Question

what is the magnitude of the electric field at a distance of 8.97e-10m from the nucleus

◆巾Corse Contents , HOMEWORK , Set 1 ( due Thurs » Nuclear charge, electric O Timer □ Notes Evaluate What is the total charge of the uranium nucleus? (The neutral uranium atom has 92 electrons.) 1.47x10-17 c You are correct. Previous Tries Feedback Print What is the magnitude of its electric field at a distance of 8.97x10-10 m from the nucleus? Remember to use the charge in coulombs. Incorrect. Tries 8/20 Previous Tries What is the magnitude of the force on an electron at that distance? What would the magnitude of the force be if the distance of the electron from the nucleus were doubled? Post Discussion Send Feedback

Answer 1

(b) Magnitude of electric field at a distance 'r' from a charge 'q' is given as

E = kq/r^2, where k = 9*10^9 Nm^2/C^2

In this case, q = 1.47*10^-17 C, r = 8.97*10^-10 m,

So, required electric field due to the nucleus is

E = 9*10^9*1.47*10^-17/(8.97*10^-10)^2 N/C

= 1.64*10^11 N/C

(c) Magnitude of force on an electron of charge 'e'

F = eE = (1.67*10^-19*1.64*10^11) N

= 2.74*10^-8 N . (as e = 1.67*10^-19 C)

(d) When distance is doubled, we have,

Electric field, E' = kq/(2r)^2 = kq/4r^2 = E/4

Hence, new force on the electron becomes

F' = eE/4 = 2.74*10^-8/4 N = 6.85*10^-9 N