A particle of mass `10 ^(-3) ` kg and charge ` 5 mu C ` centers into a uniform electric field of ` 2 xx 10^(5) N C^(-1) ` moving with a velocity of `20 ms^(-1) ` in a direction opposite to that of the field. Calculate the distance it would travel before coming to rest.

Explanation

Here mass of particle m `= 10 ^(-3) kg , `charge `q= 5 muC= 5 xx 10^(-6) C, ` initial velocity of particle `=u =20 ms ^(-1) , ` final velocity v= 0 and electric field E = `-2xx 10 ^(5) N C^(-1) `. (The electric field has been taken -ve because its direction is opposite to that of direction of motion of charged particle. )
` therefore ` Acceleration or particle ` a= (F)/(m)=(qE)/(m)= ( (5xx 10^(-6))xx (-2xx 10^(5)))/(10^(-3)) =-10 ^(3) ms ^(-2) `
From equation `v^(2) -u^(2) = 2as`, the distance travelled by charged particle.
` " " s=( v^(2) -u^(2))/(2a) =((0)^(2) -(20)^(2))/(2(-10^(3)) =0.2 m `