A Point Charge in an Electric Field — Question 10

Section 22-7 · Physics pages 3610–3611

Question 10 A rotating prescribed field can return velocity to zero while transporting a particle A particle of mass m and charge q > 0 starts from rest at the origin in a spatially uniform prescribed field E(t) = E0(cos ωt, sin ωt), with ω > 0. Assume the electric force dominates and motion remains nonrelativistic. Define a0 = qE0/m. E(t) Schematic; labels, not drawing lengths, specify the data. (a) Integrate to obtain both velocity and position components. (b) Evaluate them after one period T = 2π/ω and check the net work from the kinetic-energy change.
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Question 10 - Solution A rotating prescribed field can return velocity to zero while transporting a particle E(t) (a) Method and calculation vx = a0 ω sin ωt, vy = a0 ω (1 −cos ωt). x = a0 ω2 (1 −cos ωt), y = a0 ω2 (ωt −sin ωt). (b) Interpretation and further calculation v(T) = 0, r(T) = ( 0, 2πa0 ω2 ) . W0→T = ∆K = 0. The velocity vector completes a cycle, while its nonzero time average produces a net displacement. Physical check The source driving this time-dependent field is external; the calculation concerns mechanical work on this particle over one complete pulse.

Question 10

A rotating prescribed field can return velocity to zero while transporting a particle

A particle of mass m and charge q>0q>0 starts from rest at the origin in a spatially uniform prescribed field 𝑬(t)=E0(cos⁡ωt,sin⁡ωt)\mathbf E(t)=E_0(\cos\omega t,\sin\omega t), with ω>0\omega>0. Assume the electric force dominates and motion remains nonrelativistic. Define a0=qE0/ma_0=qE_0/m.

Diagram shown in the original SVG worksheet.

  1. Integrate to obtain both velocity and position components.

  2. Evaluate them after one period T=2π/ωT=2\pi/\omega and check the net work from the kinetic-energy change.

Question 10 - Solution

A rotating prescribed field can return velocity to zero while transporting a particle

Diagram shown in the original SVG worksheet.

(a) Method and calculation

vx=a0ωsin⁡ωt,vy=a0ω(1−cos⁡ωt).v_x=\frac{a_0}{\omega}\sin\omega t,\quad v_y=\frac{a_0}{\omega}(1-\cos\omega t).x=a0ω2(1−cos⁡ωt),y=a0ω2(ωt−sin⁡ωt).x=\frac{a_0}{\omega^2}(1-\cos\omega t),\quad y=\frac{a_0}{\omega^2}(\omega t-\sin\omega t).

(b) Interpretation and further calculation

𝒗(T)=𝟎,𝒓(T)=(0,2πa0ω2).\mathbf v(T)=\mathbf0,\quad\mathbf r(T)=\left(0,\frac{2\pi a_0}{\omega^2}\right).W0→T=ΔK=0.W_{0\to T}=\Delta K=0.The velocity vector completes a cycle, while its nonzero time average produces a net displacement.

Physical check

The source driving this time-dependent field is external; the calculation concerns mechanical work on this particle over one complete pulse.