At some instant the velocity components of an electron moving between two charged parallel plates are `v_x=1.5xx10^5m//s` and `v_y=3.0xx10^6 m//s`.Suppose that the electric field between the plates is given by `E = (120 N//C) hatj`
. (a) What is the acceleration of the electron?
(b) What will be the velocity of the electron after its x-coordinate has changed by 2.0 cm?

Answer

A, B

Explanation

a. `a=(qE)/m=-((1.6xx10^-19)(120hatj))/((9.1xx10^-31))`
`=-(2.1xx10^13hatj)m//s^2`
`b. t=X/v_x=(2xx10^-2)/(1.5xx10^5)=4/3xx10^-7s`
`v=u+at`
`=(1.5xx10^5hati+3.0xx10^6hatj)`
`-(2.1xx10^13xx4/3xx10^-7)hatj`
`=(1.5hati+2hatj)xx10^5m//s`