117.
(a) With
V =
1000 V, we solve
V
=
q
4
πε
o
R
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
6.95
x
10
9
1
2
(3.7
x
10
8
)
=
38 seconds.
(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
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.
 Spring '10
 Rebello,NobelS
 Charge

Click to edit the document details