A particle of mass `2 xx 10^-5 kg` moves horizontally between two horizontal plates of a charged parallel plate capacitor between which there is an electric field of `200 NC^-1` acting upward. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both `vec B and vec v.` If g is `9.8 ms^-2` and the charge on the particle is `10^-6 C,` then find the velocity of charge particle so that it continues to move horizontally

Options

A. `2 ms^-1`
B. `20 ms^-1`
C. `0.2 ms^-1`
D. `100 ms^-1`

Answer

a

Explanation

Net force on the particle should be zero.

`qE=10^-6xx200=2xx10^-4N`
`mg=2xx10^-5xx9.8=1.96xx10^-4N`
Since `qEgtmg`, so magnetic force `qvB` should act down
ward to balance the forces
`qE=mg+qvB implies 2xx10^-4=1.96xx10^-4+10^-6vxx2`
`implies v=2m//s`