PHYSICS 2 • ELECTRIC POTENTIAL

Electric Potential from Point Charges — Compute electric potential from point charges using superposition

Learn how scalar electric potentials from individual charges combine algebraically to describe multi-charge systems.

Historical Context & Motivation

The concept of electric potential arose from a fundamental need to describe the energy landscape surrounding electric charges without having to track every vector component of the electric field. While Coulomb's law (1785) gave physicists a powerful tool for calculating forces between charges, the calculations became cumbersome for systems of many charges because forces are vectors—each must be decomposed into components, summed independently along each axis, and then recombined. The introduction of a scalar quantity that captures the same information dramatically simplified multi-charge problems and opened the door to modern circuit theory and electrostatics.

1785
Coulomb's Inverse-Square Law
Charles-Augustin de Coulomb experimentally established that the electrostatic force between two point charges varies as the inverse square of their separation distance, providing the quantitative foundation for all of electrostatics.
1800
Volta's Pile and the Idea of Potential
Alessandro Volta constructed the first electrochemical battery, demonstrating a sustained "electromotive force" and motivating the formal concept of electric potential difference as the work per unit charge.
1828
Green's Mathematical Framework
George Green introduced the potential function in his privately published essay, establishing the mathematical machinery—including the superposition principle—that connects potential to field through differentiation.
1867
Thomson and Maxwell Formalize Electrostatics
William Thomson (Lord Kelvin) and James Clerk Maxwell integrated electric potential into a comprehensive electromagnetic theory, cementing its role as a scalar field from which the vector electric field is derived via the gradient.

The central question this lesson addresses is deceptively simple: given a collection of point charges scattered through space, how do we find the electric potential at any arbitrary point? Because potential is a scalar, we can exploit the superposition principle by simply adding individual potentials algebraically—no vector decomposition required. This makes multi-charge potential calculations far more tractable than the corresponding electric field calculations and provides a natural gateway to understanding energy storage, capacitance, and circuit behavior.

Core Principles & Definitions

Before computing potentials from point charges, it is essential to establish the foundational ideas that underpin the calculation. Electric potential is intimately connected to the concepts of work, energy, and the conservative nature of the electrostatic field. The following principles form the conceptual scaffolding for everything that follows in this lesson.

1

Electric Potential (V)

The electric potential at a point in space is the work done per unit positive test charge by an external agent to bring the charge from infinity (the reference point where V = 0) to that location, without acceleration. It is measured in volts (1 V = 1 J/C).
2

Scalar Nature of V

Unlike the electric field E⃗, which is a vector, the electric potential V is a scalar quantity. This means it has magnitude and sign but no direction, making algebraic summation straightforward.
3

Superposition Principle

The total electric potential at any point equals the algebraic sum of the potentials produced by each individual charge. Positive charges contribute positive V; negative charges contribute negative V. No vector components are needed.
4

Reference at Infinity

The standard convention sets V = 0 at r → ∞. All potential values are measured relative to this reference. This choice is consistent with defining potential energy as zero when charges are infinitely separated.
5

Sign Convention

A positive point charge creates a positive potential that decreases with distance. A negative point charge creates a negative potential that increases (becomes less negative) with distance. Signs are incorporated directly into the calculation.
KEY TAKEAWAY
Think of electric potential as an elevation map for charges. Just as a topographic map assigns a single height number to every point on the terrain—without needing to specify a direction—the potential assigns a single voltage value to every point in space. A positive charge creates a "hill" in this landscape, and a negative charge creates a "valley." When multiple charges are present, the total landscape is simply the sum of all the individual hills and valleys—superposition is just adding heights at each location on the map.

Visualizing Potential from Point Charges

A single positive point charge creates a radially symmetric potential that falls off as 1/r. The diagram below shows how the potential varies along a line connecting two point charges of different signs, illustrating the superposition concept visually. By examining how the individual contributions (dashed curves) combine to form the total potential (solid curve), you can develop geometric intuition for how charge arrangements shape the electric potential landscape.

The red dashed curve (V₊) shows the positive potential from +Q diverging near the charge location. The blue dashed curve (V₋) shows the negative potential from −Q. The solid violet curve (Vtotal) is the algebraic sum of the two contributions, illustrating superposition. Note that between the charges, the total potential passes through zero where the positive and negative contributions exactly cancel.

