core32 min studyLesson 11 in Physics

Electrostatic potentials and boundary conditions

Solve a field problem by specifying its charges and boundaries, and understand why a valid solution is unique.

Before you begin

On this page

The question for today

What you’ll understand

Translate between an electrostatic potential and its electric field.
Distinguish prescribed potential from prescribed normal derivative.
Use uniqueness to justify a boundary-value solution.

A voltage is not a substance stored at one point. It describes the change in potential energy per unit test charge between positions. In electrostatics, the field can be reconstructed from one scalar function, but its differential equation alone is incomplete. The shape of the region and the electrical conditions on its boundary select which solution describes an experiment.

A local equation needs global information

When magnetic fields are time independent, Faraday's law gives $\nabla\times\mathbf E=0$ . In a simply connected region, a single-valued scalar potential therefore exists with $\mathbf E=-\nabla\phi$ . A positive charge experiences force toward decreasing potential. Adding the same constant to every potential changes neither this gradient nor any voltage difference. The reference zero is a convention, whereas differences are operationally measurable.

Combining this representation with Gauss's law gives Poisson's equation in vacuum. The charge density is an input, not an integration constant to choose afterward. A charge-free region obeys Laplace's equation even when its field is large: distant electrodes can impose substantial fields there. Zero charge density means zero divergence, and does not imply zero electric field or constant potential throughout the region.

\mathbf E=-\nabla\phi,\qquad \nabla^2\phi=-\frac{\rho}{\epsilon_0}

These SI equations assume electrostatics in vacuum. Potential is measured in volts, charge density in C/m³, and its gradient in V/m. A spatially varying dielectric requires a different constitutive boundary problem.↔ Scroll wide equations horizontally.

What an electrode actually fixes

A connected ideal conductor in electrostatic equilibrium is an equipotential. Otherwise its mobile charges would continue moving tangentially. A voltage source can prescribe that potential relative to another conductor; this is Dirichlet boundary information. The normal electric field just outside the conductor then determines the surface charge required to maintain it. Prescribing a voltage does not generally prescribe that charge independently.

Neumann information instead specifies the normal derivative of potential, and hence the normal field. A pure Neumann problem needs a compatibility condition between total enclosed charge and boundary flux, and determines potential only up to a constant. These are different experimental arrangements. Arbitrarily specifying both potential and its normal derivative everywhere can overconstrain the problem rather than making the prediction more accurate.

\phi(0)=0,\quad \phi(d)=V\quad\Longrightarrow\quad \phi(x)=\frac{Vx}{d},\quad E_x=-\frac{V}{d}

This solution holds between ideal infinite parallel plates with no volume charge and no dependence on the other coordinates. Real plate edges produce fringing fields; the approximation is useful far from those edges when their separation is small compared with their width.↔ Scroll wide equations horizontally.

Uniqueness turns a guess into a proof

Suppose two smooth solutions have the same charge density and the same potential on a closed boundary. Their difference $u$ satisfies Laplace's equation and vanishes on that boundary. Integration by parts makes the volume integral of $|\nabla u|^2$ equal a boundary term that is zero. A nonnegative integrand can then have zero integral only when the two fields coincide, under the usual regularity assumptions.

This argument explains the method of images. Fictitious charges outside the physical solution region are a mathematical device: if their potential satisfies the real equation and all real boundary values within the chosen region, uniqueness identifies the answer there. The fictitious charges do not replace the conductor on the other side of its surface. Check the domain before treating any image construction as a global physical charge distribution.

\int_\Omega |\nabla u|^2\,dV=\oint_{\partial\Omega}u\,\partial_nu\,dA-\int_\Omega u\,\nabla^2u\,dV=0

For equal Dirichlet data, u vanishes on the boundary and ∇²u vanishes in the interior. Appropriate conditions at infinity replace a finite enclosing boundary for an exterior problem.↔ Scroll wide equations horizontally.

Work it through

Voltage, force direction, and external work

Ideal plates at $x=0$ and $x=2.0\,\mathrm{mm}$ are held at 0 V and 100 V. Move a $+2.0\,\mathrm{nC}$ test charge slowly between them. Find the field and work supplied by an external agent.

Convert the spacing to $d=2.0\times10^{-3}\,\mathrm m$ . The charge-free, one-dimensional solution is linear because $d^2\phi/dx^2=0$ and the two endpoint potentials fix its slope.
Differentiate: $E_x=-100/(2.0\times10^{-3})=-5.0\times10^4\,\mathrm{V/m}$ . The field points toward the zero-volt plate; the positive charge is moved against its electric force.
The change in potential energy is $q\Delta\phi=(2.0\times10^{-9})(100)=2.0\times10^{-7}\,\mathrm J$ . With negligible kinetic-energy change, the external agent supplies this positive work; the field does the opposite work.

The result

The uniform field is −50 kV/m in the x direction, and external work is +0.20 μJ. This calculation treats the test charge as too small to alter the electrode configuration appreciably.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Differentiate a candidate potential

Inspect y = ax² and its tangent as a and x change. Interpreting y as a normalized potential makes minus the displayed slope a normalized field. Its nonzero second derivative represents a uniform source in one dimension; this lab does not solve the parallel-plate boundary problem.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forA potential profile determines a field through its slope, while a boundary problem determines the profile.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The difference solves Laplace's equation with zero boundary values. The energy integral forces its gradient to vanish; the zero boundary value then fixes its constant. The common field can nevertheless be nonzero.

Your turn

Try it yourself

A charge-free one-dimensional gap of width 5 mm has electrode potentials −20 V and +30 V. Determine the field and explain whether changing both electrodes by +200 V changes it.

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

Give me a hint

Calculate the potential difference before dividing by the spacing.

Compare with the solution

The potential difference is 50 V, so $E_x=-50/0.005=-10^4\,\mathrm{V/m}$ . Shifting both electrodes by +200 V adds a constant to the potential in this ideal isolated boundary problem and leaves its gradient unchanged. A nearby grounded object would introduce additional fixed boundary data, so the common shift would not generally leave the full laboratory field unchanged.

Keep the idea

Take it with you

Charge density and boundary information jointly determine electrostatics.
Dirichlet and Neumann data represent different physical constraints.
Uniqueness validates a candidate solution only in its stated domain.

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