Applying Gauss' Law: Planar Symmetry — Question 4

Question 4

A uniformly charged slab produces harmonic motion inside its volume

An infinite insulating slab occupies $|x|<a$ and has fixed uniform density $\rho>0$. There are no other fields. A weak particle of mass m and charge $-q$, with $q>0$, starts at rest at $x=A$, where $0<A<a$. Ignore collisions and radiation.

1. Derive the interior and exterior slab fields from symmetry and Gauss law.

2. Derive the particle trajectory and period, and show that the entire orbit remains within the region where the interior expression applies.

Question 4 - Solution

A uniformly charged slab produces harmonic motion inside its volume

(a) Method and calculation

$$x>0:\quad 2E_x(x)S=2\rho xS/\epsilon_0\Rightarrow E_x(x)=\rho x/\epsilon_0\quad(|x|<a).$$$$E_x(x)=\frac{\rho a}{\epsilon_0}\operatorname{sgn}(x)\quad(|x|>a).$$

(b) Interpretation and further calculation

$$m\ddot x=-qE_x=-\frac{q\rho}{\epsilon_0}x,\quad\omega=\sqrt{\frac{q\rho}{m\epsilon_0}}.$$$$x(t)=A\cos\omega t,\quad T=2\pi\sqrt{\frac{m\epsilon_0}{q\rho}},\quad|x(t)|\le A<a.$$

Physical check

The amplitude restriction matters: a larger excursion would enter the constant-field exterior and would not remain purely harmonic.