Saved Bookmarks
| 1. |
A particle of mass 10 ^(-3)kg and charge5 mu Ccenters into a uniform electric field of2 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. |
|
Answer» Solution :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 ` |
|