A particle of mass `10^(-3)`kg and charge `5muC` is thrown at a speed of `20ms^(-1)` against a uniform electric field of strength `2xx10^(5)NC^(-1)`. The distance travelled by particle before coming to rest is

Options

A. 0.1 m
B. 0.2 m
C. 0.3 m
D. 0.4 m

Answer

B

Explanation

`F=qE=5xx10^(-6)xx2xx10^(5)=1N`
Since, the particle is thrown against the field
`therefore" "a=-F//m=-1/10^(-3)=-10^(3)ms^(-2)`
`As" "v^(2)-u^(2)=2as`
`therefore" "0^(2)-(20)^(2)=2xx(-10^(3))xxsors=0.2m`