P25_087 - 87. (First problem of Cluster) (a) The field...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 87. (First problem of Cluster) (a) The field between the plates is uniform; we apply Eq. 25-42 to find the magnitude of the (horizontal) field: |E | = ∆V /D (assuming ∆V > 0). This produces a horizontal acceleration from Eq. 23-1 and Newton’s second law (applied along the x axis): ax = | Fx | q |E | q ∆V = = m m mD where q > 0 has been assumed; the problem indicates that the acceleration is rightward, which constitutes our choice for the +x direction. If we choose upward as the +y direction then ay = −g , and we apply the free-fall equations of Chapter 2 to the y motion while applying the constant (ax ) acceleration equations of Table 2-1 to the x motion. The displacement is defined by ∆x = +D/2 and ∆y = −d, and the initial velocity is zero. Simultaneous solution of ∆x = ∆y = 1 ax t 2 2 1 v0y t + ay t2 , 2 v0x t + leads to d= and gD gmD2 = . 2 ax 2q ∆V (b) We can continue along the same lines as in part (a) (using Table 2-1) to find v , or we can use energy conservation – which we feel is more instructive. The gain in kinetic energy derives from two potential energy changes: from gravity comes mgd and from electric potential energy comes q |E |∆x = q ∆V /2. Consequently, 1 1 mv 2 = mgd + q ∆V 2 2 which (upon using the expression for d above) yields v= mg 2 D 2 q ∆V + q ∆V m . (c) and (d) Using SI units (so q = 1.0 × 10−10 C, m = 1.0 × 10−9 kg) we plug into our results to obtain d = 0.049 m and v = 1.4 m/s. ...
View Full Document

Ask a homework question - tutors are online