Review / Chapter 3 — Conductors

Faraday Cage & Cavities

Chapter 3 — Conductors

Conductors perfectly screen their interiors from external fields — stand inside a metal cage during a lightning strike and you'll be fine. But the interior isn't always field-free; it depends on whether there are charges inside the cavity. This page pulls together the cavity story and the "Faraday cage" result.

The Faraday cage result

Suppose you have a closed conducting shell (of any shape) with no charge inside the hollow cavity. Then no matter what external charges or fields surround the conductor, the field inside the cavity is zero.

Proof sketch (uniqueness theorem): inside the cavity, \(\nabla^2 V = 0\) (no charge). The cavity wall is at some constant potential \(V_0\). A candidate solution is \(V(\vec r) \equiv V_0\), which satisfies Laplace's equation and the boundary condition. By uniqueness it's the solution. Constant potential means \(\vec E = 0\).

Conductor in an external field with a hollow cavity; the cavity interior is field-free
A conductor sitting in an external field. The induced charge on the outer surface cancels the external field inside the conductor. Inside the empty cavity, the field is zero.

Charge inside the cavity

Now put a charge \(+q\) inside a cavity of a neutral conductor. Two things happen:

So the outside world sees an effective charge \(+q\) on the outer surface, while the cavity world sees only the original \(+q\) plus its induced \(-q\) on the wall. The inside and outside are electrostatically decoupled by the conductor.

A clean statement

The field inside a cavity depends only on the charges inside the cavity. The field outside the conductor depends only on the charges outside plus the total charge on the outer surface.

Earnshaw revisited

You can't use conductors to trap a charge stably either. In a charge-free region, \(\nabla^2 V = 0\); solutions have no interior minima. Any "trap" configuration using conductors alone fails — the charge will always find a saddle direction to escape. (This is why magnetic or time-dependent fields are needed for ion traps.)

Practical consequences

Practice Problems

Problem 1easy
A neutral hollow metal sphere sits in a uniform external field. What is the field inside the hollow? Where does the induced charge sit?
Hint
Faraday cage argument.
Solution

The cavity is empty of charge and enclosed by an equipotential; \(\vec E = 0\) inside by uniqueness. Induced charges sit on the outer surface of the shell; the inner surface is bare.

Answer: \(\vec E = 0\) inside; induced charges live only on the outer surface.

Problem 2medium
A neutral conductor has two cavities of radii \(r_1\) and \(r_2\). Point charges \(q_1\) and \(q_2\) sit at the centers of the two cavities. A third charge \(q_0\) sits far outside (distance \(d \gg R\)). Find (a) the force on \(q_1\), (b) the charges on each of the cavity walls and the outer surface, (c) the approximate force on the conductor.
Hint
Charges inside cavities are screened from everything external. The outer-surface charge responds to outside influences.
Solution

(a) Force on \(q_1\): By the cavity-screening argument, the only field that reaches \(q_1\) comes from charges inside its cavity. With \(q_1\) at the center of a spherical cavity, the induced \(-q_1\) on the cavity wall is uniform and exerts zero force on \(q_1\). \(F_1 = 0\). Same for \(q_2\).

(b) Wall 1 has induced \(-q_1\); wall 2 has induced \(-q_2\). By conservation, the outer surface has \(+q_1 + q_2\) total. The presence of \(q_0\) reshuffles that charge on the outer surface but doesn't change the total (conductor is neutral except for these induced redistributions).

(c) Force on the conductor: approximately the Coulomb force between \(q_0\) and the effective outer-surface charge \(q_1 + q_2\), treating the conductor as a point charge at \(d \gg R\): \(F \approx \frac{1}{4\pi\epsilon_0}\frac{q_0(q_1 + q_2)}{d^2}\).

Answer: \(F_1 = F_2 = 0\); walls: \(-q_1\), \(-q_2\); outer surface: \(+q_1 + q_2\); \(F_\text{cond} \approx kq_0(q_1+q_2)/d^2\).

Problem 3medium
A charge \(+q\) sits off-center inside a neutral spherical conducting shell. Is the induced charge distribution on the inner wall uniform? What about the outer surface?
Hint
Screening applies in only one direction.
Solution

Inner wall: non-uniform — denser near \(q\). The induced distribution exactly cancels the field of \(q\) outside the cavity. Outer surface: uniform (by symmetry, since the conductor's outer boundary is a sphere and no external charges are present). The outer surface carries \(+q\) spread uniformly — the conductor hides where \(q\) is inside.

Answer: inner wall non-uniform, outer surface uniform at \(\sigma = q/(4\pi R^2)\).

Problem 4hard
Using uniqueness, argue that a charge cannot be held in stable static equilibrium at a point in a region containing only conductors (and no other charges at that point).
Hint
What would stability require about \(\nabla^2 V\)? Is that compatible with Laplace's equation?
Solution

Stability at a point \(P\) would require that \(V\) (for a positive test charge) has a local minimum at \(P\) — meaning \(\nabla^2 V > 0\) there. But in a charge-free region, \(\nabla^2 V = 0\) (Laplace's equation). Contradiction — so no strict minimum exists. At best you get saddles, which aren't stable. Conclusion: no static electrostatic trap for point charges.

Answer: Laplace's equation forbids interior minima of \(V\), so no stable equilibrium.