PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electrostatic fields and potential · 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
Four equal fixed point charges +Q sit at (±b,0,0) and (0,±b,0), all in the z=0 plane. A test charge +q is placed at the origin, where by symmetry E=0. Someone argues: "By symmetry the field vanishes at the origin. If the test charge is nudged along x, y, or z, the symmetric arrangement of positive charges pushes it straight back, so the origin is a genuine point of stable equilibrium in three dimensions, a static electric trap." The same person then adds: "Since V has a minimum for the positive test charge at the origin, we can also specify ∂V/∂n=0 on a small sphere centred there, independently of the equipotential value V=V0 found on that sphere. Having both pins the equilibrium down as the solution of a proper boundary-value problem."
((a)) Is the first claim (stability in all directions) correct? Justify the answer using the mean-value property and Earnshaw's theorem, without computing any curvature.
((b)) What is wrong with the second claim, about specifying V and ∂V/∂n independently on the small sphere?