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There is another useful system of units,

Question:

There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by

$\mathbf{F}=\frac{\mathrm{Qq}}{r^{2}} \hat{\boldsymbol{r}}$

where the distance r is measured in cm (= 10–2 m), F in dynes (=10–5 N) and the charges in electrostatic units (es units), where

1es unit of charge = {1/[3]} × 10‒9 C

The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c = 2.99792458 × 108 m/s. An approximate value of c then is c = [3] × 108 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.

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 = x C, where x is a dimensionless number. Show that this gives

$\frac{1}{4 \pi \in_{0}}=\frac{10^{-9}}{x^{2}} \frac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{C}^{2}}$

With $x=\frac{1}{[3]} \times 10^{-9}$, we have

$\frac{1}{4 \pi \epsilon_{0}}=[3]^{2} \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$

or, $\frac{1}{4 \pi \epsilon_{0}}=(2.99792458)^{2} \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$ (exactly)

Solution:

(i) $F=\frac{Q q}{r^{2}}=1 d y n e=\frac{[1 \text { esu of charge }]^{2}}{[1 \mathrm{~cm}]^{2}}$

l esu of charge $=1(d y n e)^{1 / 2}(\mathrm{~cm})$

Hence, $[1$ esu of charge $]=[\mathrm{F}]^{1 / 2} \mathrm{~L}=\left[\mathrm{MLT}^{-2}\right]^{1 / 2} \mathrm{~L}=\mathrm{M}^{1 / 2} \mathrm{~L}^{3 / 2} \mathrm{~T}^{-1}$

$[1$ esu of charge $]=M^{1 / 2} L^{3 / 2} T^{-1}$

Thus charge in cgs unit is expressed as fractional powers $(1 / 2)$ of $\mathrm{M}$ and $(3 / 2)$ of $\mathrm{L}$.

(ii) Consider the coloumb force on two charges, each of magnitude 1 esu of charge separated by a distance of $1 \mathrm{~cm}$ :

The force is then 1 dyne $=10^{-5} \mathrm{~N}$.

This situation is equivalent to two charges of magnitude $x \mathrm{C}$ separated by $10^{-2} \mathrm{~m} .$

This gives:

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{x^{2}}{10^{-4}}$

which should be 1 dyne $=10^{-5} \mathrm{~N}$. Thus

$\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{x^{2}}{10^{-4}}=10^{-5} \Rightarrow \frac{1}{4 \pi \varepsilon_{0}}=\frac{10^{-9}}{x^{2}} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$

With $x=\frac{1}{[3] \times 10^{9}}$, this yields

$\frac{1}{4 \pi \varepsilon_{o}}=10^{-9} \times[3]^{2} \times 10^{18}=[3]^{2} \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$

With $[3] \rightarrow 2.99792458$, we have

$\frac{1}{4 \pi \varepsilon_{0}}=8.98755 \ldots \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$ exactly.

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