Integrate the potential above a rod’s midpoint
You will be able to: Evaluate the perpendicular-bisector potential and differentiate it.
How does a rod’s midpoint potential connect to its vertical field?
Above the center of a charged strip, mirrored elements have the same distance. Their potentials add even though their horizontal field components cancel. Symmetry simplifies the limits, not the sign of scalar addition.
A useful starting point: A rod’s potential adds scalar charge contributions →
Words and symbols before equations
- Half-length a
- The uniform rod runs from −a to +a.
- Height y
- Positive observation distance along the perpendicular bisector.
- Inverse hyperbolic sine asinh
- A compact notation with asinh(t) = ln(t + √(1 + t²)).
- Observation derivative
- Differentiate with respect to y while keeping the rod’s charge and geometry fixed.
What this picture assumes
Uniform finite rod from x = −a to +a; probe at (0,y), y > 0. V(∞) = 0; Eₓ = 0 on the perpendicular bisector. Scalar contributions add even though horizontal field components cancel.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Rod length 2 m. At y=1 m: V=15.86 V; Eᵧ=12.73 N/C; Eₓ=0 by symmetry.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
The distance to a source element at x is √(x² + y²). Therefore V(y) = kλ∫₋ₐᵃ dx/√(x² + y²). Both halves contribute with the same sign.
The result is V = 2kλ asinh(a/y) = 2kλ ln[(a + √(a² + y²))/y]. Either form uses a dimensionless argument. It approaches zero as y tends to infinity.
Differentiate with respect to y: Eᵧ = −dV/dy = 2kλa/[y√(a² + y²)], matching the earlier vector-field integration. The x component is zero on the bisector by symmetry, not because V is zero.
A worked example, step by step
A rod from −1 to +1 m has λ = +1 nC/m. Find V at y = 1 m and the local Eᵧ.
- Use a = y = 1 m in the bisector formula.
- V = 18 ln(1 + √2) = 15.86 V.
- Eᵧ = 18/[1√2] = 12.73 N/C.
- V is positive and decreases with increasing y, so the field points upward.
The mirrored potentials add. Only particular field components cancel.
Does zero horizontal field imply zero potential?
Compare with an explanation
No. It reflects symmetric cancellation of vector components, while potentials add.
Predict. Change one thing. Explain.
Move y upward at fixed rod. Compare decreasing V with the positive upward field. At large y, compare V with kQ/y.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Rod length 2 m. At y=1 m: V=15.86 V; Eᵧ=12.73 N/C; Eₓ=0 by symmetry.
Uniform finite rod from x = −a to +a; probe at (0,y), y > 0. V(∞) = 0; Eₓ = 0 on the perpendicular bisector. Scalar contributions add even though horizontal field components cancel.
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 questionFor the worked geometry, double λ to 2 nC/m. Find V and Eᵧ, and explain why the two quantities scale by the same factor.
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Compare with the answer and four-point rubric
- 1 point: Both the potential and field integrals are linear in λ.
- 1 point: V = 36 ln(1 + √2) = 31.73 V.
- 1 point: Eᵧ = 36/√2 = 25.46 N/C.
- 1 point: Doubling every charge element doubles each contribution at unchanged distances.
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 1What is the full rod length?
2a.
RECALL 2Which quantity must stay fixed when differentiating V(y)?
The source charge distribution and rod geometry.
RECALL 3Why is the logarithm argument dimensionless?
Its numerator and denominator both have length units.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Integrate the potential above a rod’s midpoint
- V = 2kλ asinh(a/y).
- V = 2kλ ln[(a + √(a²+y²))/y].
- Eᵧ = −dV/dy = 2kλa/[y√(a²+y²)].
Remember: The mirrored potentials add. Only particular field components cancel.
Conditions: Uniform finite rod from x = −a to +a; probe at (0,y), y > 0. V(∞) = 0; Eₓ = 0 on the perpendicular bisector. Scalar contributions add even though horizontal field components cancel.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 2 (official Unit 9) · Objectives 9.2.A, 9.2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.2, objectives 9.2.A, 9.2.B. 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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