Physics / Electric Fields & Equipotentials
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EXPERIMENT 45 · PHYSICS LABORATORY

Electric Fields & Equipotentials theory

Understand electric fields & equipotentials: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
V(x) = E x + b \iff y = m x + b \implies E = m, \quad b \to 0.000 \text{ V}

Keep the physical quantities, graph variables and units consistent.

01The Electrostatic Potential, Conservative Vector Fields, and Gradient Operator

In classical electrostatics, the fundamental force between stationary electric charges is governed by Coulomb's law. Because the electrostatic force exerted on a test charge q₀ by any static charge distribution is a central, inverse-square force (F ∝ 1/r²), the curl of the electrostatic field vector vanishes identically throughout space: ∇ × E = 0. By Stokes' theorem, this irrotational property guarantees that the line integral of E around any closed curve is zero (∮ E · dℓ = 0), proving that electrostatic fields are strictly conservative vector fields. Consequently, the vector electric field E can be expressed as the negative spatial gradient of a scalar electrostatic potential function V(r): E = −∇V = −[(∂V/∂x) î + (∂V/∂y) ĵ + (∂V/∂z) k̂]. The negative sign carries profound physical significance: it dictates that the electric field vector always points in the direction of maximum spatial rate of decrease of electric potential—analogous to a gravitational acceleration vector pointing down the steepest gradient of a topographical elevation contour. For an arbitrary path between two points A and B, the potential difference is path-independent: V_B − V_A = −∫_A^B E · dℓ. In one dimension, this reduces to the fundamental potential gradient expression: E_x = −dV/dx. The physical unit of electric field is newtons per coulomb (N C⁻¹), which is dimensionally identical to volts per meter (V m⁻¹).

\nabla \times \vec{E} = 0 \iff \oint \vec{E} \cdot d\vec{\ell} = 0, \qquad \vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right), \qquad E_x = -\frac{dV}{dx}
Why does the electrostatic field vector E point in the direction of the steepest decrease of electric potential V?

By definition, E = −∇V. The gradient operator ∇V points in the direction of maximum spatial increase of V. The negative sign ensures that the electric field E points toward regions of lower electric potential, directing positive charges toward lower potential energy.

02Equipotential Surfaces and Spatial Orthogonality of Field Lines

An equipotential surface is a geometric locus of points in space at which the electrostatic potential V has a single constant value: V(x, y, z) = C. If a differential displacement dℓ is made along an equipotential surface, the infinitesimal change in electric potential is identically zero: dV = 0. According to the chain rule of multivariable calculus, the total differential of potential is given by dV = (∇V) · dℓ = −(E · dℓ). Setting dV = 0 yields the fundamental condition: E · dℓ = |E| |dℓ| cos θ = 0. For any non-zero electric field (|E| ≠ 0) and non-zero tangential displacement (|dℓ| ≠ 0), this condition demands cos θ = 0, which requires the angle θ between the electric field vector E and the tangent vector dℓ to be exactly 90° (θ = π/2 radians). Therefore, electric field lines must intersect equipotential surfaces at right angles everywhere throughout three-dimensional space. Furthermore, the magnitude of the electric field is inversely proportional to the perpendicular spacing Δs between adjacent equipotential surfaces separated by a fixed potential increment ΔV: |E| ≈ |ΔV| / Δs. Where equipotential surfaces crowd closely together (small Δs), the electric field is strong and intense; where equipotential surfaces are widely spaced (large Δs), the electric field is weak. Because conductors are equipotential volumes in static equilibrium, electric field lines must also emerge strictly perpendicular to conducting surfaces.

dV = -\vec{E} \cdot d\vec{\ell} = 0 \implies \cos\theta = 0 \iff \theta = 90^\circ, \qquad |\vec{E}| = -\frac{\Delta V}{\Delta s} \implies |\vec{E}| \propto \frac{1}{\Delta s}
What is the physical justification for why electric field lines must always intersect equipotential surfaces at right angles?

Because the potential is constant along an equipotential surface (dV = 0), zero work is done moving a charge along it (dW = −q₀ E · dℓ = 0). Since dW = q₀ |E| |dℓ| cos θ = 0, the angle θ between the electric field E and the surface tangent dℓ must be 90°.

03The Current Analogy and Laplace's Equation in Conducting Media

Direct experimental measurement of electrostatic fields in insulating air or vacuum is virtually impossible with standard laboratory probes, because electrostatic charges are minuscule (~nC) and dissipate rapidly through humidity and air ionization, while physical voltmeter probes would instantly drain the static charge. To circumvent this limitation, physicists utilize the current-flow analogy first developed by Gustav Kirchhoff. Consider a thin, uniform two-dimensional conducting sheet of electrical conductivity σ and thickness t (surface sheet resistance R_s = 1 / (σ t) in units of Ω/□). Under steady-state direct current flow, charge conservation requires the divergence of the volume current density J to vanish at every point not containing an external current source: ∇ · J = 0. Substituting Ohm's law in point form, J = σ E = −σ ∇V, into the continuity equation yields: ∇ · (−σ ∇V) = 0. For a homogeneous and isotropic sheet where conductivity σ is uniform and position-independent, σ factors out of the divergence, yielding Laplace's equation for the electric potential: ∇²V = ∂²V/∂x² + ∂²V/∂y² = 0. This is mathematically identical to Laplace's equation governing electrostatic potential in a charge-free dielectric vacuum (∇²V = 0, derived from Gauss's law ∇ · E = 0 with E = −∇V). Because Laplace's equation with identical boundary conditions has a unique solution (by the uniqueness theorem of partial differential equations), the potential distribution V(x, y) and streamline contours in a steady conducting sheet are identical to the electrostatic potential and field lines of static charges on electrodes of the same geometry.

