Part 2 · Chapter 10

Electrostatic Boundary-Value Problems

Some geometries refuse a tidy formula — a charge near a grounded plane, a field inside a slotted box. Two ideas crack them open: replacing a conductor by a fictitious image charge, and splitting a hard partial equation into easy one-variable pieces.

Electromagnetic Field Theory Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • What makes a problem a boundary-value problem and the role of the uniqueness theorem.
  • The method of images for a point charge above a grounded plane.
  • How to find the induced surface charge and the force on the real charge.
  • The image of a line charge near a plane.
  • The (unequal) image of a charge near a grounded sphere.
  • The idea of separation of variables and Fourier-series solutions of Laplace's equation.
Section 10-1

The Boundary-Value Idea

Chapter 9 gave us Laplace's and Poisson's equations and the uniqueness theorem. This chapter is about the techniques for solving them when the geometry is awkward. Two stand out at undergraduate level: the method of images, a clever trick for charges near conductors, and separation of variables, the systematic workhorse for rectangular and other regions.

Both lean entirely on the uniqueness theorem. If we can produce any potential that satisfies Laplace's equation in the region of interest and matches the boundary conditions, it must be the correct potential — no matter how we found it, even if we invented imaginary charges along the way.

Section 10-2

Method of Images: the Plane

Consider a point charge \(+Q\) a height \(h\) above an infinite grounded conducting plane (\(V=0\)). The conductor's induced surface charge makes the field complicated — but the plane forces one simple condition: \(V=0\) everywhere on it. The method of images finds a charge arrangement, in free space, that reproduces that same condition.

V = 0 +Q h −Q h (image)
Replace the grounded plane with a mirror-image −Q a distance h below it

The trick: remove the plane and place an image charge \(-Q\) at the mirror position, a distance \(h\) below where the plane was. Two equal and opposite charges symmetric about the plane produce \(V=0\) on it automatically (every point is equidistant from both). Above the plane, the potential of the real-plus-image pair therefore equals the true potential:

Potential above a grounded plane
\[ V = \frac{Q}{4\pi\varepsilon_0}\!\left(\frac{1}{r_+} - \frac{1}{r_-}\right) \]

Here \(r_+\) and \(r_-\) are the distances from the field point to the real charge and the image. The image is purely a calculational fiction — it has no physical existence and the result is valid only in the region above the plane.

Section 10-3

Induced Charge & Force

Once the potential is known, the rest follows. The induced surface charge on the plane comes from \(\rho_S = -\varepsilon_0\,\partial V/\partial n\) at the surface; directly beneath the charge it is densest:

Induced surface charge density
\[ \rho_S = \frac{-Q h}{2\pi (x^2+y^2+h^2)^{3/2}}, \qquad Q_{\text{induced}} = \int \rho_S\,dS = -Q \]

The total induced charge is exactly \(-Q\) — every field line from \(+Q\) terminates on the plane. The force on the real charge is just its attraction to the image, an inverse-square pull toward the plane:

🔑
Force on a charge above a grounded plane
\[ F = \frac{Q^2}{4\pi\varepsilon_0 (2h)^2} = \frac{Q^2}{16\pi\varepsilon_0 h^2} \]

The charge feels exactly the force of an image \(-Q\) at separation \(2h\), always attractive — pulling it toward the conductor. This image force is why a charge is drawn to any nearby grounded surface.

Section 10-4

Line Charge Near a Plane

The same mirror trick handles an infinite line charge \(\rho_L\) parallel to a grounded plane at height \(h\). The image is a line charge \(-\rho_L\) at depth \(h\), and the potential above the plane uses the logarithmic line-charge potential of Chapter 4:

Line charge above a grounded plane
\[ V = \frac{\rho_L}{2\pi\varepsilon_0}\,\ln\frac{r_-}{r_+} \]

This is the model for a power line over the earth or a microstrip trace over a ground plane — the image accounts for the conductor exactly, and from it the line's capacitance to ground follows directly, a result we reuse for transmission lines in Part 6.

Section 10-5

Charge Near a Sphere

A point charge \(+Q\) at distance \(d\) from the centre of a grounded sphere of radius \(a\) also has an image — but, unlike the plane, the image is smaller and sits inside the sphere. Matching \(V=0\) on the spherical surface requires:

Image of a charge near a grounded sphere
\[ Q' = -\frac{a}{d}\,Q, \qquad d' = \frac{a^2}{d} \]
V=0 +Q d Q' d' = a²/d
Near a grounded sphere the image is reduced and lies inside, at d′ = a²/d

The potential outside is then the free-space potential of \(Q\) plus this image \(Q'\). The same construction gives the induced charge and the (attractive) force, and underlies the capacitance of a sphere near a plane.

Section 10-6

Separation of Variables

When no image will do — a potential specified on the walls of a rectangular trough, say — the systematic method is separation of variables. We seek a solution that is a product of single-variable functions, \(V(x,y)=X(x)\,Y(y)\). Substituting into Laplace's equation splits it into two ordinary equations linked by a separation constant:

The separated equations
\[ \frac{1}{X}\frac{d^2X}{dx^2} = -\frac{1}{Y}\frac{d^2Y}{dy^2} = -k^2 \]

Each piece solves to sines/cosines in one direction and exponentials (or hyperbolics) in the other. Summing these product solutions and fitting the boundary potentials with a Fourier series gives the full answer:

Fourier-series solution (trough example)
\[ V(x,y) = \sum_{n=1}^{\infty} c_n \sin\!\frac{n\pi x}{a}\,\sinh\!\frac{n\pi y}{a} \]
Two tools, two regimes. Reach for images when a charge sits near a simple conductor (plane, sphere, or two planes meeting at a special angle) — the answer is a short closed form. Reach for separation of variables when the potential is prescribed on the edges of a finite region and there is no charge inside — the answer is an infinite series. Recognising which situation you face is half the battle; the uniqueness theorem guarantees that whichever route reaches a fitting solution, it is the right one.
Section 10-7

Worked Examples

1 Force on a charge above a plane

Problem. A \(+10\ \text{nC}\) charge sits \(5\ \text{cm}\) above a grounded plane. Find the force on it.

