A Point Charge in an Electric Field — Question 8

Section 22-7 · Physics pages 3606–3607

Question 8 A particle can strike a plate even when its predicted exit height is zero A particle enters midway between parallel boundaries at y = ±h with upward velocity u > 0. Its constant vertical acceleration is −a with a > 0. The horizontal flight time through the field region is T, chosen so the unobstructed trajectory would have y(T) = 0. h Schematic; labels, not drawing lengths, specify the data. (a) Find a in terms of u,T, the time of maximum height and that height. (b) For u = 10.0 m/s, T = 2.00 ms and h = 4.00 mm, determine whether the particle clears the region. Find the first impact time if not.
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Question 8 - Solution A particle can strike a plate even when its predicted exit height is zero h (a) Method and calculation y(t) = ut −1 2at2, y(T) = 0 ⇒a = 2u/T. tmax = u/a = T/2, ymax = uT/4 = 5.00 mm > h. (b) Interpretation and further calculation a = 1.00 × 104 m/s2, ut −1 2at2 = h. thit = u − √ u2 −2ah a ≈.552786 ms. The particle hits the upper boundary before the predicted field-free exit; the later ideal trajectory is no longer physically realized. Physical check Clearance requires checking the entire path, including interior extrema, rather than only its endpoints.

Question 8

A particle can strike a plate even when its predicted exit height is zero

A particle enters midway between parallel boundaries at y=±hy=\pm h with upward velocity u>0u>0. Its constant vertical acceleration is −a-a with a>0a>0. The horizontal flight time through the field region is T, chosen so the unobstructed trajectory would have y(T)=0y(T)=0.

Diagram shown in the original SVG worksheet.

  1. Find a in terms of u,T, the time of maximum height and that height.

  2. For u=10.0u=10.0 m/s, T=2.00T=2.00 ms and h=4.00h=4.00 mm, determine whether the particle clears the region. Find the first impact time if not.

Question 8 - Solution

A particle can strike a plate even when its predicted exit height is zero

Diagram shown in the original SVG worksheet.

(a) Method and calculation

y(t)=ut−12at2,y(T)=0⇒a=2u/T.y(t)=ut-\tfrac12at^2,\quad y(T)=0\Rightarrow a=2u/T.tmax=u/a=T/2,ymax=uT/4=5.00mm>h.t_{\max}=u/a=T/2,\quad y_{\max}=uT/4=5.00\,\mathrm{mm}>h.

(b) Interpretation and further calculation

a=1.00×104m/s2,ut−12at2=h.a=1.00\times10^4\,\mathrm{m/s^2},\quad ut-\tfrac12at^2=h.thit=u−u2−2aha≈.552786ms.t_{\mathrm{hit}}=\frac{u-\sqrt{u^2-2ah}}a\approx.552786\,\mathrm{ms}.The particle hits the upper boundary before the predicted field-free exit; the later ideal trajectory is no longer physically realized.

Physical check

Clearance requires checking the entire path, including interior extrema, rather than only its endpoints.