Question

An electron is traveling at 2.0E5 m/s parallel to a uniform electric field of 9.11e-3 N/C strength. The electron is traveling to the right. The electric field lines point to the right. What is the speed of the electron (in m/s) after traveling 1.8 meters to the right?

Answer 1

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:

$F_e=qE=ma$      (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:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(9.11*10^{-3}N/C)}{9.1*10^{-31}kg}\\\\a=1.6*10^9\frac{m}{s^2}$

Next, you use the following formula to calculate the final speed of the electron:

$v^2=v_o^2-2ax$      (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:

$v=\sqrt{v_o^2-2ax}=\sqrt{(2.0*10^5m/s)^2-2(1.6*10^9m/s^2)(1.8m)}\\\\v=1.85*10^5\frac{m}{s}$

The speed of the electron after it has traveled 1.8m is 1.85*10^5 m/s