Three is another useful system of units, besides the SI/mksA system, called the cgs (centimeter-gram -second) system, Coulumb's law is givenby F = (Qq)/(r^(2)) 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 (es units), where1 esunit of charge =(1)/([3]) xx 10^(-9) C The number [3] actually ariesfrom the speed of light in vacumm which is now taken to be exactly given by c = 2.99792458xx10^(8) m//s. An approximatevalue of c thenis c = [3] xx10^(8) m//s. (i) Show that the coulomb law in cgs units yields 1 esu of charge = 1 (dyn e)^(1//2)cm. Obtain the dimensious of units of charge in terms of mass M, length L and time T. Show that it is given in termsof fractional powers of M and L. (ii) Write 1 esu of charge = xC, wherex is a dimenionless number. Show that this gives (1)/(4pi in_(0)) = (10^(-9))/(x^(2)) (N.m^(2))/(C^(2)) Withx = (1)/([3]) xx10^(-9), we have (1)/(4pi in_(0)) = [3]^(2) xx10^(9) (Nm^(2))/(C^(2)) or (1)/(4pi in_(0)) = (2.99792458)^(2) xx 10^(9) (Nm^(2))/(C^(2)) (exactly).

Three is another useful system of units, besides the SI/mksA system, c

1.	Three is another useful system of units, besides the SI/mksA system, called the cgs (centimeter-gram -second) system, Coulumb's law is givenby F = (Qq)/(r^(2)) 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 (es units), where1 esunit of charge =(1)/([3]) xx 10^(-9) C The number [3] actually ariesfrom the speed of light in vacumm which is now taken to be exactly given by c = 2.99792458xx10^(8) m//s. An approximatevalue of c thenis c = [3] xx10^(8) m//s. (i) Show that the coulomb law in cgs units yields 1 esu of charge = 1 (dyn e)^(1//2)cm. Obtain the dimensious of units of charge in terms of mass M, length L and time T. Show that it is given in termsof fractional powers of M and L. (ii) Write 1 esu of charge = xC, wherex is a dimenionless number. Show that this gives (1)/(4pi in_(0)) = (10^(-9))/(x^(2)) (N.m^(2))/(C^(2)) Withx = (1)/([3]) xx10^(-9), we have (1)/(4pi in_(0)) = [3]^(2) xx10^(9) (Nm^(2))/(C^(2)) or (1)/(4pi in_(0)) = (2.99792458)^(2) xx 10^(9) (Nm^(2))/(C^(2)) (exactly).
Answer» Solution :(i) From`F = (Qq)/(r^(2))` `1 DYN e = ((1" esu of charge")^(2))/((1 CM)^(2))` `:.` 1 esu of charge`= (1 dyn e)^(1//2) XX 1cm = F^(1//2). L = (MLT^(-2))^(1//2)` L or 1 esu of charg e ` = 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 fo charge = x C, where x is a dimenionless number Coulomb forceon two charges, each of magnitude1 esu separatedby `1 cm is 1 dyne = 10^(-5) N`. This situation is equivalent to two charges of magnitude x C separated by `10^(-2)m`. `:. F = (1)/(4pi in_(0)) (x^(2))/((10^(-2))^(2)) = 1 dyne = 10^(-5) N :. (1)/(4pi in_(0)) = (10^(-9))/(x^(2)) (N m^(2))/(C^(2))` Taking`x = (1)/(\|3\| xx10^(9))`, we get, `(1)/(4pi in_(0))= 10^(-9)xx \|3\| xx10^(18) (Nm^(2))/(C^(2)) = 9xx10^(9) (Nm^(2))/(C^(2))` If `\|3\| rarr 2.99792458`, we get`(1)/(4pi in_(0)) = 8.98755xx10^(9) Nm^(2) C^(-2)`

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