Complete question:
An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: F₁ = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force F₂.
Answer:
(a) The net force of the electron, ∑F = 4.07 x 10⁻¹⁷ N
(b) the electric force, F₂ = 4.8 x 10⁻¹⁷ N
Explanation:
Given;
initial velocity of the electron,
= +6.18 x 10⁵ m/s
final velocity of the electron,
= 2.59 x 10⁶ m/s
the distance traveled by the electron, d = 0.0708 m
The first electric force, ![F_1 = 8.87*10^{-17} \ N](https://tex.z-dn.net/?f=F_1%20%3D%208.87%2A10%5E%7B-17%7D%20%5C%20N)
(a) The net force of the electron is given as;
∑F = F₁ - F₂ = ma
where;
a is the acceleration of the electron
![a = \frac{v_f^2 -v_0^2}{2d} \\\\a = \frac{(2.59*10^6)^2 -(6.18*10^5)^2}{2(0.0708)}\\\\a = 4.468*10^{13} \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv_f%5E2%20-v_0%5E2%7D%7B2d%7D%20%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B%282.59%2A10%5E6%29%5E2%20-%286.18%2A10%5E5%29%5E2%7D%7B2%280.0708%29%7D%5C%5C%5C%5Ca%20%3D%204.468%2A10%5E%7B13%7D%20%5C%20m%2Fs%5E2)
∑F = ma = (9.11 x 10⁻³¹ kg)(4.468 x 10¹³)
∑F = 4.07 x 10⁻¹⁷ N
(b) the electric force, F₂ is given as;
∑F = F₁ - F₂
F₂ = F₁ - ∑F
F₂ = 8.87 x 10⁻¹⁷ - 4.07 x 10⁻¹⁷
F₂ = 4.8 x 10⁻¹⁷ N