Question

An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed increases to 6.60 x10^5 m/s in a distance of 5.40 cm. Assume its acceleration is constant.

Required:
a. Determine the magnitude of the force exerted on the electron.
b. Compare this force (F) with the weight of the electron (Fg), which we ignored.

Answer 1

a. F = 2.32*10^-18 N

b. The force F is 2.59*10^11 times the weight of the electron

Step-by-step explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

`v^2=v_o^2+2ax` (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

`a=(v^2-v_o^2)/(2x)=((6.60*10^5m/s)^2-(4.00*10^5m/s)^2)/(2(0.054m))\\\\a=2.55*10^(12)(m)/(s^2)`

Next, you use the second Newton law to calculate the force:

`F=ma`

m: mass of the electron = 9.11*10^-31kg

`F=(9.11*10^(-31)kg)(2.55*10^(12)m/s^2)=2.32*10^(-18)N`

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

`F_g=mg=(9.11*10^(-31)kg)(9.8m/s^2)=8.92*10^(-30)N`

The quotient between the weight of the electron and the force F is:

`(F)/(F_g)=(2.32*10^(-18)N)/(8.92*10^(-30)N)=2.59*10^(11)`

The force F is 2.59*10^11 times the weight of the electron