USAPhO 2008, Четвъртфинал — Задача 1

• (a) $K_0 = \frac{-qQ}{8\pi R\epsilon_0}$
• (b) $t = \frac{\pi}{2}\sqrt{\frac{4\pi\epsilon_0 m R^3}{-qQ}}$

Solution:

Assume that $q$ is negative and $Q$ is positive.

(a) Apply Gauss's Law to a spherical shell of radius $r$ where $r < R$. Then,

where the charge density $\rho = \frac{3Q}{4\pi R^3}$

Solving for the Electric Field at a distance $r$ from the center, we find

We now find the potential difference between the center of the sphere and the outer boundary of the charged cloud.

Substituting in our expression for $E$, we now find

Using conservation of energy,

(b) Apply Newton's Second Law to the object of charge $-q$ when it is located a distance $r$ away from the center of the sphere.

We recognize that this is the differential equation for simple harmonic motion

where $\omega = \sqrt{\frac{-q\rho}{3\epsilon_0 m}}$.

Since the charge has the minimum kinetic energy needed to reach the surface, the trip from the center of the sphere to the outer boundary is one-fourth of a cycle of SHM.

Therefore,

Substituting, in for $rho$, we find