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LESSON 09 / 16 · TOPIC 10.3

A uniform plate field bends a charged particle’s path

You will be able to: Combine a = qE/m with independent horizontal and vertical motion and check plate collisions.

Calculus-based electrostaticsFree study resourceReview editionTeacher review pending

How does a particle move sideways through a capacitor?

A tiny charged particle enters halfway between horizontal plates moving to the right. With the bottom plate positive and the top negative, the electric field points upward. A positive particle curves upward; a negative one curves downward.

A useful starting point: Area and separation set parallel-plate capacitance →

Words and symbols before equations

Specific charge q/m
Signed charge-to-mass ratio, in C/kg.
Horizontal speed vₓ
Initial speed parallel to the plates; constant when the field is vertical.
Transit time
Time needed to reach the far end if no plate is struck.
Impact
First contact with a plate, where this free-flight model stops.
Particle flight stops at exit or first impactTop negative plate: y = +10 mmBottom positive plate: y = −10 mmx: 0 → 0.10 m; y: −10 → +10 mm (different scales)
Read this model snapshot. a_y = 500 m/s². Exit at t = 5 ms, x = 0.1 m, y = 6.25 mm; v_y = 2.5 m/s.
What this picture assumes

Ideal upward field between bottom positive and top negative plates, gap 0.020 m and length 0.10 m. Entry is at the midplane with zero vertical speed. Gravity, fringing and radiation neglected; speeds are nonrelativistic. Path stops at first impact. Horizontal and vertical display scales differ and are labeled.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. a_y = 500 m/s². Exit at t = 5 ms, x = 0.1 m, y = 6.25 mm; v_y = 2.5 m/s.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the physics

Choose x right and y up, with y = 0 halfway across gap d. The field is E_y = ΔV/d upward. The electric acceleration is a_y = (q/m)E_y; its sign includes the charge sign.

With initial v_y = 0, x = vₓt and y = ½a_yt². Eliminating time gives a parabolic path y = a_yx²/(2vₓ²). The model neglects gravity, fringing, collisions before a plate, and relativistic effects.

First compare the exit time L/vₓ with the impact time √(d/|a_y|). If impact occurs sooner, stop the path at the plate instead of drawing motion through metal. Reversing charge reverses the curvature without changing horizontal speed.

A worked example, step by step

Let d = 0.020 m, L = 0.10 m, ΔV = 10 V, q/m = +1 C/kg and vₓ = 20 m/s. Does the particle exit?

  1. E_y = 10/0.020 = 500 N/C and a_y = 500 m/s².
  2. The candidate exit time is 0.10/20 = 0.005 s.
  3. At that time y = ½(500)(0.005)² = 0.00625 m, smaller than d/2 = 0.010 m.
  4. The particle exits 6.25 mm above the midplane; v_y = a_yt = 2.5 m/s.
Common mix-up

A trajectory must stop at its first plate collision. Force follows the charge sign; field direction does not.

CHECK THE IDEA

If q/m = 0, what path is predicted?

Compare with an explanation

A straight horizontal line, since this model includes no other force.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Reverse q/m and increase voltage. Identify the first setting at which the particle hits a plate before reaching the far end. Compare the exit-time prediction with the truncated path.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Particle flight stops at exit or first impactTop negative plate: y = +10 mmBottom positive plate: y = −10 mmx: 0 → 0.10 m; y: −10 → +10 mm (different scales)

a_y = 500 m/s². Exit at t = 5 ms, x = 0.1 m, y = 6.25 mm; v_y = 2.5 m/s.

Ideal upward field between bottom positive and top negative plates, gap 0.020 m and length 0.10 m. Entry is at the midplane with zero vertical speed. Gravity, fringing and radiation neglected; speeds are nonrelativistic. Path stops at first impact. Horizontal and vertical display scales differ and are labeled.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant conductor equilibrium, charge conservation, capacitance or energy 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.

1. A negative particle in an upward electric field accelerates…

Show answer and reasoning

down. F = qE reverses direction for negative q.

2. With a_y = 200 m/s² and t = 0.010 s, vertical displacement is…

Show answer and reasoning

0.010 m. y = ½at² = 0.010 m.

Original written challenge

4 points · self-check · not an official AP question

A midplane particle has a_y = 800 m/s², gap 0.020 m, plate length 0.10 m and horizontal speed 10 m/s. Find its first impact time and position.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: The top plate is 0.010 m from entry height.
  2. 1 point: Solve 0.010 = ½(800)t² to get t = 0.005 s.
  3. 1 point: x = 10(0.005) = 0.050 m.
  4. 1 point: This is before the far end at 0.10 m, so the free-flight path terminates at impact.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1What sets the curve direction?

The sign of qE_y.

RECALL 2Why does vₓ remain constant?

No horizontal force is included.

RECALL 3Which time determines the endpoint?

The smaller of plate-impact time and exit time.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

A uniform plate field bends a charged particle’s path

  • a_y = (q/m)ΔV/d.
  • x = vₓt; y = ½a_yt² when v_y0 = 0.
  • t_impact = √(d/|a_y|) for entry at the midplane.

Remember: A trajectory must stop at its first plate collision. Force follows the charge sign; field direction does not.

Conditions: Ideal upward field between bottom positive and top negative plates, gap 0.020 m and length 0.10 m. Entry is at the midplane with zero vertical speed. Gravity, fringing and radiation neglected; speeds are nonrelativistic. Path stops at first impact. Horizontal and vertical display scales differ and are labeled.

Refresh Kid · AP Physics C: Electricity and Magnetism Unit 3 (official Unit 10) · Objectives 10.3.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 10.3, objectives 10.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 3: Conductors and Capacitors, numbered Unit 10 in the official combined Physics C sequence. Topics 10.1–10.4 retain their official identifiers. Models state the electrostatic conditions, geometry approximations and whether charge or voltage stays fixed. Capacitor geometries include parallel plates, concentric spheres and long coaxial cylinders. Dielectric comparisons assume a fully filling ideal linear material. The optional 3D plate view uses explicitly different gap and lateral scales to show the small separation. 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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