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Unformatted text preview: 87. (First problem of Cluster)
(a) The ﬁeld between the plates is uniform; we apply Eq. 2542 to ﬁnd the magnitude of the (horizontal)
ﬁeld: E  = ∆V /D (assuming ∆V > 0). This produces a horizontal acceleration from Eq. 231 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 freefall equations of Chapter 2 to the y motion while applying the constant (ax )
acceleration equations of Table 21 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 21) 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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 Fall '08
 SPRUNGER
 Physics, Acceleration

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