A ball of mass `10^(-2) kg` and having charge `+ 3 xx 10^(-6)C` is tied at one end of a `1m` along thread. The other end of the thread is fixed and a charge `- 3 xx 10^(-6) C` is placed at this end. The ball can move in the circulr orbit of radius `1m` in the vertical plane. Initially, the ball is at the bottom. Find the monimum initial horizontal velocity of the ball, so that it will be able to complete the full circle.

Explanation

Let `F` be the elctric force and `T` be the tension in the string, both towards `O`, Fig.
The centripetal force required `= mv^(2)//r`
As is clear from Fig.
`T + F - mg cos theta = (mv^(2))/(r )`
or `T = (mv^(2))/(r ) - F + mg cos theta` ..(i)
`T` will be minimum, when `cos theta = min. = - 1`
i.e. `theta = 180^(@)` i.e. at highest point `H` of the vertical circle, where velocity is `v_(H)`.
From (i), `T_(min) = (mv_(H)^(2))/(r ) - F - mg` ..(ii)
For looping the loop, `T_(min) ge 0`
From (ii), `[m(v_(H)^(2))/(r ) - F - mg] ge or (mv_(H)^(2))/(r ) ge (F + mg)` ..(iii)
or `v_(H)^(2) ge (r )/(m) [F + mg] or v_(H)^(2) ge [(F)/(m) + g]` (`:' r = 1m)` ..(iv)
According to coulomb's law in electrostatics, `F = (kq_(1)q_(2))/(r^(2))`
where `K = 9 xx 10^(9)Nm^(2)c^(-2)`
`F = (9xx10^(9)(3xx10^(-6))(3xx10^(-6)))/(1^(2)) = 81 xx 10^(-3)N`
From (iv), `[v_(H)^(2)]_(min) = [(81 xx 10^(-3))/(10^(-2)) + 10] = 81.1`
If `v_(L)` is minimum velocity of projection at lower point, then applying the principle of conservation of energy, total energy at `L =` total energy at `H`
`(1)/(2) mv_(L)^(2) = (1)/(2)mv_(H)^(2) + mg(2r)`
`v(L)^(2)= v_(H)^(2) + 4gr`
`v_(L) = sqrt(18.1+4 xx10 xx 1) = sqrt(58.1) = 7.6 m s^(-1)`