{3/2} T^{-1}]$.
Thus,the powers of $M$ and $L$ are $1/2$ and $3/2$ respectively,which are fractional.
$(ii)$ Let $1 \ esu = x \ C$. The force between two $1 \ esu$ charges at $1 \ cm$ distance is $1 \ dyne = 10^{-5} \ N$.
In $SI$ units,$F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$.
Substituting $q_1 = q_2 = x \ C$ and $r = 10^{-2} \ m$:
$10^{-5} \ N = \frac{1}{4\pi \epsilon_0} \frac{x^2}{(10^{-2} \ m)^2}$.
$\frac{1}{4\pi \epsilon_0} = \frac{10^{-5} \ N \times 10^{-4} \ m^2}{x^2} = \frac{10^{-9}}{x^2} \frac{N \ m^2}{C^2}$.
Substituting $x = \frac{1}{[3]} \times 10^{-9}$,we get $\frac{1}{4\pi \epsilon_0} = \frac{10^{-9}}{(1/[3] \times 10^{-9})^2} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$.
For $1 \ esu$ of charge,$F = 1 \ dyne$ and $r = 1 \ cm$.
$1 \ dyne = \frac{(1 \ esu)^2}{(1 \ cm)^2} \implies 1 \ esu = (1 \ dyne)^{1/2} \ cm$.
Since $[F] = [M^1 L^1 T^{-2}]$ and $[L] = [L^1]$,the dimensions of $1 \ esu$ are $[M^1 L^1 T^{-2}]^{1/2} \times [L^1] = [M^{1/2} L^{3/2} T^{-1}]$.
Thus,the powers of $M$ and $L$ are $1/2$ and $3/2$ respectively,which are fractional.
$(ii)$ Let $1 \ esu = x \ C$. The force between two $1 \ esu$ charges at $1 \ cm$ distance is $1 \ dyne = 10^{-5} \ N$.
In $SI$ units,$F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$.
Substituting $q_1 = q_2 = x \ C$ and $r = 10^{-2} \ m$:
$10^{-5} \ N = \frac{1}{4\pi \epsilon_0} \frac{x^2}{(10^{-2} \ m)^2}$.
$\frac{1}{4\pi \epsilon_0} = \frac{10^{-5} \ N \times 10^{-4} \ m^2}{x^2} = \frac{10^{-9}}{x^2} \frac{N \ m^2}{C^2}$.
Substituting $x = \frac{1}{[3]} \times 10^{-9}$,we get $\frac{1}{4\pi \epsilon_0} = \frac{10^{-9}}{(1/[3] \times 10^{-9})^2} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$.
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