Coulomb’s Law - Force Between Two Point Charges

Two charge – 10c and + 10 c are placed 10 cm apart. Potential at centre of the line joining the two charge is:-

One end of a 10 cm long silk thread is fixed to a large vertical surface of a charged non-conducting plate and the other end is fastened to a small ball of mass 10 g and a charge of 4.0× 10^-6 C. In equilibrium, the thread makes an angle of 60° with the vertical. Find the surface charge density on the plate.

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).

Two identically-charged particles are fastened to the two ends of a spring of spring constant 100 N m⁻¹ and natural length 10 cm. The system rests on a smooth horizontal table. If the charge on each particle is 2.0 × 10⁻⁸ C, find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.  

What are the differences between the Coulomb force and the gravitational force?