Two small particles charged with equal positive charges Q each, are fixed apart at a distance 2a. Another small particle having a charge q lies midway between the fixed charges. Show that
(i) For small displacement (relative to a) along line joining the fixed charges, the middle charge executes SHM'if it is +ve and
(ii) For small lateral displacement, it executes SHM ifit is-ve. Compare the frequencies of oscillation in the two cases

Explanation

The two situations are shown in figure (i) Let x be the displacement of the charge +q from the mean position. Now net force acting on the charge q toward its equilirbium position is
`F=(KQq)/((a-x)^2)-(KQq)/((a+x)^2)`

`=(4KQqax)/((a^2-x^2)^2)~~(4KQqax)/(a^4)[Asxltlta]`
`~~(4KQqx)/(a^3)`
Restoring acceleration, a `=F/m=-(4KQqx)/(ma^3)`
[– ve sign shows restoring tendency]
`a=-omega^2x` [where m is the mass of the charge] As acceleration is directly proportional to displacement, hence the motion is SHM. Its time period `T_(1)` is given by
`T_(1)=(2pi)/(omega)`
`T_1=2pisqrt(((ma^3)/(4QqK)))=2pisqrt((piepsilon_(0)ma^3)/(qQ))" "...(1)`
(ii) Restoring force on –q toward Q is given by
`F=(2KQq)/((a^(2)+x^2)).(x)/(sqrt((a^(2)+x^(2))))`
`=(2KQq)/((a^(2)+x^2)^(3//2))~~(2KQqx)/(a^3)" " ["As x "ltlta]`
Restoring acceleration `=F/m=-(2KQq)/(ma^3)x`
`a=-omega^2x`
Hence the motion is SHM. Its time period `T_2` is given by
`T_2=(2pi)/(omega)`
`T_2=2pisqrt((ma^3)/(2QqK))=2pisqrt((2piepsilon_0ma^(3))/(qQ)" "...(2)`
Now, `(n_(1))/(n_(2))=(T_2)/(T_1)=sqrt(2)`