Several features of this diagram are worth emphasizing. First, notice that the potential diverges (goes to ±∞) as you approach either charge—this is the 1/r behavior of the Coulomb potential. Second, the zero-crossing between the two charges occurs at the location where V₊ = −V₋, meaning the magnitudes of the two potentials are equal. For charges of equal magnitude |Q|, this zero occurs at the geometric midpoint. Third, the total potential is asymmetric if the charges have different magnitudes—the zero crossing shifts toward the smaller charge because its contribution falls off more quickly.

Mathematical Framework

The mathematical expression for the electric potential from a single point charge follows directly from the definition of potential as work per unit charge. Beginning with Coulomb's law and integrating the electric field from infinity to a point at distance r from the charge, one obtains the fundamental result.

POTENTIAL FROM A SINGLE POINT CHARGE
V = kQ / r = Q / (4πε₀r)
where V is the electric potential in volts (V), k = 8.99 × 10⁹ N·m²/C² is Coulomb's constant, Q is the source charge (with sign) in coulombs, r is the distance from the charge to the field point in meters, and ε₀ = 8.854 × 10⁻¹² C²/(N·m²) is the permittivity of free space.

Note that the charge Q retains its sign in this expression—there is no absolute value. A positive charge produces V > 0 at all finite distances, while a negative charge produces V < 0. Also, r is always a positive scalar representing the magnitude of the displacement from the charge to the observation point.

SUPERPOSITION OF POINT-CHARGE POTENTIALS
V_total = Σᵢ Vᵢ = k Σᵢ (Qᵢ / rᵢ)
where Qᵢ is the i-th source charge (with sign) and rᵢ is the distance from the i-th charge to the observation point P. The sum is a simple algebraic sum—no vector decomposition is needed because each Vᵢ is a scalar.

The derivation of the single-charge potential proceeds as follows. The electric field from a point charge Q located at the origin is E⃗ = kQ r̂ / r². The potential difference between a point at distance r and the reference at infinity is V(r) − V(∞) = −∫(∞ to r) E⃗ · dr⃗. Evaluating the integral along a radial path gives V(r) = −∫(∞ to r) (kQ/r'²) dr' = kQ/r, confirming the expression above. Because the electrostatic field is conservative, this result is path-independent—the potential depends only on the endpoint distance r, not on the integration path chosen.

RELATIONSHIP BETWEEN POTENTIAL AND FIELD
E⃗ = −∇V
The electric field is the negative gradient of the potential. For a one-dimensional case along the x-axis, this reduces to Eₓ = −dV/dx. The potential provides a complete description of the electrostatic environment; the field can always be recovered by differentiation.
CRITICAL DISTINCTION
Do not confuse superposition of potentials (scalar addition) with superposition of electric fields (vector addition). When computing the total electric field from multiple charges, you must add x-, y-, and z-components separately. When computing the total potential, you simply add signed numbers. This is the primary computational advantage of working with potential.

Geometric Configurations & Distance Calculations

The most common source of error in superposition problems is computing the distances rᵢ from each charge to the observation point. When charges are not collinear with the field point, you must use the Pythagorean theorem or the general distance formula. The diagram below illustrates a typical two-dimensional configuration with three point charges and an observation point P, labeling all relevant distances.

Three point charges Q₁ = +3 μC at (1, 1), Q₂ = −2 μC at (4, 1), and Q₃ = +1 μC at (2, 3) are shown with an observation point P at (5, 4). The dashed lines represent the distances r₁, r₂, and r₃ from each charge to P, computed using the Pythagorean distance formula. These distances feed directly into the superposition formula Vtotal = k(Q₁/r₁ + Q₂/r₂ + Q₃/r₃).

When charges are arranged along a single axis (collinear configuration), the distances simplify to absolute differences of coordinates. However, in two- and three-dimensional arrangements, you must apply the general distance formula: rᵢ = √[(xP − xi)² + (yP − yi)² + (zP − zi)²]. A systematic approach is to tabulate each charge, its coordinates, the observation point coordinates, and the resulting distance before performing any potential calculation.

