Question

In a horizontal unifrom electirc field, a small charged disk is gently released on the top of a fixed sperical dome. The disk slides down the some without friction and breaks away from the surface of the dome at the angular position `theta = sin^(-1)(3//5)` from verical. Determine the ration of the force of gravity acting on the disk to force of its interaction with the field .

Answer 1

Motoin fron A to B
`mgR(1-cos theta)+qER sin theta =(mv^(2))/(2)`
Given `sin theta=(3)/(5)`
`:. cos theta=(4)/(5)`
`:. mgR((1)/(5))+qER((3)/(5))=(mv^(2))/(2)`
or `(mg+3qE) = (5)/(2) (mv^(2))/(R )` …(i)
Also the equation of circular motion is
`mg cos theta-N-qE sin theta=(mv^(2))/(R )`
When the paricle loses contact , `N=0, sin theta=(3)/(5)`, and `cos theta=(4)/(5)`.Therefore ,
`mg((4)/(5))-qE((3)/(5))=(mv^(2))/(R )`
or, `4mg-3qE=5(mv^(2))/(R)` ...(ii)
Form Eqs, (i) and (ii), `(mg+3qE)/(4mg-3qE)=(1)/(2)`
or, `2mg+6qE=4mg-3qE`
or , `2mg=9qE` or `(mg)/(qE)=(9)/(2)`.