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  • There is another useful system of units, besides the SI/mksA system, called the cgs (centimeter-gram-second) system. In this system Coloumb's law is given by Frr =

There is another useful system of units, besides the SI/mks system, called the cgs (centimetre-gram-second) system. In this system, Columb's law is given by

F=\frac{Qq}{r^2}

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

\frac{1}{3}\times10^{-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 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 = x\: C, where x is a dimensionless number. Show that this gives

\frac{1}{4\pi\epsilon_0}=[3]^2\times10^{9}\frac{Nm^2}{C^2}

Answers (1)

 

(i)    According to relation,

F =\frac{Qq}{r^2}= 1dyne= \frac{\left [ esu\; of\: chaarge \right ]^2}{\left [ 1cm \right ]^{2}}
So, 1 esu of charge

=\left ( 1dyne \right )^{1/2}\times 1 cm

  F^{1/2}.L=\left [ MLT^{2} \right ]^{1/2}\left [ L \right ]= \left [ M^{1/2}L^{3/2}T^{1} \right ]
\Rightarrow Hence, [1 esu of charge] =M^{1/2}L^{3/2}T^{1}

Thus, the charge in C.G.S. unit (in esu) is represented in terms of fractional power 1/2 of M and 3/2 of L.

(ii)    If two charges, each of magnitude 1 esu, are separated by 1 cm, Coulomb force on the charge is 1 dyne = 10^{-5}N.
                Let 1 esu of charge =x C, where x is a dimensionless number. We can consider the situation of two charges of magnitude x C separated by 10^{-2}m.

The force between the charge
                    F=\frac{1}{4 \pi \varepsilon _0}\frac{x^{2}}{\left ( 10^{-2} \right )^{2}}= 1dyne=10^{-5}N
                        \therefore \frac{1}{4 \pi \varepsilon _0}= \frac{10^{-9}}{x^{2}}\frac{Nm^{2}}{C^{2}}
Taking,
\\x=\frac{1}{\left | 3 \right |\times 10 ^{9}},\\ We \: get\: \: \frac{1}{4\pi \varepsilon _0}=10^{-9}\times \left | 3 \right |^{2}\times 10^{18} \\ \frac{1}{4\pi \varepsilon _0}=9\times 10^{9}\frac{Nm^{2}}{C^{2}}
 

Posted by

infoexpert21

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