Question

There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by

F = `(Qq)/r^2 hatr`

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 1 es unit of charge = `1/([3]) xx 10^-9 C`

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

`1/(4pi ∈_0) = 10^-9/x^2 (N*m^2)/C^2`

With `x = 1/([3]) xx 10^-9`, we have `1/(4pi ∈_0) = [3]^2 xx 10^9 (Nm^2)/C^2`

or, `1/(4pi ∈_0) = (2.99792458)^2 xx 10^9 (Nm^2)/C^2` (exactly).

Answer 1

(i) F = `Q_q/r^2` = 1 dyne = `([1 "esu of charge"]^2)/[1 cm]^2`

Or, 1 esu of charge = 1 (dyne)^1/2 (cm)

Hence, [1 esu of charge] = [F]^1/2 L = [MLT^–2]^1/2 L = M^1/2 L^3/2 T^–1

[1 esu of charge] = M^1/2 L^3/2 T^–1

Thus charge in cgs unit is expressed as fractional powers (1/2) of M and (3/2) of L.

(ii) Consider the coloumb force on two charges, each of magnitude 1 esu of charge separated by a distance of 1 cm:

The force is then 1 dyne = 10^–5N.

This situation is equivalent to two charges of magnitude x C separated by 10^–2m.

This gives: F = `1/(4piε_0) * x^2/10^-4` which should be 1 dyne = 10^–5 N.

Thus `1/(4 piε_0) * x^2/10^-4 = 10^-5` ⇒ `1/(4 piε_0) = 10^-9/x^2 (Nm^2)/C^2`

With `x = 1/([3] xx 10^9)`, this yields

`1/(4 piε_0) = 10^-9 xx [3]^2 xx 10^18 = [3]^2 xx 10^9 (Nm)^2/C^2`

With [3] → 2.99792458, we get

`1/(4 piε_0) = 8.98755.... xx 10^9 (Nm^2)/C^2` exactly