A particle of mass m and charge q is located midway between two fixed charged particles each having a charge q and a distance 2l apart. Prove that the motion of the particle will be SHM if it is displaced slightly along the line connecting them and released. Also find its time period.

Explanation

Let the charge q at the mid-point the displaced slightly to the left.
The force on the displaced charge q due to charge q at A.
`F_(1) = (1)/(4pi epsilon_(0))(q^(2))/((l + x)^(2))`

The force on the displaced charge q due to charge at B,
`F_(2) = (1)/(4pi epsilon_(0)) (q^(2))/((l-x)^(2))`
Net restoring force on the displaced charge q.
`F = F_(2) - F_(1)` or F
`= (1)/(4pi epsilon_(0))(q^(2))/((l-x)^(2)) - (1)/(4pi epsilon_(0))(q^(2))/((l+x)^(2))`
or `F = (q^(2))/(4pi epsilon_(0))[(1)/((l-x)^(2)) - (1)/((l+x)^(2))]`
`= (q^(2))/(4pi epsilon_(0)) (4lx)/((l^(2) - x^(2))^(2))`
Since `l gt gt x, :. F = (q^(2) lx)/(pi epsilon_(0) l^(4))` or `F = (q^(2)x)/(pi epsilon_(0) l^(3))`
We see that `F prop x` and it is opposite to the direction of displacement. Therefore, the motion is SHM.
`T = 2pisqrt((m)/(k))`, here `k = (q^(2))/(pi in_(0) l^(2))`
`= 2pisqrt((m pi in_(0) l^(2))/(q^(2)))`