Answer :

i) As mentioned in the question the force between

the two charges is given by,


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.


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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