Problem 5: Particle in a Tetrahedron
Question: Four infinite planes form a tetrahedron. Three edges meeting at vertex O have length \(l\), and the opposite face is equilateral with edge \(b\). All carry \(+\sigma\). A particle \((m, q)\) is released from rest on the symmetry axis at distance \(d\) from the lower face. Find the collision speed.
1. Net Force Analysis
The particle experiences forces from all four faces. The base face pushes it up (towards O), while the three side faces push it down (towards the base).
The net electric field magnitude depends on the angle \(\theta\) of the side faces:
$$ E_{net\_down} = 3 E_{side} \cos\theta – E_{base} $$Using the geometry of the tetrahedron, the geometry factor is:
$$ \text{Factor} = \frac{\sqrt{3}b}{\sqrt{4l^2 – b^2}} $$The direction of motion depends on whether the downward force from the sides is stronger than the upward force from the base.
2. Case Analysis
Case A: Particle hits the bottom (\( \frac{b}{\sqrt{3}} < l < b \))
If \(l < b\), the side faces are steep, and their downward vertical component dominates. The net force is down.
- Distance traveled: \(d\)
- Acceleration: \(a = \frac{q\sigma}{2m\epsilon_0} \left( \frac{\sqrt{3}b}{\sqrt{4l^2 – b^2}} – 1 \right)\)
$$ v = \sqrt{ \frac{q \sigma d}{m \epsilon_0} \left( \frac{\sqrt{3}b}{\sqrt{4l^2 – b^2}} – 1 \right) } $$
Case B: Particle hits the top corner (\( l > b \))
If \(l > b\), the side faces are flatter. The upward force from the base dominates. The particle accelerates upwards toward vertex O.
- Tetrahedron Height: \(H = \sqrt{l^2 – \frac{b^2}{3}}\)
- Distance traveled: \(H – d\)
- Acceleration: \(a = \frac{q\sigma}{2m\epsilon_0} \left( 1 – \frac{\sqrt{3}b}{\sqrt{4l^2 – b^2}} \right)\)
$$ v = \sqrt{ \frac{q \sigma}{m \epsilon_0} \left( \sqrt{l^2 – \frac{b^2}{3}} – d \right) \left( 1 – \frac{\sqrt{3}b}{\sqrt{4l^2 – b^2}} \right) } $$
Case C: Stays Immobile (\( l = b \))
If \(l = b\), the upward and downward forces cancel each other out exactly.
$$ v = 0 $$