Saved Bookmarks
| 1. |
Take the particle in question number 49 an electron projected with velocity v_(x)=4xx10^(6)ms^(-1). If electric field between the plates separated by 1 cm is 8.2xx10^(2)NC^(-1), then the electron will strike the upper plate if the length of plate is (Take m_(e)=9.1xx10^(-31)kg) |
|
Answer» 2.14cm `E=8.2xx10^(2)NC^(-1),q=e=1.6xx10^(-19)C,m_(e)=9.1xx10^(-31)kg` The electron will strike the UPPER plate at its other end of x = L as soon as its deflection. And `y=d/2=10^(-2)/2M=5xx10^(-3)m` From eqn. (III), `L=sqrt((2m_(e)v_(x)^(2)y)/(qE))=sqrt((2xx9.1xx10^(-31)xx(4xx10^(6))^(2)xx5xx10^(-3))/(1.6xx10^(-19)xx8.2xx10^(2)))` `L=3.3xx10^(-2)m` = 3.3 cm |
|