Reconcile charged-particle deflection with an energy ledger

Establish the model and evidence

A uniform electric field exerts constant force F=qE on a particle whose charge remains fixed. Dividing by its mass gives constant acceleration. This resembles a projectile under gravity, except reversing the charge reverses electric acceleration. The horizontal velocity remains unchanged here because the stipulated field is entirely vertical.

Use horizontal motion to determine how long the particle spends in the field. If its entry horizontal speed is vx and region length L, the transit time is t=L/vx. With initial vertical speed zero, y=½ayt² and vy=ayt. The geometry must be checked: a formula extended beyond a plate collision would no longer describe free motion. Every allowed setting in this bounded virtual region remains between the plates.

An independent energy route checks the same result. With potential reference V(0)=0 and field E in +y, the potential at displacement y is −Ey. The electric energy change is qΔV, and the kinetic energy change is its negative when only the electric force does work. A negative charge moves toward higher potential but still loses electric potential energy because q is negative.

This is a hypothetical CPU teaching model, not a beam experiment or a practical design. Gravity, collisions, radiation and source recoil are explicitly omitted. The source maintains the uniform field; its wider energy accounting is outside the fixed-field particle subsystem. The neutral setting must give no electric acceleration or energy change, providing a useful zero-case check.

Worked evidence

For q=+1 nC, mass 10⁻⁹ kg, E=200 N/C and vx=10 m/s across L=0.020 m: ay=200 m/s², t=0.002 s, y=0.0004 m and vy=0.4 m/s. ΔV=−0.08 V, ΔU=−0.08 nJ and ΔK=+0.08 nJ.

Predict and investigate

If charge reverses sign but keeps its magnitude, which of displacement, potential change and kinetic-energy gain reverse?

Compare +1 and −1 nC at 200 N/C, then zero charge. Retain the trajectory endpoint and both energy changes. Double field with the same charge and check linear displacement but quadratic kinetic-energy gain.

Transfer to changed conditions

A hypothetical +2 nC particle of mass 2×10⁻⁹ kg crosses a 0.030 m region at vx=15 m/s in a +y field of 300 N/C, entering with vy=0. Find y and the energy changes.

Compare the worked transfer

t=0.002 s and ay=300 m/s², so y=0.0006 m and vy=0.6 m/s. ΔV=−0.18 V, ΔU=−0.36 nJ and ΔK=+0.36 nJ. A finite plate gap must exceed this deflection for the trajectory to remain applicable.

Explain the boundary

Explain why field lines are not generally trajectories, and why a negative charge gaining speed can move toward higher potential.

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. These cases, diagrams and numerical records are original. All experiments are bounded browser calculations or synthetic data, not measured laboratory evidence. Do not construct or handle charged capacitors, high-voltage sources or particle-beam equipment for these activities.