Electric Potential from Point Charges — Compute electric potential from point charges using superposition - Physics 2

SECTION 1

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.

SECTION 2

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.

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).

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.

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.

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.

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.

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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.

SECTION 3

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 (V_total) 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.

SECTION 4

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.

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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.

SECTION 5

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 V_total = 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 = √[(x_P − x_i)² + (y_P − y_i)² + (z_P − z_i)²]. A systematic approach is to tabulate each charge, its coordinates, the observation point coordinates, and the resulting distance before performing any potential calculation.

Charge	Value (μC)	Position	Distance 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

Distance calculations for the three-charge configuration shown above.

SECTION 6

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

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Step 1 — Identify Given Values

We 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.
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Step 2 — Calculate Distances

r₁ = √[(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
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Step 3 — Compute Individual Potentials

V₁ = 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
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Step 4 — Apply Superposition

V_total = V₁ + V₂ + V₃ = 5,394 + (−5,688) + 2,843 = 2,549 V.

V_total ≈ 2,550 V (or 2.55 kV)
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Step 5 — Interpret the Result

The 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).

SECTION 7

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.

Feature	Electric Potential (V)	Electric Field (E⃗)
Type	Scalar — magnitude and sign only	Vector — magnitude and direction
Superposition	Algebraic sum of signed numbers	Vector sum (component-by-component)
Multi-charge computation	Simpler — one calculation per charge	Harder — requires component decomposition
Physical meaning	Energy per unit charge (J/C)	Force per unit charge (N/C)
Gives direction of force?	Not directly (requires gradient)	Yes, directly
Best suited for	Energy calculations, equipotential mapping	Force calculations, field line visualization

Comparison of potential and field approaches to electrostatic problems.

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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.

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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.

SECTION 8

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.

Concept	This Lesson (Discrete Charges)	Advanced Extension
Superposition formula	V = k Σ (Qᵢ / rᵢ)	V = k ∫ (dq / r) for continuous ρ, σ, or λ
Potential energy	U = qV (energy of test charge q in existing potential)	U = k Σᵢ<ⱼ (QᵢQⱼ / rᵢⱼ) (assembly energy)
Recovering E⃗	E⃗ = −∇V (compute gradient numerically or analytically)	Same relationship; Poisson's equation ∇²V = −ρ/ε₀
Equipotential surfaces	Set V = constant and solve for loci	Boundary conditions for Laplace/Poisson equations

How discrete point-charge potential connects to advanced electrostatic theory.

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.

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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.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

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.

Show answer

Yes, the electric potential is zero at the midpoint between the two charges, because the positive potential from +Q and the negative potential from −Q have equal magnitudes there and cancel algebraically (V = kQ/r − kQ/r = 0). However, the electric field is not zero at the midpoint; both charges produce fields pointing in the same direction (from + toward −) at that location, so the fields add rather than cancel. The electric field is zero outside the pair, on the side of the positive charge, where the field vectors from the two charges point in opposite directions and can cancel. This highlights a critical conceptual distinction: zero potential does not imply zero field, because potential is a scalar sum while the field is a vector sum.

Question 2: BASIC 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.

Show answer

The distance from Q₁ to the observation point is r₁ = 1.0 m, and the distance from Q₂ to the observation point is r₂ = |3.0 − 1.0| = 2.0 m. Applying superposition: V = kQ₁/r₁ + kQ₂/r₂ = (8.99 × 10⁹)(4.0 × 10⁻⁶)/1.0 + (8.99 × 10⁹)(−6.0 × 10⁻⁶)/2.0 = 35,960 V + (−26,970 V) = 8,990 V ≈ 9.0 kV. The result is positive because the nearby positive charge dominates despite the negative charge having a larger magnitude, since the positive charge is closer to the field point.

Question 3: INTERMEDIATE

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.

Show answer

The centroid of an equilateral triangle is equidistant from all three vertices. The distance from each vertex to the centroid is r = L/√3 = 0.50/√3 ≈ 0.2887 m, where L is the side length. By superposition: V = k(Q₁/r + Q₂/r + Q₃/r) = (k/r)(Q₁ + Q₂ + Q₃). The sum of charges is (+2.0 + 2.0 − 4.0) × 10⁻⁶ = 0. Therefore V = (k/r)(0) = 0 V. This elegant result occurs because all charges are equidistant from the centroid, so the superposition reduces to a single factor times the net charge. Any charge configuration with zero net charge and equal distances to the field point will produce zero potential at that point.

Question 4: APPLIED

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.

Show answer

First, convert all positions to meters: oxygen at (0, 0), H₁ at (9.6 × 10⁻¹¹, 0), H₂ at (−5.9 × 10⁻¹¹, 7.6 × 10⁻¹¹), and P at (3.0 × 10⁻¹⁰, 0). Compute distances:

r_O

= 3.0 × 10⁻¹⁰ m; r_H1 = (3.0 − 0.96) × 10⁻¹⁰ = 2.04 × 10⁻¹⁰ m; r_H2 = √[(3.0 + 0.59)² + (0.76)²] × 10⁻¹⁰ = √[12.888 + 0.578] × 10⁻¹⁰ ≈ 3.67 × 10⁻¹⁰ m. Now apply superposition: V = k[2e/

r_O

+ (−e)/r_H1 + (−e)/r_H2] = (8.99 × 10⁹)(1.602 × 10⁻¹⁹)[2/(3.0 × 10⁻¹⁰) − 1/(2.04 × 10⁻¹⁰) − 1/(3.67 × 10⁻¹⁰)] = (1.44 × 10⁻⁹)[6.67 × 10⁹ − 4.90 × 10⁹ − 2.72 × 10⁹] = (1.44 × 10⁻⁹)(−0.95 × 10⁹) ≈ −1.4 V. The small negative value reflects the near-cancellation of the positive oxygen contribution by the two hydrogen contributions, consistent with the molecule's near-neutral character at moderate distances.

Question 5: CRITICAL 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.

Show answer

(a) By symmetry, every charge is at distance R from the center. Since potential is a scalar, V_center = Σᵢ kQ/R = NkQ/R. (b) Replacing one +Q with −Q changes one term from +kQ/R to −kQ/R, a net shift of −2kQ/R. The new potential is V = NkQ/R − 2kQ/R = (N − 2)kQ/R. (c) The potential at the center depends only on each charge's magnitude and its distance from the center—not on its angular position—because V is a scalar. Since all positions on the circle are at the same distance R, replacing any charge yields the same potential. However, the electric field would not share this property: the field is a vector, and the direction of the contribution from each charge depends on its angular position. Replacing a charge at the top of the circle versus the right side would produce different resultant field vectors at the center, even though the magnitudes of the individual contributions are the same. This elegantly demonstrates why scalar superposition is simpler than vector superposition.

SUMMARY

Lesson Summary

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.