\vec{J} = \sigma \vec{E} = -\sigma \nabla V, \qquad \nabla \cdot \vec{J} = 0 \implies \nabla^2 V = \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0, \qquad R_s = \frac{1}{\sigma t}
How does steady current flow in Teledeltos conducting paper simulate static electrostatic fields in a dielectric vacuum?

Both steady current in a uniform conductor (∇ · J = 0, J = σE) and electrostatics in a charge-free dielectric (∇ · E = 0, E = −∇V) satisfy Laplace's equation ∇²V = 0. By the uniqueness theorem, identical electrode boundary conditions produce identical potential distributions V(x, y).

04Uniform Field Geometry between Parallel Plates and Edge Fringing Effects

When two long, parallel, planar conducting electrodes separated by a distance d are charged to a potential difference V₀, the electric field in the central inter-electrode region is uniform. In Cartesian coordinates where the plates are parallel to the y-axis (at x = 0 with V = 0, and at x = d with V = V₀), symmetry dictates that the potential depends only on the longitudinal coordinate x: V = V(x). Laplace's equation reduces to the ordinary differential equation d²V/dx² = 0. Integrating twice yields the linear solution V(x) = C₁ x + C₂. Applying the boundary conditions V(0) = 0 and V(d) = V₀ gives C₂ = 0 and C₁ = V₀ / d, establishing the linear potential profile: V(x) = (V₀ / d) x. The corresponding electric field is completely uniform and directed along the x-axis: E = −(dV/dx) î = −(V₀ / d) î. Its magnitude is constant: E = V₀ / d. The equipotential lines are a series of equidistant, mutually parallel straight lines parallel to the electrodes, spaced by Δx = ΔV / E. Near the open edges of the plates (where |y| approaches L/2), the boundary condition of infinite plates fails: electric field lines bulge outward into space in curved arcs, and equipotential lines bow outward away from the center. This edge fringing effect reduces the local field strength and distorts linearity. To obtain high metrological accuracy, potential measurements must be collected along the central longitudinal axis (y = 0), where fringing field components cancel by bilateral symmetry.

\frac{d^2 V}{dx^2} = 0 \implies V(x) = \left(\frac{V_0}{d}\right) x = E x, \qquad E = -\frac{dV}{dx} = \frac{V_0}{d} = 100.00 \text{ V m}^{-1}, \qquad \Delta x = \frac{\Delta V}{E}
Why is the electric field strictly uniform between parallel plates, and why must measurements be taken along the central axis?

Between parallel plates, Laplace's equation d²V/dx² = 0 has the linear solution V(x) = (V₀/d)x, yielding a constant gradient E = V₀/d. Measurements are taken along the central axis because edge fringing fields bend and distort the field lines near the electrode ends.

05Point Charge Dipoles, Cylindrical Symmetry, and Non-Uniform Field Topologies

Unlike parallel plates, most electrode geometries produce spatially non-uniform electric fields where the potential gradient varies with position. Two configurations of foundational importance are the electric dipole and coaxial concentric cylinders. For an electric dipole consisting of two point electrodes carrying equal and opposite potentials (+V₀/2 and −V₀/2) separated by distance 2a, the electric potential at position (x, y) is obtained by superposition: V(x, y) = (q / 4πε₀) [1 / √((x − a)² + y²) − 1 / √((x + a)² + y²)]. In the conducting sheet analogy, the equipotentials form families of non-concentric circles known as Apollonian circles, which enclose the electrode pins and crowd closely near the electrode contact points, indicating very high electric field strengths: E ∝ 1/r² near the poles and E ∝ 1/r³ along the broadside axis. For concentric circular electrodes consisting of an inner cylindrical pin of radius r₁ at potential V₁ and an outer concentric ring of radius r₂ at potential V₂, cylindrical symmetry requires V = V(r). Laplace's equation in polar coordinates becomes: (1/r) d/dr [r (dV/dr)] = 0. Integrating yields the logarithmic potential profile: V(r) = V₁ + (V₂ − V₁) [ln(r / r₁) / ln(r₂ / r₁)]. Differentiating with respect to radial distance r isolates the radial electric field: E(r) = −dV/dr = −[(V₂ − V₁) / ln(r₂ / r₁)] (1 / r). This reveals that the electric field strength falls off inversely with radius (E ∝ 1/r), demonstrating why equipotential rings crowd tightly around the central pin and spread farther apart near the outer cylinder.

