Question

An electron initially 3.00 m from a nonconducting infinite sheet of uniformly distributed charge is fired toward the sheet. The electron has an initial speed of 390 m/s and travels along a line perpendicular to the sheet. When the electron has traveled 2.00 m , its velocity is instantaneously zero, and it then reverses its direction.

Required:
a. What is the surface charge density on the sheet?
b. Given the same initial velocity, from what distance should the electron be fired if it is to just reach the sheet?

Answer 1

Answer:

a.    σ = 3.82*10^-18C/m^2

b.    d = 2.00m

Explanation:

a. In order to calculate the surface charge density of the sheet, you first calculate the acceleration of the electron on its motion.

You use the following formula:

$v^2=v_o^2-2ad$                (1)

v: final speed of the electron = 0m/s

vo: initial speed of the electron = 390m/s

a: acceleration of the electron = ?

d: distance traveled by the electron = 2.00m

You solve the equation (1) for a, and replace the values of all parameters:

$a=\frac{v_o^2-v^2}{2d}\\\\a=\frac{(390m/s)^2}{2(2.00m)}=3.8*10^4\frac{m}{s^2}$

Next, you calculate the electric field that exerts the electric force on the electron, by using the second Newton law, as follow:

$F_e=qE=ma$               (2)

q: charge of the electron = 1.6*10^-19C

E: electric field of the sheet = ?

m: mass of the electron = 9.1*10^-31kg

You solve the equation (2) for E:

$E=\frac{ma}{q}=\frac{(9.1*10^{-31}kg)(3.8*10^{4}m/s^2)}{1.6*10^{-19}C}\\\\E=2.16*10^{-7}\frac{N}{C}$

Next, you use the following formula to calculate the surface charge density, by using the value of its electric field:

$E=\frac{\sigma}{2\epsilon_o}$          (3)

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

σ: surface charge density of the sheet

You solve for σ:

$\sigma=2\epsilon_o E=2(8.85*10^{-12}C^2/Nm^2)(2.16*10^{-7}N/C)\\\\\sigma=3.82*10^{-18}\frac{C}{m^2}$

The surface charge density of the sheet id 3.82*10^-18C/m^2

b. To calculate the required distance for the electron reaches the sheet, you take into account that the electron acceleration is the same in all places near the sheet, then by the result of the previous point, you can conclude that the electron must be fired from a distance of 2.00m.