Answer:
v = 1.85*10^5 m/s
Explanation:
In order to calculate the speed of the electron after it has traveled 1.8m, you first take into account that the electric field generates a desceleration on the electron, because the direction of the electron and electric field are the same.
You use the Newton second law, to calculate the deceleration of the electron:
(1)
q: charge of the electron = 1.6*10^-19C
m: mass of the electron = 9.1*10^-31kg
E: magnitude of the electric field = 9.11*10^-3N/C
a: deceleration = ?
You solve the equation (1) for a, and replace the values of the other parameters:
![a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(9.11*10^{-3}N/C)}{9.1*10^{-31}kg}\\\\a=1.6*10^9\frac{m}{s^2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BqE%7D%7Bm%7D%3D%5Cfrac%7B%281.6%2A10%5E%7B-19%7DC%29%289.11%2A10%5E%7B-3%7DN%2FC%29%7D%7B9.1%2A10%5E%7B-31%7Dkg%7D%5C%5C%5C%5Ca%3D1.6%2A10%5E9%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Next, you use the following formula to calculate the final speed of the electron:
(2)
v: final speed of the electron = ?
vo: initial speed of the electron = 2.0*10^5 m/s
x: distance traveled by the electron = 1.8m
You solve the equation (2) for v and replace the values of the other parameters:
![v=\sqrt{v_o^2-2ax}=\sqrt{(2.0*10^5m/s)^2-2(1.6*10^9m/s^2)(1.8m)}\\\\v=1.85*10^5\frac{m}{s}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bv_o%5E2-2ax%7D%3D%5Csqrt%7B%282.0%2A10%5E5m%2Fs%29%5E2-2%281.6%2A10%5E9m%2Fs%5E2%29%281.8m%29%7D%5C%5C%5C%5Cv%3D1.85%2A10%5E5%5Cfrac%7Bm%7D%7Bs%7D)
The speed of the electron after it has traveled 1.8m is 1.85*10^5 m/s