Electric Potential
Add source potentials as scalars and connect labeled equipotentials to field direction and strength.
AP Physics 2 Topic 10.5 · College Board CED printed p. 50, 51 / PDF p. 57, 58 · See this topic in the course outline
Potential tells you energy per charge at a place. A contour map of potential can tell you which way the field points before you put any test charge there.
For another explanation, read Calculating Potentials. Use the corrected point-charge scalar sums to practice potential.
For another explanation, read Equipotentials, Field lines, and Gradients. Use the corrected contour maps and algebraic field estimates here; calculus-gradient material is an optional extension beyond the AP core.
Before you start
- Potential energy and signed scalar addition
- Field-vector maps
A scalar map from the source charges
Electric potential is potential energy per unit test charge due to the source configuration: . Its unit is joule per coulomb, called the volt. With zero potential at infinity for isolated point sources, . Use each source’s sign and its positive distance to P. Potential is a scalar; no component arrows or directions enter this sum.
The test charge is absent from the source-potential calculation. Once a small test charge q is added, its interaction energy with the sources is qV. A zero potential can occur where the field is nonzero because signed scalar cancellation and vector cancellation are different. Use four or fewer particles in normal AP calculations; more are allowed with high symmetry. General extended-charge calculations are outside the required mathematics.
Potential difference and connected conductors
Between locations A and B, . A 6 V difference means 6 J of energy change per coulomb of positive charge moved from lower to higher potential. Chemical processes in a battery separate charge and maintain a terminal potential difference as the battery supplies energy; emf is an energy-per-charge quantity measured in volts, despite its name.
When conductors touch, mobile electrons redistribute until the connected surfaces reach the same potential in electrostatic equilibrium. This does not require equal net charges: a larger isolated spherical conductor can hold more charge at the same potential than a smaller one. No field in the conducting material means no potential change through it, so it is an equipotential. The numerical common potential depends on its surroundings and reference.
Turn a voltage map into field arrows
Equipotential lines, also called isolines, connect points of equal V. There is no potential change along one, so the field has no tangential component there. At ordinary nonzero-field points, E⃗ is perpendicular to each isoline and points toward decreasing potential. Conversely, draw equipotentials perpendicular to known field arrows. They do not cross unless the potential labeling describes the same contour.
For a uniform field along x, . More generally this gives the average component along the displacement; a magnitude is used along the field direction, or as an estimate over a short normal contour separation. Do not divide by an arbitrary sideways distance and call the result the full field. For equally spaced potential values, closer contour spacing indicates larger field magnitude.
| x (m) | V (V) |
|---|---|
| 0.00 | 12 |
| 0.02 | 8 |
| 0.04 | 4 |
| 0.06 | 0 |
Plotting these data gives slope −200 V/m, hence E⃗ = +200 N/C in the x direction. In a two-dimensional map of this region, contours are vertical straight lines; horizontal arrows point right. A positive charge initially accelerates right, and a negative one left. With a sideways initial velocity, the trajectory bends; it need not stay on one field line.
The midpoint has zero potential but a field
Place +3.0 nC at x = −0.10 m and −3.0 nC at x = +0.10 m. Find V and E⃗ at x = 0. Use V = 0 at infinity.
Both distances are 0.10 m, so .
Why this step? Potential contributions keep charge sign and add as scalars.
The positive source’s field points right; the negative source’s field also points right. Each magnitude is 2.70 × 10³ N/C.
Why this step? The point lies away from the positive source and toward the negative source.
The net field is 5.39 × 10³ N/C right.
Why this step? Vector contributions add here even though scalar potentials cancel.
Answer and check
V = 0 V and E⃗ = 5.39 × 10³ N/C right. A positive test charge has zero interaction energy at this reference point, yet a nonzero force.
Read direction and magnitude from contours
Three straight parallel vertical equipotentials are labeled 10 V at x = 0 cm, 6 V at x = 2 cm, and 2 V at x = 4 cm. Estimate the field and the potential-energy change for q = −1.0 μC moved from the first to the last line.
Create the labeled contour map and a matching local field-vector map, using the same arrow-length scale throughout the region. The starting contour construction is shown below.
Place vertical contours at x = 0, 2 and 4 cm and label them 10, 6 and 2 V. Identify the 2 cm separation perpendicular to the lines; displacement along a line has ΔV = 0.
Why this step? A contour is a collection of equal-potential points. Its normal separation, rather than a sideways path length, supplies the field estimate.
Potential falls as x increases, so the field points right and is normal to the vertical lines.
