The Electrostatic Trap That Can't Hold

physicselectromagnetism/electrostaticsdifficulty 5/5#electrostatics#earnshaw#saddle#stability#false-attractor#domain-collision

Two equal point charges $+q$ sit on the $x$-axis at $x=\pm a$. A small test charge $+Q$ is placed at
the origin — the one point where, by symmetry, the electric field vanishes.

(a) Confirm the field is zero at the origin. (b) A tempting claim: *"the test charge is
hemmed in by two like charges, so the origin is a stable trap."* Is it? Analyse small displacements
along $x$ and along $y$ separately. (c) Show your result is forced by a single fact about the
potential in charge-free space — so no arrangement of fixed charges could ever trap $+Q$
((1)).

++ +q+q ++ +q+q QQ unstableunstable stablestable

The field of the two source charges (the origin is the central null):

-3 -2 -1 1 2 3 -2 -1.5 -1 -0.5 0.5 1 1.5 2 x y

An equipotential of the pair (the "peanut" enclosing both charges):

-3 -2 -1 1 2 3 -2 -1 1 2 x y

Laplace's equation forbids the trap:

$$ \nabla^2\phi = 0 \quad\text{in charge-free space} $$(1)

Check your answer

Show answer & solution

Answer

(a) $ \vec E(0,0) = 0 \quad\text{(the two fields are equal and opposite along } x). $

(b)

Not stable. Displace along $x$: the nearer charge dominates and pushes $+Q$ further out — the origin
is a maximum of potential along $x$ (unstable). Displace along $y$: both charges push it back —
a minimum along $y$ (stable). Equal-and-opposite curvatures ⇒ a saddle, not a trap.

(c) $ \partial_{xx}\phi + \partial_{yy}\phi + \partial_{zz}\phi = 0 \;\Rightarrow\; \text{no local minimum of } \phi. $

Solution

Necessary insight. Stability is about the curvature of the potential, and in charge-free
space the curvatures can't all have the same sign — (1) ties them together. That one
fact decides the whole problem before any force is computed.

(a) By symmetry the $x$-components of the two fields cancel and the $y$-components are zero on the
axis, so $\vec E(0,0)=0$.

(b) Write $\phi$ near the origin as $\phi_0 + \tfrac12(\phi_{xx}x^2 + \phi_{yy}y^2 + \phi_{zz}z^2)$.
Direct expansion (or the field plot above) shows $\phi_{xx}<0$: a push outward along the axis, so
the origin is a hilltop along $x$. Transverse to the axis $\phi_{yy}>0$ and $\phi_{zz}>0$: restoring.
So the null is a saddle — stable transversely, unstable along the axis.

(c) The false attractor is "surrounded ⇒ trapped." But $\nabla^2\phi = \phi_{xx}+\phi_{yy}+\phi_{zz}=0$
between the charges, so the three curvatures sum to zero and cannot all be positive: $\phi$ has **no
local minimum anywhere in free space. This is Earnshaw's theorem** — no static arrangement of
charges can stably trap another charge by electrostatics alone. (It is why ion traps must use
oscillating or magnetic fields.)

Grading

  • 2 pts (a) Field null located at the midpoint by symmetry.
  • 5 pts (b) Stability analysed in both directions; the saddle identified (not just "unstable").
  • 3 pts (c) Connected to $\nabla^2\phi = 0$ (Earnshaw), not argued case-by-case only.

Total: 10 pts