A charged particle of mass `m` having charge `q` remains in equilibrium at height `d` above a fixed charge `Q`. Now the charged particle is slightly displaced along the line joining the charges, show that it will executes `S.H.M.` and find the time period of oscillation.

Explanation

If we displace the particle down , then the electric repulsion between the charges increases due to decreased distance between them. The particle will accelerate upward. Using Newton.s second law we can write equation as follows :
F-mg =ma
We can use Coulomb.s law to rewrite the above equations as follows :
`1/(4piepsilon_0)(Qq)/(d-x)^2-mg=ma`
`rArr (Qq)/(4piepsilon_0d^2)1/(1-x/d)^2-mg=ma`
`rArr ma=(Qq)/(4piepsilon_0d^2)(1-x/d)^2 -mg` …(i)
If x < < d then `(1-x/d)^(-2) approx 1+(2x)/d`
Equation (i) can be re-written as follows :
`rArr ma=(Qq)/(4piepsilon_0d^2)(1+(2x)/d)-mg`
`rArr ma=(Qq)/(4piepsilon_0d^2)+(2xQq)/(4piepsilon_0d^3)-mg` ...(ii)
At distance d above the point O the particle is in equilibrium , hence we can write :
`(Qq)/(4piepsilon_0d^2)=mg`...(iii)
Using equation (ii) and equation (iii) we get :
`a=(Qq)/(2piepsilon_0md^3)x`...(iv)
We can compare equation (iv) with standard equation of S.H.M, which is
`rArr a=omega^2 x =(Qq)/(2piepsilon_0md^3)x`
From the above equations we have :
`omega=sqrt((Qq)/(2piepsilon_0md^3))`
Here, `omega` is the angular frequency of S.H.M. The time period of oscillation can be written in terms of angular frequency as follows :
`T=(2pi)/omega`
Hence, the time period of oscillation can be written as :
`T-2pisqrt((2piepsilon_0md^3)/(Qq))`