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)

1*esu*^{2}=1dyne x 1 *cm*^{2}

1esu =

1esu= 1 (dyne)^{1/2} cm

Dimensional formula of esu=[ *M*^{1}*L*^{1}*T*^{-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^{-5}N

According to question 1 esu charge = x C

So, In MKS unit

F= N

As we know 1 dyne = 10^{-5}N

∴ N = 10^{-5}N

With x =

∴ N =

This yield

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