Two circular rings A and B each of radius `a=30cm` are placed co-axially with their axes horizontal in a uniform electric field `E=10^(5)N//C` directed vertically upward as shown in the figure. The distance between the centers of these rings A and B is `h=40cm`. Ring A has a positive charge `q_(1)=10muC` while ring B has a negative charge of magnitude `q_(2)=20muC`. A particle of mass m = 100 gm and carrying a positive charge `q=10muC` is released from rest at the centre of the ring A. Calculate its velocity when it has moved a distance 40 cm.

Explanation

Weight of the particle `W=mg=(100)/(1000)xx10=1` Newton
Force due to electrostatic field `E=qE=10xx10^(-6)xx10^(5)=1` Newton.
Thus the weight of the particle is balanced by the electrostatic force.
The particle experiences net force in the through the centers of ring A and B and is accelerated towards the centre of ring B. At the centre of ring B.
KE = Loss of PE
`implies (1)/(2)mv^(2)=U_(A)-U_(B)=(1)/(4piin_(0))(qq_(1))/(a)+(1)/(4piin_(0))(q(-q_(2)))/(sqrt(a^(2)+h^(2)))-(1)/(4piin_(0))(qq_(1))/(sqrt(a^(2)+h^(2)))-(1)/(4piin_(0))(q(-q_(2)))/(a)`
`(1)/(2)mv^(2)=(q)/(4piin_(0))[q_(1)((1)/(a)-(1)/(sqrt(a^(2)+h^(2))))+q_(2)((1)/(a)-(1)/(sqrt(a^(2)+h^(2))))]`
`=(q)/(4piin_(0))[(q_(1)+q_(2))((sqrt(a^(2)+h^(2))-a)/(asqrt(a^(2)+h^(2))))]`
`implies v=sqrt((2q(q_(1)+q_(2)))/(4piin_(0)m)[(sqrt(a^(2)+h^(2))-a)/(asqrt(a^(2)+h^(2)))])`
For the given value `v=6sqrt(2)m//s`