A ring can have maximum potential where its field is zero
You will be able to: Find the ring-axis potential and recover the axial field by differentiation.
Why does a charged ring have nonzero center potential but zero center field?
Every part of a positive ring contributes positive potential at its center. Those scalar contributions add. The corresponding field vectors point in different directions and cancel at the center.
A useful starting point: Integrate the potential above a rod’s midpoint →
Words and symbols before equations
- Ring radius R
- The fixed radius of a uniformly charged thin ring in the xy plane.
- Axis z
- The signed coordinate perpendicular to the ring plane through its center.
- Total charge Q
- The sum of all ring elements dq.
- Axial stationary point
- A point where dV/dz is zero; this describes one component of the gradient.
What this picture assumes
Uniform thin ring in xy plane, probe on z axis, V(∞) = 0. Axial graphs are functions of position, not trajectories. A maximum or minimum along z alone does not establish three-dimensional stability.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- At z=0 m: V=18 V and E_z=0 N/C. Ring radius R=1 m, Q=2 nC.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
At any point on the axis, every ring element has the same distance √(R² + z²). Thus V = ∫k dq/√(R² + z²) = kQ/√(R² + z²), choosing V(∞) = 0.
Differentiate: E_z = −dV/dz = kQz/(R² + z²)³ᐟ². At z = 0, V = kQ/R but E_z = 0. Uniform ring symmetry also cancels the transverse components there.
For positive Q, the axial V curve has a maximum at the center. That is an axial statement, not a claim of stable confinement in three dimensions. Far along the axis, V approaches kQ/|z|.
A worked example, step by step
A ring has Q = +2 nC and R = 1 m. Calculate V and E_z at z = 0 and z = 1 m.
- At the center, V = kQ/R = 18 V.
- The center field is zero by symmetric cancellation.
- At z = 1 m, V = 18/√2 = 12.73 V.
- E_z = 18/(2√2) = 6.36 N/C along +z, consistent with a falling potential.
A zero field can occur at a nonzero potential. A one-dimensional stationary point does not prove three-dimensional stability.
Is V(+z) equal to V(−z)?
Compare with an explanation
Yes, because the distance depends on z². Their axial fields have opposite signs.
Predict. Change one thing. Explain.
Move the probe through z = 0. Predict the even symmetry of V and odd symmetry of E_z. Compare potential height with slope.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At z=0 m: V=18 V and E_z=0 N/C. Ring radius R=1 m, Q=2 nC.
Uniform thin ring in xy plane, probe on z axis, V(∞) = 0. Axial graphs are functions of position, not trajectories. A maximum or minimum along z alone does not establish three-dimensional stability.
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 Q = −1 nC, R = 1 m, find V at z = 0 and 1 m, and state the sign of E_z at z = 1 m.
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Compare with the answer and four-point rubric
- 1 point: V(0) = −9 V.
- 1 point: V(1) = −9/√2 = −6.36 V.
- 1 point: V increases as z increases near 1 m.
- 1 point: E_z is negative, pointing back toward the negatively charged ring.
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 does the scalar integral simplify?
Every ring element is equally far from an axial point.
RECALL 2What gives E_z from V?
The negative derivative with respect to z.
RECALL 3Does center V = kQ/R imply E = kQ/R² there?
No. Field vectors cancel at the center.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A ring can have maximum potential where its field is zero
- V(z) = kQ/√(R²+z²).
- E_z = kQz/(R²+z²)³ᐟ².
- V(0) = kQ/R; E(0) = 0 for the uniform ring.
Remember: A zero field can occur at a nonzero potential. A one-dimensional stationary point does not prove three-dimensional stability.
Conditions: Uniform thin ring in xy plane, probe on z axis, V(∞) = 0. Axial graphs are functions of position, not trajectories. A maximum or minimum along z alone does not establish three-dimensional stability.
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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