10. Strategy
Solution
Use the principle of superposition and Eq. (17-9).
Sum the electric fields at the center due to each charge.
k q
kq 
E  Ea  Eb  Ec  Ed  Ea  Eb  Ec  Eb  Ea  Ec   2a  2c  toward c
 r
r 

8.988 109 N  m 2 C2

(9.0 106 C  3.0 106 C) toward c  5.4 108 N C toward c
2
 (0.020 m)2 (0.020 m) 2 
2




Do the same for the potential at the center.
V 
kQi k
 (3q  q  q  q)  0
ri
r
17. Strategy Use Eq. (17-9).
Solution
V 
Find the electric potential at the third corner, B.
kQi k
8.988 109 N  m2 C2
 (QA  QB ) 
(2.0 109 C  1.0 109 C)  9.0 V
ri
r
1.0 m
22. (a) Strategy
Solution
Use Eq. (16-4b).
Find the electric force that acts on the particle.
F  qE  (4.2 109 C)(240 N C to the right)  1.0 to the right
(b) Strategy The work done on the particle is equal to the electric force times the
displacement of the particle.
Solution
Compute the work done on the particle.
W  Fd  (1.0 N)(0.25 m)  0.25 J
(c) Strategy The electric field points in the direction of decreasing potential, so Va  Vb and
Va  Vb  0. Use Eq. (17-10).
Solution
Compute the potential difference.
Va  Vb  Ed  (240 N C)(0.25 m)  60 V
26.Strategy
Equipotential surfaces are perpendicular to electric field lines at all points. For
equipotential surfaces drawn such that the potential difference between adjacent surfaces is
constant, the surfaces are closer together where the field is stronger. The electric field always
points in the direction of maximum potential decrease.
Solution
Outside the cylinder, E is radially directed away from the axis of the cylinder. The
equipotential surfaces are perpendicular to E at any point, so they are
28.(a) Strategy
cylinders .
Equipotential surfaces are perpendicular to electric field lines at all points. For
equipotential surfaces drawn such that the potential difference between adjacent surfaces is
constant, the surfaces are closer together where the field is stronger. The electric field
always points in the direction of maximum potential decrease.
Solution
The electric field lines are radial. They begin on the point charge and end on the
inner surface of the shell. Then they begin again on the surface and extend to infinity.
(b) Strategy
Solution
Use Eqs. (16-5) and (17-8), and the principle of superposition.
For r  r1, E is that due to the point charge, E  kq r 2 . For
r1  r  r2 , E  0, since this is inside a conductor. For r  r2 , E once again is that due to
the point charge, kq r 2 . For r  r1, V  kq r (point charge). For r1  r  r2 , V  kq r1 ,
since V is continuous, and it is constant in a conductor. For r  r2 ,
V
kq  kq kq 
  
r1  r
r2 
(to preserve continuity). The graphs of the electric field magnitude and potential:
34. Strategy The electron must have enough kinetic energy at point A to overcome the potential
decrease between A and C. Use conservation of energy and Eq. (17-7).
Solution
Find the required kinetic energy.
K A  U  eV  (1.602  1019 C)(  60.0 V  100.0 V)  2.56 1017 J
35.Strategy and Solution
Since positive charges move through decreases in potential, and since the
potential and potential energy are greatest at A, the proton will spontaneously travel from point
A to point E. So, K A  0 .
37.Strategy and Solution
(a) Electrons travel opposite the direction of the electric field, so E is directed
(b) For a uniform electric field, Fy  eE 
t 
vy
ay

v y md
eV
eV
eV
 ma y , so a y 
. Thus,
d
md
.
(c) Since the electron gains kinetic energy, its potential energy
48.Strategy
upward .
decreases .
Use the definition of electric flux and Gauss’s law.
Solution
(a) The Gaussian surface is a cylinder whose axis is parallel to a radius vector (of the sphere)
through it. One end is just within the conductor and the other is just outside it. The ends
have area A, with a radius much smaller than that of the sphere, so the electric field is
approximately uniform. Find E just outside the conductor.
E  EA cos  
Q
e0
Cylindrical surface: E  EA cos90  EA(0)  0
End inside conductor: E  EA cos  (0) A cos  0, since the electric field is zero
inside a conductor.
End just outside the conductor: E  EA cos 0  EA 
Q
e0
, so E 
Q

 .
Ae0 e0
(b) Consider an area A of the surface of an arbitrary conductor. If A is small enough such that
its surface is approximately flat, then the electric field will be nearly uniform just outside
the surface. Comparing an area of the same size on a spherical conductor with the same
charge density to that of the arbitrary conductor, we see that the electric field just outside
either conductor should be  e0 ; as long as A is small enough that it is approximately
flat, then this holds for any conductor.
56.Strategy
Use Eq. (17-16).
Solution
Find the average dielectric constant of the tissue in the limb.
eA
dC
(0.030 m)(0.59 1012 F)
C   0 , so  

 5.0 .
d
e0 A [8.854 1012 C2 (N  m 2 )](4.0 104 m 2 )
62. (a) Strategy The capacitance after the slab is removed is equal to the capacitance with the
slab divided by the dielectric constant.
Solution
C0 
C


(b) Strategy
Solution
Compute the capacitance.
6.0 F
 2.0 F
3.0
Use Eqs. (17-10) and (17-17).
Find the potential difference across the capacitor.
E0   E
E0d   Ed
V0   V  3.0(1.5 V)  4.5 V
(c) Strategy
Use the definition of capacitance, Eq. (17-14).
Solution
Compute the charge on the plates.
Q  CV  (2.0 F)(4.5 V)  9.0 C
(d) Strategy
Solution
U
Use Eq. (17-18c).
Compute the energy stored in the capacitor.
Q
(9.0 106 C)2

 20 J
2C 2(2.0 106 F)
2