Why this step? The field direction is toward lower potential, regardless of the test-charge sign.
. Equal contour spacing for equal voltage steps supports a uniform-field estimate.
Why this step? Use displacement perpendicular to the contours.
Draw several horizontal field arrows to the right, at different heights and in both intervals. Give them equal lengths on one scale: each represents 200 N/C. They cross the contours at right angles and point from 10 toward 6 toward 2 V. A leftward arrow would incorrectly point toward increasing V.
Why this step? The zero potential change along a contour eliminates the tangential field component; equal voltage drops over equal normal distances make equal field magnitudes.
.
Why this step? The negative charge’s energy increases as it moves to lower potential.
Answer and check
E = 200 N/C right; ΔU = +8.0 μJ. Moving along one of the vertical equipotentials would instead give ΔU = 0.
Contours predict acceleration, then velocity predicts the path
An equipotential contour connects points with the same electric potential and describes the sources’ scalar potential. The field is perpendicular to an equipotential, points toward decreasing potential, and has no component along that contour. In a uniform region, Eₓ = −ΔV/Δx. Draw that source-field map first. Then multiply by the moving charge to find force; changing the charge sign reverses its acceleration, not the source map.
A particle initially moving along a contour can still accelerate across it. For example, vertical contours give horizontal field and a sideways initial velocity can give a curved path. A local field arrow is not a complete motion prediction: use the sign, mass, initial velocity and relevant field region. Equal-potential endpoints have zero net electric work, even when their intervening path does not lie entirely on one contour.
10.5.B.110.5.B.210.5.B.2.i10.5.B.2.ii10.5.B.2.iii10.5.B.2.iv
A contour map produces a curved trajectory
A uniform region has vertical contours 18 V at x = 0, 12 V at x = 0.020 m, and 6 V at x = 0.040 m. A particle with q = +2.0 nC and m = 4.0 × 10⁻⁶ kg starts at x = 0.010 m, y = 0 and moves upward at 0.050 m/s with zero x velocity. Ignore all other forces and use only the uniform region.
- Each 6 V drop across 0.020 m gives Eₓ = +300 N/C. Draw equal rightward arrows normal to the equally spaced vertical contours.
- Force is +6.0 × 10⁻⁷ N right; aₓ = qEₓ/m = +0.15 m/s². There is no y force, so v_y remains 0.050 m/s.
- At t = 0.20 s, x = 0.010 + (1/2)(0.15)(0.20)² = 0.0130 m and y = 0.0100 m, inside the given region. The initial tangent is upward and the path bends right.
- Potential drops by 300 × 0.0030 = 0.90 V, so ΔU = −1.8 × 10⁻⁹ J. The acquired x kinetic energy is (1/2)m(0.15 × 0.20)² = +1.8 × 10⁻⁹ J. The unchanged y kinetic store cancels in the difference.
Answer and check
Equal rightward 300 N/C field arrows; upward-entry trajectory bends right to (0.0130,0.0100) m at 0.20 s. ΔK = −ΔU = 1.8 nJ.
Make the model respond
Focused Physics 2 model
Two charges: field vectors and potential
Predict first. Can the electric potential be zero while the electric field is nonzero?
E = Σ kq r̂/r² (vector sum); V = Σ kq/r (scalar sum); V(∞) = 0
- E₁
- E₂
- Net E
On a narrow screen, scroll within a figure to read its labels. The table and explanation provide the same physics in text.
Net |E| = 32.36 N/C while V = 0 V. A zero scalar potential does not imply a zero vector field.
Read the model data and comparisons
| Quantity | Value | Unit | Meaning or limit |
|---|---|---|---|
| Point x | 0 | m | |
| Point y | 0.8 | m | |
| Distance r1 | 1 | m | |
| E1,x | 16.18 | N/C | |
| E1,y | 21.57 | N/C | |
| V1 | 26.96 | V | |
| Distance r2 | 1 | m | |
| E2,x | 16.18 | N/C | |
| E2,y | -21.57 | N/C | |
| V2 | -26.96 | V | |
| Net Ex | 32.36 | N/C | |
| Net Ey | 0 | N/C | |
| Net |E| | 32.36 | N/C | |
| Net V | 0 | V |
A positive source produces a field away from itself; a negative source produces a field toward itself. Add x and y components to obtain the vector field. Add signed potentials directly because potential is a scalar. E describes force per positive test charge and V describes potential energy per charge. The displayed arrows are fields, not forces on the source charges.
