Capacitance, Energy & Method of Images (IPhO E&M)

IPhO E&M lesson on capacitance as geometry, energy methods for forces, and the method of images for grounded conductors.

International Physics Olympiad preparation

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Capacitance is not “another formula”. It is a geometric property of conductors and dielectrics, and the fastest IPhO solutions often come from energy and boundary conditions rather than direct integration. The method of images is the cleanest example: you replace a conductor boundary by fictitious charges so the potential automatically satisfies the boundary condition.

Identify the conductors (or nodes) and what is held fixed: charge Q or potential difference V.
Find either V(Q) (via vector E and Δ V = -∫ vector E · d vector ℓ) or Q(V).
Get C = Q/V (two-conductor) or C = Q/V relative to infinity (isolated conductor).
For forces, avoid direct Coulomb sums: use energy with the correct constraint (fixed Q vs fixed V).
For grounded conductors, consider images: propose a potential that satisfies Laplace/Poisson plus the boundary condition, then use uniqueness.

1. Definitions (Must Know)

Capacitance of a two-conductor system:
C = Q/(Δ V)
where Q is the charge on one conductor and Δ V is the potential difference.
Common energy identities:
U = (1/2)QV = Q²/2C = (1/2)CV².
Energy density in a linear dielectric:
u = (1/2)ε E².
Force from energy at fixed Q (1D coordinate x):
Fₓ = -(d/dx)(Q²/2C(x)).
Force from energy at fixed V:
Fₓ = (1/2)V²dC/dx.
Method of images (core idea): choose fictitious “image charges” so that the potential satisfies the conductor boundary condition (typically V = 0 for a grounded conductor). By the uniqueness theorem, that potential is the physical solution in the accessible region.

2. Key Ideas (What Earns Marks)

Capacitance is geometry. If you can get vector E(r) from Gauss and then integrate to get Δ V, you can get C in one line.
Energy beats force integrals. Many “find the force” questions collapse to differentiating U(Q) or an effective U(V).
Constraints matter. The sign mistake in energy methods almost always comes from silently switching between fixed Q and fixed V.
Images are boundary-condition hacks. You do not compute induced surface charge first. You guess a potential with images, then you can compute forces/induced charge if needed.
Images only work for special boundaries. Infinite grounded plane, grounded sphere, some wedges. If the boundary is not one of these, do not force it.

3. Detailed Explanations

A. Capacitance by E then V.
For a highly symmetric geometry:

Use Gauss to find E(r).
Integrate Δ V = -∫ vector E · d vector ℓ.
Use C = Q/Δ V.

Classic example (coax): for charge per length λ on the inner conductor, E(r) = λ/(2πε r) in the dielectric, so

Δ V = ∫ₐ^b λ/(2πε r) dr = λ/2πε ln(b/a), C/L = λ/(Δ V) = 2πε/(ln(b/a)).

B. Where does the charging energy go?
Charging an initially uncharged capacitor C up to voltage V stores field energy U = 1/2 CV². If an ideal battery of emf V is connected quasi-statically, the battery supplies

W_batt = ∫₀^Q V dQ = VQ = CV²,

so half ends in the field and half is dissipated (in the resistor, in radiation, or in whatever non-ideal mechanism actually makes the current stop).

C. Force from energy (fixed Q vs fixed V).
Suppose a mechanical coordinate x changes the capacitance C(x).

If the capacitor is isolated so Q is fixed:
U(Q,x) = Q²/2C(x) ⇒ Fₓ = -(∂ U)/(∂ x) = (Q²/2C(x)²)dC/dx.
If the capacitor is held at constant V by a battery, the field energy 1/2 C V² is not the correct mechanical potential energy because the battery exchanges energy as Q = CV changes. The correct force result is
Fₓ = (1/2)V²dC/dx,
which points toward increasing C (larger overlap, dielectric pulled in, smaller gap).

D. Method of images: grounded plane (the workhorse).
A point charge q at distance a above an infinite grounded conducting plane can be replaced (in the region above the plane) by the real charge and an image charge -q the same distance below the plane. On the plane, the distances to q and -q are equal, so the potential cancels to 0 automatically.

