--- title: "Calculating electric potential from a charge distribution" description: "Potential is a scalar, so contributions add without regard to direction. This usually makes potential easier to compute than the field. For a continuous body, calculate potential with an integral over" canonical: https://lightmysky.com/learn/science/calculating-electric-potential-from-a-charge-distribution-mt_B4WiVkpH4j source: https://lightmysky.com/learn/science/calculating-electric-potential-from-a-charge-distribution-mt_B4WiVkpH4j.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Calculating electric potential from a charge distribution Potential is a scalar, so contributions add without regard to direction. This usually makes potential easier to compute than the field. For a continuous body, calculate potential with an integral over the charge. Subject: Science · Area: Electricity & Magnetism · Ages 18 to 19 Page: https://lightmysky.com/learn/science/calculating-electric-potential-from-a-charge-distribution-mt_B4WiVkpH4j ## Ready when they can - Adds the potentials of several point charges at a given place - Sets up and evaluates the potential integral for a line or ring of charge - Explains why the choice of zero potential is free and where it is normally put ## Lesson: Voltage adds like numbers Electric potential V is potential energy U per unit test charge q: V = U/q, so U = qV. Its unit is the volt: 1 V = 1 J/C. In a fixed electrostatic field, a positive charge released from rest where E is nonzero initially accelerates toward lower V if only the electric force acts. It loses potential energy as it gains kinetic energy. This describes its initial motion, not every later part of its path. For a point charge q, V = kq/r when zero potential is set at infinity. Here r is the distance from the charge, and k is about 8.99 × 10^9 N m²/C². Keep the sign of q: a negative source charge gives negative V with this reference. At r = 0 the point-charge formula is undefined. For several fixed point charges, add their signed contributions: V = kΣ(qi/ri). Each ri is the distance from that charge to the point. At a square's center, all four distances are equal, so V = k(q1 + q2 + q3 + q4)/r. For square side a, r = a/√2. Convert charge to coulombs before calculating: 1 nC = 10^-9 C. *(drawing: With k = 9.0 × 10^9, corner charges +1, +2, -1 and +4 nC, each 0.30 m from the center, give these contributions.)* For a finite continuous distribution, replace the sum by V = k∫dq/r, with zero at infinity. The distance r runs from each charge element dq to the observation point. For a line with charge per length λ, dq = λ dl. At a ring's center every element is distance R away, so V = (k/R)∫dq = kQ/R, where Q is the ring's total signed charge. *(drawing: Every ring element has the same distance R to the center. The charge integral gives the total charge Q.)* For one fixed charge distribution, a new zero adds the same constant C everywhere: Vnew = Vold + C. Every difference VB - VA is unchanged. Choose a reference where potential is finite and defined, not at a point charge. For finite localized charges, zero at infinity is available. Ground can also be a reference. Choosing a reference does not mean physically grounding a conductor; a grounding connection can change its charge. *(drawing: The electric work from A to B is -qΔV. A new reference leaves that work unchanged.)* Each field component is minus the local slope of potential: Ex = -∂V/∂x, and likewise for y and z. Hold y and z fixed when finding the x slope. For V = A + bx, Ex = -b. The field points toward the steepest decrease of V. A spatial constant C has zero slope, so V and V + C give the same field everywhere. Since F = qE, the same charge with the same initial position and velocity has the same motion under this electric force. **Recap.** Potential contributions add as signed numbers. Choose a reference where potential is finite and defined. The electric field is minus the local potential slope. *Lesson and revised questions written for this dataset. Electric-potential facts and methods were studied in OpenStax University Physics Volume 2, sections 7.2, 7.3 and 7.4.* ## Practice 8 questions on this page, each with its working shown. ## Needs first - [Conductors, Shielding and Surface Charge](https://lightmysky.com/learn/science/conductors-shielding-and-surface-charge-mt_2zmzTE-dgL) - [Electric Potential and the Uniform Field Between Plates](https://lightmysky.com/learn/science/electric-potential-and-the-uniform-field-between-plates-mt_LKTfeYxNde) ## Opens up - [Gauge Freedom and the Potentials as the Working Variables](https://lightmysky.com/learn/science/gauge-freedom-and-the-potentials-as-the-working-variables-mt_BovDzq3WZm) - [Recovering the Field from the Potential Gradient](https://lightmysky.com/learn/science/recovering-the-field-from-the-potential-gradient-mt_EqCLB9bqG1)