Assumptions and limits. Stationary point charges in vacuum, at x = ±0.6 m; observation point is in their plane. Potential reference is infinity. Distances at or below 10⁻⁹ m from a nonzero source are excluded. Arrow lengths share one local scale; use the table to compare magnitudes across different control settings.
Try it from memory
Write your answer and reason before opening the worked answer. Your self-assessment is a study decision; it is not an automatically graded score.
Question 1
Two unequal metal spheres touch in isolation. Must their charges become equal? What condition stops the electron redistribution?
Worked answer and reason
Their potentials become equal, not necessarily their charges. Redistribution stops when no potential difference drives further motion.
Shape and surroundings relate charge to potential; equal charge sharing needs additional symmetry.
Question 2
Equal 2 V contour steps are 1 cm apart at A and 4 cm apart at B. Compare field magnitudes and the work on a charge moved along a contour.
Worked answer and reason
The estimated field at A is four times that at B. Along an equipotential ΔV = 0, so electric work −qΔV is zero.
Normal contour crowding encodes field size; there is no field component along the contour.
JavaScript enables saved drafts and self-assessment. You can always work on paper.
Catch the tempting mistake
- Potential is a vector because electric field is a vector.
- Potential is signed energy per charge; add it as a scalar.
- A charged conductor must have zero potential because its internal field is zero.
- Zero internal field implies one constant potential; its value depends on the reference and configuration.
Put it to work
Give the physical reason as well as the calculation. The worked solutions are on a separate page.
- Problem 9 · scalar superposition and conductor contact
- Problem 10 · equipotential-to-field translation
- Problem 15 · experimental potential map and uncertainty
- Problem 17 · source symmetry and energy synthesis
Explain and apply the model
Try these problems before opening the solutions. Show the physical model, your calculation or sketch, and the reason your conclusion follows.
Read a field and predict actual motion
In a uniform static region, vertical equipotentials are 30 V at x = 0, 20 V at x = 0.020 m, and 10 V at x = 0.040 m. A nonperturbing particle has q = −1.0 nC, m = 2.0 × 10⁻⁶ kg and starts at (x,y) = (0.030 m,0) with velocity (0,+0.10 m/s). Ignore gravity and all other forces. The uniform region includes its motion for 0.20 s.
(a) Construct the labeled contour map and a matching field-vector map at several points. Calculate E and state its direction, magnitude uniformity, perpendicularity to contours, and component along a contour.
10.5.B.110.5.B.2.i10.5.B.2.ii10.5.B.2.iii10.5.B.2.iv10.3.A.3.ii
(b) Predict the initial force and acceleration. Calculate the particle’s x and y coordinates at 0.20 s. Add its initial tangent and qualitative curved trajectory to your map, and explain why it does not follow a rightward field arrow.
(c) Use the potential map to calculate ΔU and ΔK for that interval. Check ΔK from the velocity change and explain why moving between any two points on the same contour gives zero net electric work.
Written working workspace
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Scientific calculator workspace
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For the teacher
- Have students make both scalar and vector predictions at a dipole midpoint.
- Ask for contour labels and normal distance before estimating E.
Source objectives and scope
- 10.5.A — Describe the electric potential due to a configuration of charged objects.
- 10.5.B — Describe the relationship between electric potential and electric field.
As the methods to calculate the electric potential due to extended charges exceed the scope of the course, AP Physics 2 only expects that students calculate the electric potential of configurations of four or fewer particles (or more in situations of high symmetry).
Suggested skills are teaching guidance. The course applies all science practices across its objectives.
Science references for this teaching
- College Board AP Physics 2 Course and Exam Description, Unit 10 (PDF pages 50–61) — Official topic order, learning objectives, all essential-knowledge clauses, algebra-based scope boundaries; equations visually inspected in local source pages 50,54,56–61.
- OpenStax University Physics Volume 2, 8.5 Molecular Model of a Dielectric — Independent verification that bound polarization leaves a dielectric net neutral and its induced field opposes the free-charge field. All lesson prose and problems are original.
- Corrected original reading: Charge and Matter — Use the corrected charge and Coulomb-law explanations to compare signed charges and pair forces.
- Corrected original reading: Conductors and Insulators — Use the corrected charge-mobility discussion to explain contact, polarization, and grounding with a system boundary.
- Corrected original reading: The Electric Field — Use the corrected point-source and vector-superposition discussion to practice fields and forces. General quantitative fields within insulators are outside the AP core.
