Applying Gauss' Law: Planar Symmetry — Question 7

Question 7

A neutral metal insert changes gap width while leaving fixed-sheet field unchanged

Infinite insulating sheets at $x=0$ and $x=d$ carry fixed $+\sigma$ and $-\sigma$, with $\sigma>0$. A neutral conducting slab of thickness t is centered between them, with $0<t<d$. Assume equilibrium and zero exterior field.

1. Find the field in each vacuum gap and the induced density on each metal face.

2. A weak positive charge q starts from rest just to the right of the left sheet and travels across the left vacuum gap. Find its kinetic-energy gain immediately before reaching the metal, neglecting image-force disturbance.

Question 7 - Solution

A neutral metal insert changes gap width while leaving fixed-sheet field unchanged

(a) Method and calculation

$$E_{\mathrm{gap}}=\sigma/\epsilon_0,\quad E_{\mathrm{metal}}=0.$$$$\sigma_{\mathrm{metal,left}}=\epsilon_0(0-E_{\mathrm{gap}})=-\sigma,\quad\sigma_{\mathrm{metal,right}}=+\sigma.$$

(b) Interpretation and further calculation

$$g=(d-t)/2,\quad\Delta K=q\int_0^gE_{\mathrm{gap}}\,dx.$$$$\Delta K=\frac{q\sigma(d-t)}{2\epsilon_0}.$$The field magnitude is fixed by the sheet charge, while the metal shortens the available vacuum acceleration distance.

Physical check

The probe motion ends at the metal surface; no field-free transit through conducting material is assumed.