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103. (a) With
V =
1000 V, we solve
0
/4
,
Vq
R
πε
=
where
R
= 0.010 m for the net charge
on the sphere, and find
q
=
1.1
×
10
−
9
C.
Dividing this by
e
yields 6.95
×
10
9
electrons
that entered the copper sphere.
Now, half of the 3.7
×
10
8
decays per second resulted in
electrons entering the sphere, so the time required is
9
8
6.95 10
38 s
(3.7 10 / s) / 2
×
=
×
(b) We note that 100 keV is 1.6
×
10
−
14
J (per electron that entered the sphere).
Using the
given heat capacity, we note that a temperature increase of
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Charge

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