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 =

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 9 1 10 C [3] − = ×


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 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


9 2 2 2 0 1 10 N.m 4x C π − = ∈


With , we have 9 1 x 10 [3] − = ×


2 2 9 2 0 1 Nm [3] 10 4C π = × ∈


Or, (exactly) 2 2 9 2 0 1 Nm [2.99792458] 10 4C π = × ∈


i) As mentioned in the question the force between


the two charges is given by,


F=


Let us suppose the two charges are separated by 1cm and have same charge of 1esu


Therefore, F= =1dyne (As mentioned in question)


1esu2=1dyne x 1 cm2


1esu =


1esu= 1 (dyne)1/2 cm


Dimensional formula of esu=[ M1L1T-2 ]1


Now we will use dimensional analysis.


Then,





Comparing both sides we get the value of x,y,z as x=1/2, y=3/2 ,z=-1


As we can see, the power of M and L are in Fraction.


b)Lets assume the previous case that the charges of 1 esu magnitude are separated by 1 cm,


Then the force is 1 dyne = 10-5N


According to question 1 esu charge = x C


So, In MKS unit


F= N


As we know 1 dyne = 10-5N


∴ N = 10-5N


With x =


∴ N =


This yield


1