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 F=(Qq)/(r^(2))hatr where the distance r is measured in cm(=10^(-2)mu), F in dynes (=10^(-5)N) and the charges in electrostatic units (es units), where 1 es unit of charge =(1)/([3])xx10^(-9)C. The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c=2.99792458xx10^(8)m//s. An approximate value of c, then is c=3xx10^(8)m//s. (i). Show that the coulomb's 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 1esu of cahrge =xC, where x is a dimensionless number. Show that this gives (1)/(4epsi_(0))=(10^(-9))/(x^(2))(Nm^(2))/(C^(2)). With x=(1)/([3])xx10^(-9), we have (1)/(4piepsi_(0))=[3]^(2)xx10^(9)(Nm^(2))/(C^(2)),(1)/(4piepsi_(0))=(2.99792458)^(2)xx10^(9)(Nm^(2))/(C^(2)) (exactly)

There is another useful system of units, besides the SI/MKS. A system,

1.	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 F=(Qq)/(r^(2))hatr where the distance r is measured in cm(=10^(-2)mu), F in dynes (=10^(-5)N) and the charges in electrostatic units (es units), where 1 es unit of charge =(1)/([3])xx10^(-9)C. The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by c=2.99792458xx10^(8)m//s. An approximate value of c, then is c=3xx10^(8)m//s. (i). Show that the coulomb's 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 1esu of cahrge =xC, where x is a dimensionless number. Show that this gives (1)/(4epsi_(0))=(10^(-9))/(x^(2))(Nm^(2))/(C^(2)). With x=(1)/([3])xx10^(-9), we have (1)/(4piepsi_(0))=[3]^(2)xx10^(9)(Nm^(2))/(C^(2)),(1)/(4piepsi_(0))=(2.99792458)^(2)xx10^(9)(Nm^(2))/(C^(2)) (exactly)
Answer» Solution :(i) Fromthe relation, `F=(Qq)/(r^(2))=1` dyne `=[(1" esu of charge"]^(2))/([1" "cm^(2)])` so, 1 esu of charge `=(1" dyne")^(1//2)xx1cm=F^(1//2),L=[MLT^(-2)]^(1//2)L` `implies1` esu of charge `=M^(1//2)L^(3//2)T^(-1)` Thus, esu of charge is represented in terms of FRACTIONAL POWERS `(1)/(2)` of M and ` (3)/(2)` of L. (ii). Let 1 esu of charge =xC, where x is a dimensionless NUMBER. Coulomb FORCE on TWO charges, each of magnitude 1 esu separated by 1 cm is dyne `=10^(-5 )N`. this situation is equivalent to two charges of magitude x C separated by `10^(-2)m`. `thereforeF=(1)/(4piepsi_(0))(x^(2))/((10^(-2))^(2))=1" dyne"=10^(-5)N` `therefore(1)/(4piepsi_(0))=(10^(-9))/(x^(2))(Nm^(2))/(C^(2))` Taking, `x=(1)/(\|3\|xx10^(9))` we get, `(1)/(4piepsi_(0))=10^(-9)xx\|3\|^(2)xx10^(18)(Nm^(2))/(C^(2))=9xx10^(9)(Nm^(2))/(C^(2))` ltbr if `\|3\|to2.99792458` we get `(1)/(4piepsi_(0))=8.98755xx10^(9)Nm^(2)C^(-2)`

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