Answer :

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


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