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प्रश्न
Does the force on a charge due to another charge depend on the charges present nearby?
उत्तर
Coulomb's Law states that the force between two charged particle is given by `F = (q_1q_2)/(4pi∈_0r^2)`,
where
q1 and q2 are the charges on the charged particles
r = separation between the charged particles
`∈_o`= permittivity of free space
According to the Law of Superposition, the electrostatic forces between two charged particles are unaffected due to the presence of other charges.
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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).
