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advanced27 min studyLesson 233 in Physics

Image charges replace a boundary, not the conductor

Use uniqueness to validate images and distinguish force from assembly energy.

Before you begin
On this page

The question for today

What you’ll understand

  • Verify a planar image construction in its physical domain.
  • Compute induced surface charge and force.
  • Explain the factor one half in the physical image interaction energy.

An image charge is a mathematical source placed outside the region where the physical solution is required. Its value comes from satisfying conductor boundary conditions, not from claiming that a point charge secretly lives inside the metal. The distinction matters most when calculating energy: the force follows directly from the image field, whereas the energy of the physical charge-conductor system is not the ordinary two-charge interaction energy.

Verify the candidate before using it

Place a point charge q at height a above an infinite grounded plane z=0. In the upper half-space, add a fictitious charge −q at the reflected position. Every plane point has equal distances to the two charges, so their potentials cancel. The candidate has the correct real singularity, no extra upper-half-space singularity, zero plane potential, and decay at infinity. Uniqueness then identifies it as the physical potential in that half-space.

The construction does not give the field inside the conductor: there the physical electrostatic field is zero. Nor does it automatically solve an isolated neutral plane or a conductor held at a different potential. Ground can exchange charge with the metal. A spherical grounded conductor requires a different image position and magnitude, while a neutral isolated sphere needs an additional charge constraint. Boundary conditions decide the construction; visual resemblance alone cannot.

V(r)=q4πϵ0(1∣r−az^∣−1∣r+az^∣),z>0V(\mathbf r)=\frac{q}{4\pi\epsilon_0}\left(\frac{1}{|\mathbf r-a\hat{\mathbf z}|}-\frac{1}{|\mathbf r+a\hat{\mathbf z}|}\right),\qquad z>0
q is charge in C, a>0 is height in m, and V is in volts. The ideal infinite plane is grounded, vacuum occupies z>0, and the potential decays at infinity.↔ Scroll wide equations horizontally.

The induced charge is distributed

Differentiate the upper-half-space potential normally to obtain the field just outside the conductor. Multiplying its outward normal component by vacuum permittivity gives the actual induced surface charge density. This density is concentrated beneath the real charge but extends over the whole plane. Integrating the concentric rings yields total induced charge −q. The image represents the potential of that spread-out charge only in the physical upper region.

For the force on the real point charge, exclude its divergent self-field. The remaining field is precisely the image field evaluated at the real position. Their separation is 2a, giving an attractive inverse-square force for either sign of q. A common error uses the height a as their separation, producing a factor of four mistake. Another adds the real self-field, which cannot define a finite force on an ideal point charge.

σ(R)=−qa2π(R2+a2)3/2,Fz=−q216πϵ0a2\sigma(R)=-\frac{qa}{2\pi(R^2+a^2)^{3/2}},\qquad F_z=-\frac{q^2}{16\pi\epsilon_0a^2}
R is radial distance along the plane in m, σ is surface charge in C/m², and Fz is force in N. The plane’s outward normal into vacuum is +z.↔ Scroll wide equations horizontally.

Why energy has a half

Bring the real charge quasistatically from infinity while the grounded conductor adjusts. The induced distribution grows with the charge during an equivalent charging construction. The physical interaction energy is therefore one half q times the induced potential at the charge. Differentiating this energy with respect to height gives the full image force, because changing height also changes the location of the image. Treating the image as a fixed independent physical charge gives the wrong energy accounting.

The waves lab illustrates cancellation by superposition only. Reset and set phase difference to 180° and amplitude ratio to 1; predict a zero sum. Reset and set phase difference to 180° and amplitude ratio to 0.5; predict a residual half-amplitude wave. This tests how precisely matched contributions enforce a zero value. Its waves travel together at a fixed frequency; they are not Coulomb potentials, a spatial mirror construction, or a calculation of induced charge.

U=12qVind(a)=−q216πϵ0a,Fz=−dUdaU=\frac12qV_{\rm ind}(a)=-\frac{q^2}{16\pi\epsilon_0a},\qquad F_z=-\frac{dU}{da}
U is physical interaction energy in J relative to infinite separation, with the point self-energy omitted. The conductor remains grounded throughout the quasistatic family.↔ Scroll wide equations horizontally.

Work it through

A nanocoulomb above ground

Use k=1/(4πε₀)=9.0×10⁹ N·m²/C². A 2 nC charge is 0.10 m above a grounded plane. Find its force and interaction energy.

  1. The image is −2 nC at height −0.10 m, so the distance entering Coulomb’s force is 0.20 m. The force points toward the plane because the two represented sources have opposite signs.

  2. Compute kq²/(2a)²=9.0×10⁹×4×10⁻¹⁸/0.04=9.0×10⁻⁷ N. This is the magnitude of the actual force on the real charge; it requires no extra factor of one half.

  3. The induced potential at the charge is −kq/(2a)=−90 V. Half q times this value is −9.0×10⁻⁸ J. Equivalently U=−kq²/(4a), whose negative height derivative reproduces the force.

The result

The force is −9.0×10⁻⁷ N in the z direction and the physical interaction energy is −9.0×10⁻⁸ J.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Test exact cancellation

Reset and set phase difference=180°, amplitude ratio=1; predict cancellation. Reset and set phase difference=180°, amplitude ratio=0.5; predict a half-amplitude residual.

Test exact cancellationTwo waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Current readings: Sum at x = 0: 0.866; Resultant amplitude: 1.732; Phase difference: 60°. All parameters are adjustable with the labeled controls below.−3.14−1.5701.573.14−3.4−1.701.73.4x · radiansAmplitude
SumFirst componentSecond component
Sum at x = 0
0.866
Resultant amplitude
1.732
Phase difference
60°
t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThese equal-frequency traveling sine waves illustrate linear cancellation. They do not compute electrostatic images, surface charge, or image energy.

Pause & reason

Check your understanding

Which expression gives the physical interaction energy of a point charge above the grounded plane?

Your turn

Try it yourself

Keeping the charge fixed, increase its height from 0.10 m to 0.20 m. Find the new force and the external work required to move it slowly, using the worked-example values.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Force scales as inverse height squared; physical interaction energy scales as minus inverse height.

Compare with the solution

At double height the force is one quarter as large, −2.25×10⁻⁷ N. The energy rises from −9.0×10⁻⁸ J to −4.5×10⁻⁸ J. External quasistatic work is this positive difference, +4.5×10⁻⁸ J. The work is positive because the motion opposes attraction; it is not the magnitude of the initial force times the displacement.

Keep the idea

Take it with you

  • Image constructions are valid only in the specified physical domain.
  • The image field gives the full force after the real self-field is excluded.
  • Physical charge-conductor energy carries a factor one half.

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