Distance calculations for the three-charge configuration shown above.
ChargeValue (μC)PositionDistance to P (m)
Q₁+3(1, 1)√[(5−1)² + (4−1)²] = √(16+9) = 5.00
Q₂−2(4, 1)√[(5−4)² + (4−1)²] = √(1+9) ≈ 3.16
Q₃+1(2, 3)√[(5−2)² + (4−3)²] = √(9+1) ≈ 3.16

Worked Example: Three-Charge Superposition

Using the three-charge configuration from Section 5, let us compute the total electric potential at point P = (5, 4) step by step. This problem demonstrates the complete superposition procedure, from distance calculation through final summation.

Find V_total at P = (5, 4) m
1
Step 1 — Identify Given ValuesWe have three point charges: Q₁ = +3 μC = +3 × 10⁻⁶ C at (1, 1), Q₂ = −2 μC = −2 × 10⁻⁶ C at (4, 1), and Q₃ = +1 μC = +1 × 10⁻⁶ C at (2, 3). The observation point is P = (5, 4). Coulomb's constant k = 8.99 × 10⁹ N·m²/C².
2
Step 2 — Calculate Distancesr₁ = √[(5 − 1)² + (4 − 1)²] = √[16 + 9] = √25 = 5.00 m. r₂ = √[(5 − 4)² + (4 − 1)²] = √[1 + 9] = √10 ≈ 3.162 m. r₃ = √[(5 − 2)² + (4 − 3)²] = √[9 + 1] = √10 ≈ 3.162 m.
r₁ = 5.00 m, r₂ = r₃ ≈ 3.162 m
3
Step 3 — Compute Individual PotentialsV₁ = kQ₁/r₁ = (8.99 × 10⁹)(3 × 10⁻⁶)/5.00 = 26.97 × 10³/5.00 = 5,394 V. V₂ = kQ₂/r₂ = (8.99 × 10⁹)(−2 × 10⁻⁶)/3.162 = −17.98 × 10³/3.162 = −5,688 V. V₃ = kQ₃/r₃ = (8.99 × 10⁹)(1 × 10⁻⁶)/3.162 = 8.99 × 10³/3.162 = 2,843 V.
V₁ ≈ 5,394 V, V₂ ≈ −5,688 V, V₃ ≈ 2,843 V
4
Step 4 — Apply SuperpositionV_total = V₁ + V₂ + V₃ = 5,394 + (−5,688) + 2,843 = 2,549 V.
V_total ≈ 2,550 V (or 2.55 kV)
5
Step 5 — Interpret the ResultThe net potential at P is positive, meaning a positive test charge placed at P would have positive potential energy relative to infinity. The negative charge Q₂ partially cancels the contributions of Q₁ and Q₃, but the two positive charges together dominate. If you were to release a positive test charge from P, it would naturally move toward lower potential (away from the positive charges and toward the negative charge).

Potential vs. Electric Field: Strengths & Limitations

While both electric potential and electric field fully describe the electrostatic environment around charges, each representation has distinct computational and conceptual advantages. Understanding when to use which quantity is a hallmark of physical maturity in electrostatics.

Comparison of potential and field approaches to electrostatic problems.
FeatureElectric Potential (V)Electric Field (E⃗)
TypeScalar — magnitude and sign onlyVector — magnitude and direction
SuperpositionAlgebraic sum of signed numbersVector sum (component-by-component)
Multi-charge computationSimpler — one calculation per chargeHarder — requires component decomposition
Physical meaningEnergy per unit charge (J/C)Force per unit charge (N/C)
Gives direction of force?Not directly (requires gradient)Yes, directly
Best suited forEnergy calculations, equipotential mappingForce calculations, field line visualization
KEY TAKEAWAY
Working with potential instead of the electric field is analogous to using energy methods instead of force methods in mechanics. Just as you can solve many projectile and conservation problems faster with energy (a scalar) than with Newton's second law (a vector equation), you can often solve multi-charge electrostatic problems more efficiently using potential. The trade-off is that potential does not directly tell you the direction of the force—you need to take the gradient to recover the field. Choose your representation based on what the problem asks for.
COMMON PITFALL
Students sometimes try to find a "direction" for electric potential or draw vectors for it. Remember: V has no direction. It is a scalar field. The quantity that has direction is the electric field E⃗ = −∇V. If a problem asks for the force on a charge, you either need to compute E⃗ directly or first find V and then take its gradient.

Connection to Potential Energy & Continuous Distributions

The point-charge superposition formula is the discrete foundation upon which more advanced electrostatic concepts are built. Two immediate extensions deserve mention: the connection to electric potential energy of a charge configuration, and the generalization to continuous charge distributions where the sum becomes an integral.