The force on the real charge equals the Coulomb force due to the image:

F = (1/4πε₀)q²/(2a)² = q²/(16πε₀ a²),

directed toward the plane.

4. Common Mistakes

Using U = CV² instead of U = 1/2 CV².
Using the fixed-Q force formula in a fixed-V setup (or vice versa).
Writing C = ε A/d in a geometry where fringing or non-parallel surfaces dominate.
Treating the image charge as a “real charge on the conductor” rather than a mathematical device for the potential in the allowed region.
Forgetting that images require the correct boundary condition (usually grounded, V = 0).

5. Exam Tips

Write “fixed Q” or “fixed V” on your diagram before using energy.
For conductors, use potential language: conductors are equipotentials; field lines hit the surface normally.
Dimension check: [C] = F, [U] = J, [u] = J m⁻³.
For images, verify the boundary condition explicitly (one sentence is enough): “On the conductor surface, the proposed V is zero/constant.”

6. Worked Examples

Example 1 (charging a capacitor: energy accounting).
An uncharged capacitor C is connected to an ideal battery of emf V through a resistor. After a long time, how much energy is stored in the capacitor, and how much energy was delivered by the battery?

Solution sketch

Final charge is Q = CV.

Energy stored in the electric field:

U = (1/2)CV².

Battery energy delivered:

W_batt = VQ = CV².

The remainder W_batt-U = 1/2 CV² is dissipated (typically as heat in the resistor).

Example 2 (capacitance per unit length of a coaxial capacitor).
Two long coaxial conductors have radii a and b with b larger than a, filled with dielectric of permittivity ε. Find C/L.

Solution sketch

Put charge per length λ on the inner conductor. By Gauss:

E(r) = λ/(2πε r).

Potential difference:

Δ V = ∫ₐ^b E(r) dr = λ/2πε ln(b/a).

Therefore

C/L = λ/(Δ V) = 2πε/(ln(b/a)).

Example 3 (images: force and induced charge on a grounded plane).
A point charge q is a distance a above an infinite grounded conducting plane. Find the force on q and the total induced charge on the plane.

Solution sketch

Replace the plane by an image charge -q at the mirror position. The separation between q and -q is 2a, so the force magnitude on q is

F = (1/4πε₀)q²/(2a)² = q²/(16πε₀ a²),

directed toward the plane.

The induced surface-charge density follows from the normal field just above the conductor. In cylindrical coordinates, with radial distance ρ from the point directly below q,

σ(ρ) = ε₀E_z(0⁺) = -qa/(2π(ρ² + a²)^(3/2)).

Integrating over the whole plane gives

Q_ind = ∫₀^∞ σ(ρ) 2πρ dρ = -qa∫₀^∞(ρ dρ)/((ρ² + a²)^(3/2)) = -q.

This avoids the invalid claim that a closed surface enclosing the real charge has zero net flux: its hemisphere and disk contributions must be treated together under Gauss’s law.

7. Mind Stretchers

Mind stretcher: point charge outside a grounded conducting sphere

A point charge q is placed a distance d from the center of a grounded conducting sphere of radius R, along a line through the center. Assume d is larger than R. Use the image method to find the image charge and the force on q.

Solution idea: for a grounded sphere, a single image charge on the same line works.

The image charge is

q' = -qR/d,

located at distance

r' = R²/d

from the center (toward the real charge).

The separation between the real charge and the image is

s = d-r' = d-R²/d = (d²-R²)/d.

The force magnitude on q is the Coulomb force due to q':

F = (1/4πε₀)|qq'|/s² = (1/4πε₀)(q² R d)/((d²-R²)²),

directed toward the sphere (attraction).

8. Practice

Re-derive U = 1/2 CV² by integrating V dQ.
Do one force-by-energy problem at fixed Q and one at fixed V (write the constraint explicitly).
Do at least one grounded-plane image problem and one grounded-sphere image problem.

Syllabus and review details

No official syllabus alignment is listed for this lesson.