- #1
Alan I
- 15
- 0
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 rproton 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 Vp = 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...