- Corrected original reading: The Electric Potential and Potential Energy — Use the corrected algebraic point-charge energy discussion to connect configuration changes with external and electric work.
- Corrected original reading: Calculating Potentials — Use the corrected point-charge scalar sums to practice potential.
- Corrected original reading: Equipotentials, Field lines, and Gradients — Use the corrected contour maps and algebraic field estimates here; calculus-gradient material is an optional extension beyond the AP core.
- Corrected original reading: Capacitors — Use the corrected parallel-plate geometry discussion to compare C, Q, and voltage.
- Corrected original reading: Stored Energy and Dielectrics — Use the corrected energy and dielectric discussion with an explicit fixed-Q or fixed-voltage constraint.
Glossary and deeper reading · 16 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Electric charge
Electric charge is a conserved property that determines electric interactions. It can be positive or negative and is measured in coulombs; ordinary net charges arise from an imbalance of charged particles. For example, transferring electrons between initially neutral objects leaves one negatively charged and the other positively charged.
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Scalar
A scalar is a quantity described by a value and unit without a spatial direction. A scalar can have a sign when its definition or reference permits one; scalar does not mean always positive. For example, mass is a scalar, whereas the velocity of that mass also needs a direction.
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Potential energy
Potential energy describes energy associated with the configuration of interacting parts of a system. Its numerical zero is a chosen reference, while differences in potential energy determine physical energy changes. For example, separating Earth and a ball changes their gravitational potential energy even though the energy belongs to their interaction.
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Vector
A vector is a quantity with a magnitude and a direction, represented by an arrow or by components along chosen axes. Vectors add by adding corresponding components rather than simply adding their magnitudes. For example, an eastward displacement followed by a northward displacement produces a resultant pointing northeast. The zero vector has zero magnitude and no unique direction.
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Potential difference
Potential difference is the change in electric potential between two locations. It represents electric potential-energy change per unit charge and is measured in volts. For example, the potential difference across a resistor describes the energy transferred per coulomb passing through it.
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Electromotive force
Electromotive force is energy supplied per unit charge by a source or induction process. Despite its name it is measured in volts, rather than newtons, and can exist even when no current flows. For example, an ideal battery supplies an emf while an induced emf can be produced by changing magnetic flux.
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Electrical conductor
An electrical conductor contains charge carriers that can move through the material. In electrostatic equilibrium, mobile charges redistribute until the electric field within the conducting material is zero. For example, excess charge on an isolated metal object resides on its surface after the charges settle.
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Electrostatic equilibrium
Electrostatic equilibrium is a condition in which the charge distribution has settled and there is no sustained motion of free charges. Within the material of an ideal conductor, the electric field is zero and the connected conductor is at one potential. For example, charges on a metal surface rearrange until any tangential surface field that would drive them has vanished.
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Equipotential
An equipotential is a set of locations with the same electric potential. Moving a charge along it produces no electric potential-energy change, and electric field vectors are perpendicular to its surface. For example, concentric spherical surfaces around an isolated point charge are equipotentials.
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Displacement
Displacement is the vector change in position from the start of an interval to its end. It depends on those endpoints rather than the length or shape of the path traveled. For example, walking around a closed track and returning to the start gives zero displacement but a nonzero distance traveled.
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Magnitude
Magnitude is the nonnegative size of a vector quantity. It does not specify direction, and it differs from a signed component. For example, a westward velocity can have a negative x component while its speed, the velocity magnitude, stays positive.
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Velocity
Velocity describes how position changes with time and includes direction. Instantaneous velocity describes one moment, while average velocity is displacement divided by elapsed time. For example, a car turning a corner can change velocity while maintaining the same speed.
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Field line
A field line is a drawn curve whose tangent shows the local direction of a vector field. Closer spacing conventionally represents greater field strength, and lines are a representation rather than physical threads. For example, electric field lines point away from a positive point charge and toward a negative one.
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Force
A force describes an interaction that can change an object’s momentum. It is a vector measured in newtons, so both size and direction matter when forces are added. For example, a rope can pull a cart while the floor exerts friction in another direction.
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Acceleration
Acceleration is the vector rate of change of velocity with time. It can change an object’s speed, direction, or both, and it is measured in meters per second squared. For example, a satellite moving at constant speed in a circular orbit accelerates toward the center.
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Gravity
Gravity is the attractive interaction between masses in the Newtonian model. For spherical bodies or point-mass approximations, its strength decreases with the square of the separation between their centers. For example, Earth’s gravity provides the inward force that keeps a satellite in orbit.