Solution. Use \(F = Q^2/16\pi\varepsilon_0 h^2\):

Working
\[ F = \frac{(10\times10^{-9})^2}{16\pi(8.854\times10^{-12})(0.05)^2} = 9.0\ \mu\text{N (toward plane)} \]
2 Potential point above a plane

Problem. For \(+Q\) at height \(h\), find \(V\) at a point \(h\) directly above the charge (so \(2h\) above the plane).

Solution. Distances are \(r_+=h\) (to real) and \(r_-=3h\) (to image):

Working
\[ V = \frac{Q}{4\pi\varepsilon_0}\!\left(\frac{1}{h} - \frac{1}{3h}\right) = \frac{Q}{6\pi\varepsilon_0 h} \]
3 Peak induced charge density

Problem. For \(+Q\) at height \(h\), find \(\rho_S\) on the plane directly beneath the charge.

Solution. Set \(x=y=0\) in the induced-charge formula:

Working
\[ \rho_S = \frac{-Qh}{2\pi h^3} = \frac{-Q}{2\pi h^2} \]
4 Sphere image

Problem. A \(+6\ \text{nC}\) charge is \(20\ \text{cm}\) from the centre of a grounded sphere of radius \(5\ \text{cm}\). Find the image charge and its location.

Solution. Apply \(Q'=-(a/d)Q\) and \(d'=a^2/d\):

Working
\[ Q' = -\frac{0.05}{0.20}(6) = -1.5\ \text{nC}, \qquad d' = \frac{0.05^2}{0.20} = 1.25\ \text{cm} \]
5 Line charge image

Problem. A line charge \(\rho_L\) sits at height \(h\) above a grounded plane. State the image and the field at the plane directly below.

Solution. Image is \(-\rho_L\) at depth \(h\); the two fields add downward:

Working
\[ E = 2\cdot\frac{\rho_L}{2\pi\varepsilon_0 h} = \frac{\rho_L}{\pi\varepsilon_0 h}\ (\text{normal to plane}) \]
6 Product solution check

Problem. Verify that \(V = \sin(kx)\sinh(ky)\) satisfies Laplace's equation.

Solution. Add the second derivatives in \(x\) and \(y\):

Working
\[ \nabla^2 V = -k^2\sin(kx)\sinh(ky) + k^2\sin(kx)\sinh(ky) = 0 \;\checkmark \]
Review

Chapter Summary

✓ Boundary-value idea

Solve Laplace/Poisson for awkward geometries; uniqueness licenses any fitting solution.

✓ Plane image

Grounded plane ↔ image \(-Q\) at the mirror point; makes \(V=0\) on the plane.

✓ Force

\(F=Q^2/16\pi\varepsilon_0 h^2\), attractive; total induced charge is \(-Q\).

✓ Line image

Image \(-\rho_L\) at depth \(h\); models a wire over ground — capacitance to earth.

✓ Sphere image

\(Q'=-(a/d)Q\) at \(d'=a^2/d\) inside the sphere — reduced and displaced.

✓ Separation

Product solutions \(X(x)Y(y)\) summed as a Fourier series fit prescribed wall potentials.

Practice

Problems

For each item, decide first whether an image or a series is the right tool, then carry it through. Difficulty rises down the list.

  1. A \(+4\ \text{nC}\) charge is \(8\ \text{cm}\) above a grounded plane. Find the force on it.
  2. Find the potential \(2\ \text{cm}\) above a \(+Q=5\ \text{nC}\) charge that sits \(2\ \text{cm}\) above a grounded plane.
  3. Compute the peak induced surface charge density beneath a \(+10\ \text{nC}\) charge at height \(4\ \text{cm}\).
  4. Show by integration that the total induced charge on the plane equals \(-Q\).
  5. A line charge \(\rho_L=3\ \text{nC/m}\) runs \(6\ \text{cm}\) above a grounded plane. Find \(\vec{E}\) at the plane directly below.
  6. A \(+8\ \text{nC}\) charge is \(15\ \text{cm}\) from the centre of a grounded sphere of radius \(6\ \text{cm}\). Find the image charge and its position.
  7. Find the force between the charge and the grounded sphere in the previous problem.
  8. Two grounded planes meet at right angles; a charge sits in the quadrant between them. How many images are needed, and where?
  9. Verify that \(\cos(kx)e^{-ky}\) is a valid product solution of Laplace's equation.
  10. A rectangular trough has three grounded walls and a fourth held at \(V_0\). Set up (do not fully evaluate) the Fourier-series solution.
  11. Explain physically why the sphere's image charge is smaller than the real charge, while the plane's is equal.
  12. Argue why the method of images cannot be used for a charge near a dielectric–dielectric interface in the same simple form as for a conductor.
Tip: the method of images is really a search for a phantom charge that satisfies the boundary by symmetry. For a plane, mirror symmetry demands an equal-and-opposite twin. For a sphere, the geometry demands a smaller, off-centre partner. Whenever a problem says "grounded conductor" plus "a charge nearby", try an image first — and remember the answer is valid only in the real region, never behind the conductor.