Proton falling through tiny hole on charged shell

[Alan I]

Homework Statement

Suppose a hollow metal shell (outer radius 25.3 cm, inner radius 5.2 cm) carries charge Q = -7.99 pC. There is a tiny hole in the sphere, so small that it does not affect the charge distribution or the electric field created by the charge.

An proton is released from rest at distance 4.96 m from the sphere, it falls toward the sphere, through the tiny hole, and reaches the center of the sphere. Find the speed of the proton, in m/s, when it reaches the center of the sphere.

Assume: V(∞) = 0, and that there is no friction anywhere.

Homework Equations

[/B]
V = kQ/r

ΔU = qV

ΔK=-ΔU

The Attempt at a Solution

I took r_proton to be 0.253 + 4.96 = 5.213 (outer radius + distance to proton) since I'm assuming charge acting as if concentrated at the center.

Then V_p = k * (-7.99 x 10^-12)/5.213 = -0.01378

ΔU = qV ⇒ ΔU = (1.6 x 10^-19) * (-0.01378) = -2.2 x 10^-21

ΔK=-ΔU

⇒ 1/2mv² - 0 = 2.2 x 10^-21

⇒v = 1.62 x 10³ χ I'm suspecting something must be wrong with my approach...

 

[DEvens]

Yeah, this is the speed if it had fallen from infinity to the radius it started at.

But you have it at speed zero at the starting distance, and falling towards the charged sphere. So it moves through a potential difference you did not include.

Extra credit: How does the force of gravity compare to the electrical force? Is it necessary to include this? How can you very quickly determine whether or not you need to bother with gravity?