There is another useful system of units, besides the $\mathrm{SI/MKS}$. A system, called the $\mathrm{CGS}$ (centimeter-gramsecond) system. In this system Coloumb’s law is given by $\vec F = \frac{{Qq}}{{{r^2}}} \cdot \hat r$ where the distance $r$ is measured in $cm\left( { = {{10}^{ - 2}}m} \right)$ , $\mathrm{F}$ in dynes $\left( { = {{10}^{ - 5}}N} \right)$ and the charges in electrostatic units $(\mathrm{es\,unit}$), where $1$ $\mathrm{esu}$ 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 coloumb law in $\mathrm{CGS}$ units yields $1$ $\mathrm{esu}$ of charge = $= 1\,(dyne)$ ${1/2}\,cm$. Obtain the dimensions of units of charge in terms of mass $\mathrm{M}$, length $\mathrm{L}$ and time $\mathrm{T}$. Show that it is given in terms of fractional powers of $\mathrm{M}$ and $\mathrm{L}$ .
$(ii)$ Write $1$ $\mathrm{esu}$ of charge $=xC$, where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi { \in _0}}} = \frac{{{{10}^{ - 9}}}}{{{x^2}}}\frac{{N{m^2}}}{{{C^2}}}$ with $x = \frac{1}{{[3]}} \times {10^{ - 9}}$ we have, $\frac{1}{{4\pi { \in _0}}} = {[3]^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ or $\frac{1}{{4\pi { \in _0}}} = {\left( {2.99792458} \right)^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ (exactly).
There is another useful system of units, besides the $\mathrm{SI/MKS}$. A system, called the $\mathrm{CGS}$ (centimeter-gramsecond) system. In this system Coloumb’s law is given by $\vec F = \frac{{Qq}}{{{r^2}}} \cdot \hat r$ where the distance $r$ is measured in $cm\left( { = {{10}^{ - 2}}m} \right)$ , $\mathrm{F}$ in dynes $\left( { = {{10}^{ - 5}}N} \right)$ and the charges in electrostatic units $(\mathrm{es\,unit}$), where $1$ $\mathrm{esu}$ 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 coloumb law in $\mathrm{CGS}$ units yields $1$ $\mathrm{esu}$ of charge = $= 1\,(dyne)$ ${1/2}\,cm$. Obtain the dimensions of units of charge in terms of mass $\mathrm{M}$, length $\mathrm{L}$ and time $\mathrm{T}$. Show that it is given in terms of fractional powers of $\mathrm{M}$ and $\mathrm{L}$ .
$(ii)$ Write $1$ $\mathrm{esu}$ of charge $=xC$, where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi { \in _0}}} = \frac{{{{10}^{ - 9}}}}{{{x^2}}}\frac{{N{m^2}}}{{{C^2}}}$ with $x = \frac{1}{{[3]}} \times {10^{ - 9}}$ we have, $\frac{1}{{4\pi { \in _0}}} = {[3]^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ or $\frac{1}{{4\pi { \in _0}}} = {\left( {2.99792458} \right)^2} \times {10^9}\frac{{N{m^2}}}{{{C^2}}}$ (exactly).
$(i)$
$\mathrm{F}=\frac{Q q}{r^{2}}$
$\therefore 1 \text { dyne }=\frac{(1 \text { esu charge })^{2}}{(1 \mathrm{~cm})^{2}}$
$\therefore 1 \text { esu }=(1 \text { dyne })^{1 / 2} \times 1 \mathrm{~cm}$
$=\mathrm{F}^{1 / 2} \mathrm{~L}$
$\therefore$ Dimensional formula of $1 \mathrm{esu}$,
$=\left[M^{1} L^{1} \mathrm{~T}^{-2}\right]^{1 / 2} \times\left[\mathrm{L}^{1}\right]$
$=\left[M^{1 / 2} L^{3 / 2} \mathrm{~T}^{-1}\right]$
Hence, in dimensional formula of esu charge, power of $M$ is $\frac{1}{2}$ and of $L$ is $\frac{3}{2}$, which is noninteger.
$(ii)$ Suppose $1$ esu $=x \mathrm{C}$, where $x$ is a dimensionless number. The force between two charges of $1$ esu magnitude is
$10^{-5} \mathrm{~N}$ $\left(=1\right.$ dyne) when they are at distance $10^{-2} \mathrm{~m}(=1 \mathrm{~cm})$.
$\therefore \mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{x^{2}}{\left(10^{-2}\right)^{2}}$
$\therefore 10^{-5} \mathrm{~N}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{x^{2}}{\left(10^{-2}\right)^{2}}$
$\therefore 1 \text { dyne }=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{x^{2}}{\left(10^{-2}\right)^{2}}$
$\therefore \frac{1}{4 \pi \epsilon_{0}}=\frac{10^{-5} \mathrm{~N} \times 10^{-4} \mathrm{~m}^{2}}{x^{2}}$
$=\frac{10^{-9}}{x^{2}} \cdot \frac{\mathrm{N} m^{2}}{\mathrm{C}^{2}}$
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