Choose electrostatic boundary data consistently
Distinguish prescribed potential, normal derivative, and total conductor charge.
On this page
The question for today
What you’ll understand
- Translate normal field into outward-normal potential derivative.
- Check the Neumann compatibility condition.
- Explain the additive constant and distinguish local flux from total conductor charge.
A grounded electrode prescribes potential. A specified normal electric field prescribes a derivative. These statements look interchangeable because field is a potential gradient, yet they carry different information and obey different consistency conditions. In particular, prescribing normal derivatives everywhere cannot fix an absolute potential level. Learning to write the outward normal before inserting numbers prevents the sign errors that often hide this distinction in otherwise simple geometries.
Value and derivative are different experiments
Dirichlet data assign a potential value at each boundary point relative to a chosen reference. Neumann data assign the outward normal derivative there. Since electric field is minus the gradient, positive outward electric field corresponds to a negative outward potential derivative. On the left and right ends of an interval, the outward normals point in opposite directions. The same Cartesian field therefore gives normal components of opposite signs at those ends.
A conductor held at a prescribed voltage is an equipotential boundary whose surface charge is determined by the solution. An isolated conductor with known total charge is also equipotential, but its potential is initially unknown. Its total charge provides an integral constraint, not the pointwise Neumann data. Substituting a uniform charge density merely because the total charge is known changes the problem unless symmetry independently warrants uniformity.
Flux must match enclosed charge
Integrate Poisson’s equation over the region and use the divergence theorem. The integral of the prescribed normal derivative must equal minus the enclosed charge divided by permittivity. This compatibility condition is necessary for a pure Neumann solution. For example, zero derivative on a closed boundary is incompatible with a nonzero net interior charge. No choice of an additive constant can repair that mismatch because constants affect neither derivatives nor flux.
For compatible data in a connected bounded domain, the same squared-gradient argument used for Dirichlet uniqueness shows that two solutions differ by a constant. Now the boundary term vanishes because the difference derivative is zero, rather than because its value is zero. Specify a mean potential or one reference value to choose that constant. The electric field is already unique before this gauge choice, so the remaining ambiguity is not a physical uncertainty in force.
Mixed conditions and the missing reference
One can prescribe value data on one portion of a boundary and derivative data on the remainder. With a Dirichlet portion of nonzero boundary area, the additive constant is fixed and the usual homogeneous difference argument gives uniqueness. Conversely, prescribing both value and derivative independently on every boundary point generally overdetermines an elliptic problem. A measured pair may be consistent, but it cannot be chosen arbitrarily.
Use the vector lab to separate a vector from its normal projection. Reset and set angle to 0° and magnitude to 2; predict an x component of 2. Reset and set angle to 60° and magnitude to 2; predict an x component of 1. If the normal is the positive x direction these are outward field components; reversing the normal reverses their signs. This projection instrument has no closed surface, charge integration, or boundary-value solver.
Work it through
Audit two endpoint derivatives
On the one-dimensional interval 0≤x≤2 m, take ρ/ε₀=3 V/m². Prescribe outward derivatives g_left=1 V/m and g_right=−7 V/m. Check compatibility and obtain the solution with V(0)=0.
The left derivative is g_left=−V′(0), so V′(0)=−1 V/m. Integrating V″=−3 gives V′=−3x−1. At the right endpoint this equals −7 V/m, exactly the stated outward derivative.
The endpoint sum is 1−7=−6 V/m. In this one-dimensional reduction it must equal the integral of −ρ/ε₀ over the 2 m length, also −6 V/m. Equivalently multiply both sides by the fixed cross-sectional area to obtain the three-dimensional flux identity.
A second integration gives V=−1.5x²−x+C. The optional reference V(0)=0 chooses C=0. Its right potential is −8 V, which was not independently imposed by the Neumann specification.
The data are compatible. The chosen representative is V=−1.5x²−x in SI units; all constant shifts produce the same field.
The living notebook
Explore the model
Interactive lab · vectors
Read a normal projection
Reset and set angle=0°, magnitude=2; predict x=2. Reset and set angle=60°, magnitude=2; predict x=1 and y=√3. Interpret x as projection onto an explicitly chosen +x normal.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forReversing the surface normal changes the projected sign. The instrument displays one vector, not flux through a closed boundary.
Pause & reason
Check your understanding
Your turn
Try it yourself
A charge-free interval of length 4 m has outward derivatives −2 V/m at its left end and +2 V/m at its right end. Find the field and choose a potential with zero spatial mean.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Translate the left outward derivative carefully, then center a linear potential about the midpoint.
Compare with the solution
The derivative V′ is +2 V/m everywhere, so E=−2 V/m. Compatibility holds because the endpoint derivatives sum to zero. The zero-mean representative is V=(2 V/m)(x−2 m), with endpoint values −4 V and +4 V. Adding a constant would preserve the same field but violate this chosen mean.
Keep the idea
Take it with you
- Normal derivative and normal field have opposite signs.
- Pure Neumann data require an integrated charge-flux compatibility condition.
- A voltage reference fixes the remaining constant without changing the field.
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