A vacuum parallel-plate capacitor has a square cross-section of side and plate separation
PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Conductors, capacitance and boundary-value electrostatics · Conductors, capacitance and boundary-value electrostatics
The Boundary-Value Problem, Ch The field as a differential equation: Poisson, Laplace and the well-posed problem
Problem
A vacuum parallel-plate capacitor has a square cross-section of side L=5.0cm and plate separation d=2.0cm. Take z along the axis from 0 at the bottom plate to d at the top plate, and x,y across the square cross-section, 0≤x≤L, 0≤y≤L. The bottom plate (z=0) is held at V=0 and the top plate (z=d) at V0=60V. To remove edge effects, all four side walls (x=0, x=L, y=0, y=L) are lined with many narrow horizontal guard strips, insulated from one another and each held by a voltage divider at a fixed potential, so that the wall potential rises in steps from 0 at the bottom to V0 at the top. (A single continuous conducting strip could not do this: it would be one equipotential.) Treat the strips as so narrow that each side wall carries the ideal linear potential V0z/d, so that every point of the six-sided boundary of the box-shaped region between the plates has a stated potential. A proposed potential for the interior is V(x,y,z)=V0dz.
((a)) Show that V(x,y,z) satisfies Laplace's equation inside the box.
((b)) Check that V matches the stated potential on all six faces of the boundary.
((c)) Show that ∇V is never zero inside the box, so the mean-value property's prohibition on an interior maximum or minimum is automatically respected: the extreme values 0 and V0 occur only on the bottom and top plates.