Two points charges , each `Q` , are fixed at separation `2d`. A charged particle having charge `q` and mass `m` is placed between them.
(a) Now this charged particle is slightly displaced along the line joining the charges , slow that it will execute simple harmonic motion and find the time period of oscillation.
(b) If charge `q` is negative and it is displaced slightly perpendicular to the line joining the charges , repeat the part `(a)`.

Explanation

(a) Let the charged particle be displaced slightly towards right by distance `x (x lt lt d)`.
`(Q at A , q) : F_(1) = (1)/( 4 pi in_(0)) (Qq)/((d - x)^(2))` , towards left
`(Q at B , q) : F_(2) = (1)/(4 pi in_(0)) (Qq)/((d + x)^(2))` , towards right
`F_(1) gt F_(2) , F_(n e t) = F_(1) - F_(2)` , will bring it towards `O`.
Restoring force `F = - (F_(1) - F_(2))`
[negative sign shows that net force is towards origin]
` = - (1)/( 4 pi in_(0)) [(1)/((d - x)^(2)) - (1)/((d + x)^(2))]`
` = - (Qq)/(4 pi in_(0)) [ ( 4 xd)/((d^(2) - x^(2))^(2))]`
Since `x lt lt d , x^(2)` is negligible
` = - (Qqx)/( pi in_(0) d^(3))`
Acceleration of charged particle is given as
`a = (F)/(m) = - (Qq)/( pi in_(0) md^(3)) x` ....(i)
`a alpha - x`
Acceleration `alpha` - (displacement) (Case of `S.H.M.`)
`a = - omega^(2) x` .....(ii)
Comparing (i) and (ii) , we get
`omega = sqrt((Q q)/(pi in_(0) md^(3)))`
Time period `T = (2 pi)/(omega) = 2pi sqrt((pi in_(0) md^(3))/(Qq))`
(b) Force on `-q`, by each `Q is F_(0)`, as shown
Restoring force is given as
`F = - 2F_(0) cos theta`
`= - 2 (1)/(4 pi in_(0)) . (Qq)/((d^(2) + x^(2))).(x)/((d^(2) + x^(2))^(1//2))`
`= - (Qq)/( 2pi in_(0)) (x)/((d^(2) + x^(2))^(3//2))`
As `x lt lt d , x^(2)` is negligible
` = - (Qq)/( 2 pi in_(0)d^(3)) x`
`a = (F)/(m) = -(Qq)/( 2pi in_(0) md^(3)) x`
`a alpha - x` (Case of `S.H.M`)
`T = 2 pi sqrt((2 pi in_(0) md^(3))/(Qq))`