An electron starts from rest 3cm from the center of a uniformly charged sphere of radius 2 cm. If the sphere carries a total charge of 1x10^-9C, how fast will the electron be moving when it reaches the surface of the sphere? (answer: 7.26x10^6 m/s)

Question

F=k q^2/r^2
(8.99x10^9)(1x10^-9)^2/(.02)^2
v=sqrt(Fr/m)
sqrt((2.2475x10^-5N)(.02)/(9.11x10^-31kg))=7.02x10^11 m/s--not answer given

Answer 1

Your formula is incorrect. You need to take into account the varying force as the electron changes position, and the actual charge of the eectron.

The electron starts 1 cm outside the sphere, and is attracted to the surface. I suggest using conservation of energy for this problem. The electrostatic potential energy outside the sphere is
V(r) = -k e Q/r
where e is the electron charge, Q = 10^-9 C is the charge on the sphere, r is the distance from the center of the sphere, and k is the Coulomb constant. In going from r = .03 to 0.02 m, the kinetic energy gained is
(1/2) m V^2 = -kQe (1/0.03 - 1/0.02)

m is the mass of an electron.

Solve for V

Answer 2

I still get the wrong answer.

-k(-1.6x10^-19)(10^-9)/(.03-.02)=1/2(9.11x10^-31kg)v^2

v=1.78x10^7m/s

Answer 3

Granted this is obviously a very late post, I got the right answer when I did this problem:

Potential at the surface of the sphere:
V1 = Kq/r = 8.99(1)/.02 = 450V

Potential at 3.00 cm from centre of the sphere:
V2 = Kq/r = 300V

Potential difference:
ÄV = V1 - V2 = 450 -300 = 150V

KE= 1/2mv^2 =eÄV

(.5)(9.11*10^-31)v^2 =(1.6*10^-19)(150)

v^2 = 5.27*10^13
v = 7.26*10^6 m/s