Two charges -q each are fixed separated by distance 2d. A third charge q of mass m placed at the midpoint is displaced slightly by x (x < < d) perpendicular to the line joining the two fixed charged as shown in figure. Show that q will perform simple harmonic oscillation of time period. `T = [(8pi^(3) epsilon_(0)md^(3))/q^(2)]^(1//2)`

Explanation

Suppose charge at A and B are - q and O is mid point of AB and PO is x.
`therefore AB = AO + OB`
=d+d
=2d
`x lt d` and `angleAPO =theta`
m is mass of charge q.
Attractive force by each charge A and B on charge at P,
`F = (k(q)(q))/r^(2)`
where r = AP = BP
`F sin theta` components of force are of same magnitude but in opposite directions hence, their resultant is zero and Fcos6 components are in same direction,
`F. = 2F cos theta`
`=(2kq^(2))/r^(2) cos theta`
But from figure, `r = sqrt(d^(2) + x^(2))` and `cos theta = x/r`
`therefore F. = (2kq^(2))/(d^(2) + x^(2))^(2).x/(d^(2) + x^(2))^(1//2)`
`=(2kq^(2)x)/(d^(2) + x^(2))^(3//2)`
If `x lt lt d`, then x can be neglected,
`F. = (2kq^(2)x)/(d^(3))`..........(1)
`F. = Kx`
where `K = (2kq^(2))/d^(3)` is constant
`therefore F. prop x`
Hence, q can perform simple harmonic motion. Here, force is directly proportional to x and towards point O.
`omega = sqrt(K/m)`
`therefore (2pi)/T = sqrt(K/m)`
`therefore T = 2pi sqrt(m/(2kq^(2)//a^(3))`
By taking `k = 1/(4pi epsilon_(0))`
`therefore T=[(8pi^(3)epsilon_(0)md^(3))/q^(2)]^(1//2)`