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

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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. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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