Applying Gauss' Law: Spherical Symmetry — Question 10

Question 10

A particle falling into a uniformly charged sphere changes from inverse-square to harmonic motion

A fixed nonconducting sphere of radius R carries uniform positive charge Q. A weak particle of mass m and charge $-q$, with $q>0$, starts from rest at $x=2R$ and moves along a diameter. Treat the material as transparent to the particle and neglect radiation and all other forces. Define $k=1/(4\pi\epsilon_0)$.

1. Find the speed when the particle first reaches the surface at $x=R$.

2. Set time zero at that entry. Derive its interior trajectory up to the first center crossing, the crossing time and the speed at the center.

Question 10 - Solution

A particle falling into a uniformly charged sphere changes from inverse-square to harmonic motion

(a) Method and calculation

$$\tfrac12mv_R^2=\int_R^{2R}\frac{kQq}{r^2}\,dr=\frac{kQq}{2R}.$$$$v_R=\sqrt{\frac{kQq}{mR}}.$$

(b) Interpretation and further calculation

$$\ddot x+\omega^2x=0,\quad\omega^2=\frac{kQq}{mR^3},\quad x(0)=R,\quad\dot x(0)=-\omega R.$$$$x(t)=R(\cos\omega t-\sin\omega t),\quad t_c=\frac\pi{4\omega},\quad |v_c|=\sqrt2\,\omega R.$$The initial inward speed at the surface makes the center crossing faster than a quarter of the interior harmonic period.

Physical check

The work gained inside is another $kQq/(2R)$, giving total center kinetic energy $kQq/R$ and confirming the derived center speed.