How discrete point-charge potential connects to advanced electrostatic theory.
ConceptThis Lesson (Discrete Charges)Advanced Extension
Superposition formulaV = k Σ (Qᵢ / rᵢ)V = k ∫ (dq / r) for continuous ρ, σ, or λ
Potential energyU = qV (energy of test charge q in existing potential)U = k Σᵢ<ⱼ (QᵢQⱼ / rᵢⱼ) (assembly energy)
Recovering E⃗E⃗ = −∇V (compute gradient numerically or analytically)Same relationship; Poisson's equation ∇²V = −ρ/ε₀
Equipotential surfacesSet V = constant and solve for lociBoundary conditions for Laplace/Poisson equations

The transition from discrete to continuous charge distributions is conceptually straightforward: replace Qᵢ with a differential charge element dq and the sum with an integral. For a linear charge distribution with charge per unit length λ, dq = λ dl; for a surface charge density σ, dq = σ dA; and for a volume charge density ρ, dq = ρ dV. The integral V = (1/4πε₀) ∫ dq/r is simply the continuous-limit analog of the superposition sum. Mastering the discrete case in this lesson provides the essential intuition and procedural framework for these more complex integrals encountered in advanced electromagnetism courses.

🔭 LOOKING AHEAD
In subsequent topics, you will use the potential to define capacitance (C = Q/ΔV), to analyze energy stored in electric fields (U = ½CV²), and to solve boundary-value problems using Laplace's equation. The scalar superposition principle you learned here remains the conceptual backbone for all of these applications.

Practice Problems

PROBLEM 1CONCEPTUAL
Two equal and opposite charges (+Q and −Q) are separated by a distance d. Is there a point along the line connecting them where the electric potential is zero? Is there a point where the electric field is zero? Explain the distinction.
PROBLEM 2BASIC CALCULATION
A charge Q₁ = +4.0 μC is located at the origin, and a charge Q₂ = −6.0 μC is located at x = 3.0 m. Calculate the electric potential at x = 1.0 m along the x-axis.
PROBLEM 3INTERMEDIATE
Three charges are arranged at the vertices of an equilateral triangle with side length 0.50 m: Q₁ = +2.0 μC at the top vertex, Q₂ = +2.0 μC at the bottom-left, and Q₃ = −4.0 μC at the bottom-right. Find the electric potential at the centroid of the triangle.
PROBLEM 4APPLIED
In a simplified model of a water molecule, a charge of +2e sits at the oxygen position at the origin, and two charges of −e each sit at the hydrogen positions at (0.096 nm, 0) and (−0.059 nm, 0.076 nm), where e = 1.602 × 10⁻¹⁹ C. Calculate the electric potential at the point (0.30 nm, 0) due to this charge arrangement.
PROBLEM 5CRITICAL THINKING
Consider N identical positive charges, each of magnitude Q, arranged symmetrically around a circle of radius R. (a) Show that the potential at the center of the circle is V = NkQ/R. (b) Now suppose one of the N charges is replaced by a charge of −Q. Derive the new potential at the center. (c) Explain why the potential at the center is independent of which charge is replaced, and comment on whether the electric field at the center would share this property.

Lesson Summary

The electric potential from a single point charge is given by V = kQ/r, a scalar quantity measured in volts that is positive for positive charges and negative for negative charges, with the reference set at infinity. Because potential is a scalar, the superposition principle allows us to compute the total potential at any point by performing a simple algebraic sum: V_total = k Σ (Qᵢ/rᵢ), where each Qᵢ carries its sign and rᵢ is the distance from the i-th charge to the observation point.

The key procedural steps are: (1) identify all source charges and their positions, (2) compute the distance from each charge to the field point using the distance formula, (3) calculate each individual potential Vᵢ = kQᵢ/rᵢ with proper signs, and (4) sum all contributions. This approach avoids the vector decomposition required for electric field superposition, making it computationally simpler for multi-charge systems. The potential can always be connected back to the electric field through E⃗ = −∇V, and to potential energy through U = qV. These relationships form the foundation for capacitance, circuit analysis, and boundary-value problems in advanced electromagnetism.

Varsity Tutors • Physics 2 • Electric Potential from Point Charges — Compute electric potential from point charges using superposition