1
Class Examples for Chapter 23
8.
(a) The total surface area bounding the bathroom is
2
2 2.5 3.0
2 3.0 2.0
2 2.0 2.5
37 m .
A
The absolute value of the total electric flux, with the assumptions stated in the problem, is
2
3
2

 
 

(600 N/C)(37 m )
22 10 N m /C.
E A
E A
By Gauss’ law, we conclude that the enclosed charge (in absolute value) is
7
enc
0




2.0 10 C.
q
Therefore, with volume
V
= 15 m
3
, and recognizing that we
are dealing with negative charges, the charge density is
7
83
enc
3

2.0 10 C
1.3 10 C/m .
15 m
q
V
(b) We find (
q
enc
/
e
)/
V
= (2.0
10
–7
C/1.6
10
–19
C)/15 m
3
= 8.2
10
10
excess electrons
per cubic meter.
14.
Equation 236 (Gauss’ law) gives
q
enc
.
(a) Thus, the value
52
2.0 10 N m /C
for small
r
leads to
12
2
2
5
2
6
6
central
0
(8.85 10
C /N m )(2.0 10 N m /C) 1.77 10 C 1.8 10 C
q
(b) The next value that
takes is
4.0 10 N m /C
, which implies that
6
enc
3.54 10 C.
q
But we have already accounted for some of that charge in part (a), so
the result for part (b) is
q
A
=
q
enc
–
q
central
= – 5.3
10
6
C.
(c) Finally, the large
r
value for
is
6.0 10 N m /C
, which implies that
6
total enc
5.31 10 C.
q
Considering what we have already found, then the result is
total enc
central
8.9
.
A
q
q
q
C
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17.
(a) The charge on the surface of the sphere is the product of the surface charge
density
and the surface area of the sphere (which is
2
4,
r
where
r
is the radius). Thus,
2
2
6
2
5
m
4
4
8.1 10 C/m
3.7 10 C.
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 Fall '08
 IASHVILI
 Charge, Electrostatics, Electric charge, Surface

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