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 = × ∈

Question

Philoid · Accepted 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