Question

An electron is released from rest in a weak electric field given by E,--2.50 x 10.10 N/C j. After the electron has traveled a vertical distance of 1.8 μm, what is its speed?

Answer 1

`v = 1.26 * 10^8 m/s`

Step-by-step explanation:

Force on an electron while it moves through constant electric field is given as

`F = eE`

`e = 1.6 * 10^(-19) C`

`E = 2.50 * 10^(10) N/C`

`F = (2.50 * 10^(10))(1.6 * 10^(-19))`

`F = 4 * 10^(-9) N`

now by work energy theorem we can say

Work done by electric field = kinetic energy of the electron

`F . d = (1)/(2)mv^2`

`(4 * 10^(-9))(1.8 * 10^(-6)) = (1)/(2)(9.11 * 10^(-31))v^2`

`v = 1.26 * 10^8 m/s`