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PHYS 2421
Exam 1
Spring 2010
NAME___________________________________
Problem 1 (10 points). Determine the point where the total electric field is
zero
P. 21.44 Solution. By symmetry it has to be between the charges, measuring
q
q2
q1
r1 from the left charge: E k 12 k
d r1 
2 0  r1 
r1
q2
d r1 


d q1 / q2
1.2m 0.5 / 0.8
r1 d q1 / q2 r1 q1 / q2  r1 1  q1 / q2 d q1 / q2  r1 

0.53m
1  q1 / q2
1  0.5 / 0.8
Problem 2 (10 points). A hollow conducting sphere is charged up until its surface
charge density is + 6.37 106 C / m2 . Then a point charge of q 0.5C is
introduced at the center of the sphere as shown.
A) What is the new charge density on the outside surface of the sphere?
B) What is the magnitude of the electric field just outside the sphere?
C) What is the electric flux through a spherical surface just inside the inner surface
of the sphere?
Probl. 22.23 Solution:
2
A) Total charge in hollow sphere is: QTot 4r2  6.37 106 C / m2 4
0.25m 5 10 6 C , since on


the inside surface there are q charges, on the outside there will be QOut QTot q , and the new
Q q 5C 0.5C
charge density will be Out  Tot 2 
5.73 106 C / m2
2 
4r
4
0.25m
Q q
5C 0.5C
B) Gauss’ law: E  Tot 2 
6.474 105 N / C
2
12 2
2
40 r
4
8.85 10 C / N 
m 
0.25m 
  QInside
q
0.5 C
C) The electric flux is 
E
dA 
 
5.65 104 Nm 2 / C
0
0 8.851012 C 2 / N 
m2
Problem 3 (10 points). The electric field at the surface of a charged, solid copper
sphere is 3800 N/C directed towards the center of the sphere.
A) What is the potential at a point 0.175 m from the center of the sphere?
B) What is the potential at point B 0.1 m from the center of the sphere?
P. 23.44 Solution.
A) Since the point 0.175 is inside the conducting sphere, its potential is equal to the
kQ
potential at outside surface of the sphere: V  . But since the field at the surface is
R
kQ
E  2  V ER 
3800 N / C 
0.2m760 V , and it is negative because the
R
charge on the sphere is negative.
B) Since the field in the hole of the sphere is zero (there are no charges inside), the potential inside the hole
is the same as the potential inside the conductor, i.e. V 760 V .
Problem 4 (10 points). A dipole consisting of two charges of magnitude
q 1.6 10 19 C and 220 nm apart is placed between two large sheets with
equal and opposite charge densities of 125 C / m 2 .
A) What is the maximum potential energy this dipole can have?
B) What is the maximum torque the sheets can exert on the dipole and
how it should be oriented relative to the sheets to attain this value?
Probl. 21.70 
Solution.

A) U p 
E pE cos  U Max pE qdE when the dipole makes
qd
an angle of 180o respect to the field. But E / 0  U Max qdE 
0
or U Max
1.6 10


19
C
220 10 m 125C / m 4.97 10

9
8.85 10
12
2
C /N
m
2
2

19
Nm or Joules.

 

B) 
p E   pE sin 




19
9
2

qd 1.6 10 C 220 10 m 125C / m
 Max pE 

4.97 1019 Nm . This happens when
0
8.85 10 12 C 2 / N 
m2
sin 1  90 , the dipole should be aligned parallel to the sheets, i.e. perpendicular to the electric field.
0