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Question
Two particles A and B, each carrying a charge Q, are held fixed with a separation dbetween them. A particle C of mass m and charge q is kept at the middle point of the line AB. Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement? Find the time period of the oscillations if these conditions are satisfied.
Solution
For the particle to execute simple harmonic motion:
F' = mw2x
\[\Rightarrow \frac{4}{\pi \epsilon_0}\frac{qQx}{d^3} = m \left( \frac{2\pi}{T} \right)^2 x\]
\[ \Rightarrow T^2 = \frac{m \pi^3 \epsilon_0 d^3}{Qq}\]
\[\Rightarrow T= \left[ \frac{m \pi^3 \epsilon_0 d^3}{Qq} \right]^{1/2} \]
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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).
