Question

Consider an electron that is 10−10 m from an alpha particle (q = 3.2 × 10−19 C). (a) What is the electric field due to the alpha particle at the location of the electron? (b) What is the electric field due to the electron at the location of the alpha particle? (c) What is the electric force on the alpha particle? On the electron?

Answer 1

Answer:

a) 28.8×10^10N/C

b) 14.4×10^10N/C

c) Force on electron particle= 46.13×10^-9N

Force on the alpha= 146.08×10^-9N

Explanation:

Using the equation, E = kq/r^2

a) E = (8.99×10^9)×(3.2×10^-19) / (10^-10)^2

E = 28.8×10^-9N/C

b) E = kq/r^2

E = (8.99×10^9)×(1.602×10^-19)/(10^-10)^2

E= 14.4×10^10N/C

c) Felectron= E× q = (28.8×10^10)×(1.602×10^-19)

Felectron= 46.13×10^-9N

Force F on alpha = E× q= 14.4x10^10)×(3.2×10^-19)

F alpha= 146.08 ×10^-9N