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Physics 4B FINAL EXAM Chapters 23  34 Spring
1995
Name:_________________________________________
Posting Code _____ _____ _____
Solve the following problems in the space provided. Use the back of the page if needed. Each problem is
worth 10 points. You must
show your work in a logical fashion starting with the correctly applied physical
principles which are on the last page. Your score will be maximized if your work is easy to follow because
partial credit will be awarded.
1. The dipole shown at the right consists of two equal but opposite 5.00μC charges separated by 2.00cm.
Find the electric field 4.00cm from the center of the dipole
perpendicular to the dipole axis.
The field due to a point charge the point charges are,
r
E
=
k
q
2
r
ˆ
r
!
E
+
=
E
"
=
k
q
2
r
The horizontal components will cancel leaving only the
vertical components,
E
=
E
+
sin
!
+
E
"
sin
!
=
2k
q
2
r
sin
!
where
r
=
4
2
+
1
2
=
4.12cm
and
sin
!
=
1
4.12
=
0.243
. Finally,
E
=
2(9x10
9
)
5x10
!
6
0.0412
2
(0.243)
=
1.29x10
7
N / C
2. A solid metal sphere of radius 5.00cm creates an electric field of 2.25x10
6
N/C at a distance of 20.0cm from
its center. Describe the charge distribution on the sphere.
Applying Gauss's Law to an imaginary sphere 20.0cm in radius,
r
E
•
d
r
A
!
=
q
"
o
.
By the spherical symmetry of the problem, the field must be radial and constant on this imaginary sphere so,
EA
=
q
!
o
"
q
=
!
o
EA
=
!
o
E4
#
r
2
=
(8.85x10
$
12
)(2.25x10
6
)4
#
(0.200)
2
.
The charge on the sphere is,
q
=
10.0
μ
C
.
Using Gauss's Law and the same symmetry arguments for an imaginary sphere inside the metal sphere,
EA
=
q
!
o
"
q
=
!
o
EA
. However, the field inside a conductor is always zero, so the charge inside is also zero.
Therefor, all the charge on the sphere is on the surface.
5.00μC
1.00cm
4.00cm
+5.00μC
1.00cm
E
+
E

E
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3. Find the potential due to a 20.0cm long rod that has a linear charge density of 2.00μC/m at a point 3.00cm
from the end of the rod along the same axis as the rod.
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