V_{\text{dipole}} = \frac{q}{4\pi\varepsilon_0} \left(\frac{1}{r_+} - \frac{1}{r_-}\right), \qquad V_{\text{coaxial}}(r) = V_1 + \frac{V_2 - V_1}{\ln(r_2 / r_1)} \ln\left(\frac{r}{r_1}\right), \qquad E_r(r) = -\frac{dV}{dr} \propto \frac{1}{r}
Why do equipotential rings in a coaxial cylindrical electrode system become more widely spaced as radial distance r increases?

The radial electric field is inversely proportional to distance: E(r) ∝ 1/r. Because the spacing between equipotentials is inversely proportional to field strength (Δr = ΔV / E(r) ∝ r), the equipotential rings must become progressively wider apart at larger radii.

06Voltmeter Input Impedance and Circuit Loading Error Analysis

A critical source of systematic error in conducting paper potential mapping arises from voltmeter loading. When the tracking probe contacts the Teledeltos paper at position x, the internal resistance of the voltmeter Z_in forms a parallel electrical branch with the resistance of the conducting sheet to ground. If the voltmeter input resistance is comparable to the sheet resistance between the probe point and the grounded electrode, a significant fraction of the surface current diverts through the voltmeter. This shunting current distorts the local potential distribution, causing the voltmeter to read a potential V_meas that is systematically lower than the true undisturbed open-circuit potential V_true. Quantitatively, if R_th is the Thévenin equivalent resistance of the conducting paper at the probe contact point, the measured potential is given by the voltage divider relation: V_meas = V_true [Z_in / (Z_in + R_th)] = V_true [1 − R_th / (Z_in + R_th)]. The fractional loading error is therefore: δV / V_true = −R_th / (Z_in + R_th) ≈ −R_th / Z_in. For carbon-impregnated paper with surface sheet resistance R_s ≈ 2000 Ω/□, typical Thévenin resistance is on the order of R_th ≈ 1000 to 2000 Ω. Utilizing an ordinary moving-coil analog voltmeter (Z_in ≈ 20 kΩ) would produce a disastrous loading error of ~5% to 10%. In contrast, modern digital voltmeters possess an input impedance of Z_in = 10.0 MΩ (10⁷ Ω), suppressing the loading error to: δV / V_true ≈ 2000 / 10⁷ = 0.02% (negligible). Furthermore, using a light spring-loaded probe prevents localized indentation and tearing of the delicate carbon layer, which would alter local conductivity.

V_{\text{meas}} = V_{\text{true}} \left(\frac{Z_{\text{in}}}{Z_{\text{in}} + R_{\text{th}}}\right), \qquad \frac{\delta V}{V_{\text{true}}} \approx -\frac{R_{\text{th}}}{Z_{\text{in}}} \approx -\frac{2000 \; \Omega}{10^7 \; \Omega} = -0.02\%
Why must a digital voltmeter with an input impedance of at least 10 MΩ be used for electric field mapping on conducting paper?

A high input impedance (10 MΩ) ensures that virtually zero current is drawn from the conducting paper through the probe. A low-impedance meter would draw significant loading current, depressing the local potential and distorting the field distribution by up to 10%.

07Linear Regression Formulation and OLS Potential Gradient Isolation

To determine the uniform electric field strength E with maximum statistical precision, experimentalists vary the probe displacement x along the central longitudinal axis across calibrated stations (e.g. x = 1.5, 3.0, 4.5, 6.0, 7.5, and 9.0 cm) and measure the corresponding electric potential V. Designating spatial displacement x (in meters) as the independent variable and measured potential y = V (in volts) as the dependent variable yields the canonical linear regression model: y = m x + b. Comparing this with the theoretical potential relation V(x) = E x directly identifies the regression slope m as the uniform electric field strength: m = E (in V m⁻¹). The regression intercept b serves as an indispensable experimental diagnostic: theoretically b = 0.000 V, so any statistically significant non-zero intercept detects contact potential offsets between the brass electrode and silver paint, thermocouple thermal EMFs at the probe junction, or zero-offset drift in the digital voltmeter. Ordinary least-squares (OLS) regression minimizes the sum of squared potential residuals: SSE = Σ (V_i − m x_i − b)², providing an unbiased, minimum-variance estimate of the electric field strength E along with its standard error SE(m) = √[s² / Σ(x_i − x̄)²]. The coefficient of determination R² measures the degree to which the spatial potential variation is strictly linear, typically exceeding 0.9999 for well-aligned parallel plates.

V(x) = m x + b \implies m = \frac{dV}{dx} = E = 100.00 \text{ V m}^{-1}, \qquad b = \bar{V} - m \bar{x} \to 0.000 \text{ V}, \qquad R^2 = 1 - \frac{\sum (V_i - \hat{V}_i)^2}{\sum (V_i - \bar{V})^2} \ge 0.9999
When plotting electric potential V against displacement x for parallel plates, what do the slope m and intercept b signify physically?

The slope m = dV/dx represents the uniform electric field strength E between the plates (in V/m). The intercept b represents any residual zero-offset potential (contact potential, thermoelectric EMF, or meter zero error), which theoretically vanishes (b = 0).

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Connect the equations to a measurement.