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Physics · Ch 2 — Electrostatic Potential and Capacitance

Equipotential Surfaces

2.6

Equipotential Surfaces

What is an Equipotential Surface?

An equipotential surface is a surface over which the electric potential VV is constant at every point.

This means that the potential difference between any two points on the same equipotential surface is zero.


Why the Electric Field is Perpendicular to an Equipotential Surface

If the electric field E⃗\vec{E} had a component along the surface, then to move a unit positive test charge along that component, work would have to be done (since work = force × displacement).

But on an equipotential surface, no work is required to move a test charge between any two points (because ΔV=0\Delta V = 0).

Therefore, the electric field cannot have a component parallel to the surface.

Hence, the electric field is always normal (perpendicular) to the equipotential surface at every point.


Equipotential Surfaces for a Single Point Charge

For a point charge qq, the potential at a distance rr is given by:

V=14πε0qrV = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}

Since VV is constant when rr is constant, the equipotential surfaces are concentric spherical surfaces centred at the charge.

The electric field lines for a single point charge are radial lines (starting from the charge if q>0q > 0, ending at the charge if q<0q < 0).

These radial field lines are normal to the spherical equipotential surfaces at every point.


Equipotential Surfaces for a Uniform Electric Field

For a uniform electric field E⃗\vec{E} directed along the xx-axis, the potential changes only along xx.

Thus, surfaces of constant VV are planes perpendicular to the xx-axis — that is, planes parallel to the yy-zz plane.


Equipotential Surfaces for a Dipole and Two Identical Positive Charges …

Figure 2.9For a single charge q (a) equipotential surfaces are spherical surfaces centred at the charge, and (b) electric field lines are radial, starting from the charge if q > 0.
Fig. 2.9 — For a single charge q (a) equipotential surfaces are spherical surfaces centred at the charge, and (b) electric field lines are radial, starting from the charge if q > 0.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What the Figure Shows

The figure has two panels, (a) and (b), both centred on a single point charge labelled q (with a ⊕ symbol, indicating a positive charge).

  • Panel (a) shows five concentric circles around the charge. These circles represent equipotential surfaces — surfaces where the electric potential VV has the same value everywhere. The circles are drawn closer together near the charge, indicating that the potential changes more rapidly as you approach the charge.

  • Panel (b) shows eight straight arrows radiating outward from the charge in evenly spaced directions. These arrows represent electric field lines for a positive charge q>0q > 0, which point radially outward.

Physical Idea Taught

The figure illustrates two complementary visual representations of the electric field around a single point charge:

  1. Equipotential surfaces are spherical (concentric circles in 2D cross-section) because the potential VV depends only on the distance rr from the charge: V=14πε0qrV = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}. For a fixed rr, VV is constant, so each sphere of radius rr is an equipotential surface.

  2. Electric field lines are radial, starting from the charge if q>0q > 0 (or ending at it if q<0q < 0). The field is always normal (perpendicular) to the equipotential surface at every point. This is a general principle: if the field had a component along the surface, work would be required to move a test charge on that surface, contradicting the definition of an equipotential surface (no work needed).

Key Formula

The potential due to a single point charge qq at a distance rr is:

V=14πε0qrV = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}

  • VV = electric potential (in volts)
  • qq = magnitude of the point charge (in coulombs)
  • rr = distance from the charge (in metres)
  • ε0\varepsilon_0 = permittivity of free space (8.85×10−12 C2/N⋅m28.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^2) …
Figure 2.10Equipotential surfaces for a uniform electric field.
Fig. 2.10 — Equipotential surfaces for a uniform electric field.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What the figure shows

The diagram depicts three vertical, parallel planes drawn as tall parallelograms in oblique perspective. They are evenly spaced from left to right. A set of many parallel horizontal field lines runs through all three planes, with arrowheads pointing to the right. The bold label E is placed at the far right, beside the arrow tips. The planes are perpendicular to the field lines.

Physical idea

The figure illustrates that for a uniform electric field E⃗\vec{E} (constant in magnitude and direction), the equipotential surfaces are planes that are normal (perpendicular) to the field direction. In this case, the field is along the xx-axis, so the equipotential surfaces are planes parallel to the yy-zz plane. The spacing between the planes corresponds to a constant potential difference ΔV\Delta V between them.

Key formula

The relationship between the uniform field E⃗\vec{E} and the potential difference ΔV\Delta V between two equipotential surfaces separated by a perpendicular distance Δx\Delta x is:

ΔV=−Ex Δx\Delta V = -E_x \, \Delta x

where:

  • ΔV\Delta V is the potential difference between two equipotential surfaces (in volts, V)
  • ExE_x is the magnitude of the uniform electric field along the xx-axis (in volts per metre, V/m)
  • Δx\Delta x is the perpendicular distance between the two surfaces (in metres, m) …
Figure 2.11Some equipotential surfaces for (a) a dipole, (b) two identical positive charges.
Fig. 2.11 — Some equipotential surfaces for (a) a dipole, (b) two identical positive charges.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What the Figure Shows

The figure has two side-by-side panels, (a) and (b), each showing equipotential surfaces (curves of constant electric potential) in a plane containing the charges.

  • Panel (a): A dipole

    A negative charge (−q-q) is at the upper left, and a positive charge (+q+q) at the lower right. Around each charge, the equipotential surfaces are nearly circular close to the charge, but as you move away, the curves distort toward each other between the charges. This distortion reflects the combined potential of both charges.

  • Panel (b): Two identical positive charges

    Two equal positive charges (+q+q) are placed side by side. Near each charge, the equipotential surfaces are small concentric circles. Farther away, the surfaces merge into larger contours that enclose both charges, forming a peanut-shaped or figure-eight outer boundary. The region between the charges has a saddle point where the potential is locally flat.

Physical Idea

Equipotential surfaces are surfaces (here drawn as curves in a plane) where the electric potential VV is constant. The key property is that the electric field E\mathbf{E} is always perpendicular to the equipotential surface at every point. This follows because moving a test charge along an equipotential requires no work — if E\mathbf{E} had a component along the surface, work would be done, contradicting the definition.

The figure illustrates how equipotential surfaces change shape for different charge configurations:

  • For a single point charge, they are concentric spheres (circles in a plane).
  • For a dipole, the surfaces are distorted, showing the influence of both charges.
  • For two like charges, the surfaces merge at large distances, reflecting the net repulsive field.

Key Formula

The potential due to a single point charge qq at a distance rr is:

V=14πε0qrV = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}

where:

  • VV = electric potential (in volts)
  • qq = charge (in coulombs)
  • rr = distance from the charge (in metres)
  • ε0\varepsilon_0 = permittivity of free space (8.85×10−12 C2/N⋅m28.85 \times 10^{-12} \, \text{C}^2/\text{N·m}^2)

For a dipole (two charges +q+q and −q-q separated by distance 2a2a), the potential at a point far away (distance r≫ar \gg a) is:

V=14πε0pcos⁡θr2V = \frac{1}{4\pi\varepsilon_0} \frac{p \cos\theta}{r^2} …