A rod’s potential adds scalar charge contributions
You will be able to: Integrate the potential of a uniform finite rod at an axial point outside it.
Why is the potential integral simpler than the field integral?
A tiny piece of charged rod contributes a potential determined by its charge and distance. Unlike electric-field vectors, these scalar contributions need no directional projections. Their signed values simply add.
A useful starting point: Equal-potential contours reveal field direction and strength →
Words and symbols before equations
- Linear density λ
- Charge per rod length, in C/m.
- dq = λ dx
- Charge on a small source element of length dx.
- Gap a
- Positive distance from the observation point to the rod’s near end.
- Natural logarithm ln
- The function whose derivative is 1/x; logarithm arguments in physical equations must be dimensionless.
What this picture assumes
Uniform finite rod on x = 0 to L; observation at x = −a, a > 0. V(∞) = 0. Numerical midpoint elements approximate the same continuous source. The field x component is negative for positive λ.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- V=6.238 V; 10-element approximation 6.236 V. At x_obs=−a, Eₓ=-4.5 N/C; source rod is from 0 to 1 m.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Put the rod on x = 0 to L and observe at x_obs = −a. A source element at x has distance a + x. Its potential is dV = kλ dx/(a + x), with zero at infinity for this finite charge distribution.
Integrate from 0 to L: V = kλ[ln(a + x)]₀ᴸ = kλ ln[(a + L)/a]. The ratio has no units. A nonuniform λ(x) would remain inside the integral.
Far away, ln(1 + L/a) ≈ L/a, giving V ≈ kQ/a with Q = λL. To recover Eₓ, remember x_obs = −a: Eₓ = −dV/dx_obs = dV/da = −kλ[1/a − 1/(a + L)].
A worked example, step by step
A rod of length 1 m has λ = +1 nC/m. Find V at a point 1 m beyond its near end.
- The element distance is 1 + x, with source bounds 0 to 1 m.
- V = kλ∫₀¹ dx/(1 + x).
- V = 9 ln2 = 6.24 V.
- The result is positive and below the 9 V estimate that would put all charge at the near end.
The logarithm must act on a dimensionless ratio. Do not confuse source coordinate x with observation coordinate −a.
Why is there no cosine projection in this integral?
Compare with an explanation
Potential is scalar. Field components require projections; potential contributions do not.
Predict. Change one thing. Explain.
Increase the number of midpoint elements without changing the rod. Compare the approximation with the exact logarithm. Then double λ and predict the potential change.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
V=6.238 V; 10-element approximation 6.236 V. At x_obs=−a, Eₓ=-4.5 N/C; source rod is from 0 to 1 m.
Uniform finite rod on x = 0 to L; observation at x = −a, a > 0. V(∞) = 0. Numerical midpoint elements approximate the same continuous source. The field x component is negative for positive λ.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant potential, energy, work, field-gradient or line-integral relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA rod with λ = +2 nC/m extends 2 m. Find V at a point a = 1 m from its near end on the extension.
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Compare with the answer and four-point rubric
- 1 point: Set dq = λ dx and element distance 1 + x.
- 1 point: V = kλ∫₀² dx/(1 + x).
- 1 point: V = 18 ln3 V.
- 1 point: V ≈ 19.78 V, relative to infinity.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why must a be positive?
This model places the observation point outside the rod, away from its singular ideal line.
RECALL 2What sets the zero here?
Infinite distance from this finite charge distribution.
RECALL 3What does increasing numerical resolution do?
Improves the approximation to the same physical potential.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A rod’s potential adds scalar charge contributions
- V = ∫ k dq/r for finite sources with V(∞) = 0.
- Axial rod: V = kλ ln[(a + L)/a].
- Far away: V ≈ kQ/a.
Remember: The logarithm must act on a dimensionless ratio. Do not confuse source coordinate x with observation coordinate −a.
Conditions: Uniform finite rod on x = 0 to L; observation at x = −a, a > 0. V(∞) = 0. Numerical midpoint elements approximate the same continuous source. The field x component is negative for positive λ.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 2 (official Unit 9) · Objectives 9.2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.2, objectives 9.2.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 2: Electric Potential, numbered Unit 9 in the official combined Physics C sequence. Topics 9.1–9.3 retain their official identifiers. Models assume electrostatic fields and state whether source charges are fixed or free. Potential integrals use the specified rods, ring, arc and infinite line or cylinder. Infinite-line examples use a finite reference radius, not zero potential at infinity. The optional 3D equipotential surface uses height to represent volts, not a particle trajectory. Checked with the Fall 2026 clarifications. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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