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Homework 3
Ch17: Q3, 15; P.1, 11, 19, 29, 37, 45, 51
Questions:
3.
State clearly the difference (a) between electric potential and electric field,
(b) between electric potential and electric potential energy.
Solution
(a) Electric potential, a scalar, is the electric potential energy per unit charge at a point in space.
Electric field, a vector, is the electric force per unit charge at a point in space.
(b) Electric potential energy is the work done against the electric force in moving a charge from
a specified location of zero potential energy to some other location. Electric potential is the
electric potential energy per unit charge.
15. We have seen that the capacitance C depends on the size, shape, and position of the
two conductors, as well as on the dielectric constant K. What then did we mean
when we said that C is a constant in Eq. 17 – 7?
Solution
We meant that the capacitance did not depend on the amount of charge stored or on the potential
difference between the capacitor plates. Changing the amount of charge stored or the potential
difference will not change the capacitance.
Problems:
1. (I) How much work does the electric field do in moving a 7.7 C charge from ground
to a point whose potential is 55 V higher?
Solution
The work done by the electric field can be found from Eq. 17-2b.
W
Vba   ba  Wba  qVba   7.7  106 C  55 V   4.2  10 4 J
q


11. (II) What is the speed of an electron with kinetic energy (a)750-eV, and (b) 3.2-keV?
Solution
The kinetic energy of the electron is given in each case. Use the kinetic energy to find the speed.
(a) mv  KE  v 
2KE
(b) mv  KE  v 
2KE
1
2
1
2
2
2
m
m



2  750 eV  1.60 1019 J eV
31
9.1110 kg


  1.6 10
2 3.2 103 eV 1.60 1019 J eV
31
9.1110 kg
7
ms
  3.4 10
7
ms
19.
(II) Three point charges are arranged at the corners of a square of side L as shown
in Fig. 17–25. What is the potential at the fourth corner (point A), taking V  0 at a great
distance?
Solution
The potential at the corner is the sum of the potentials due to each of the charges, using Eq. 17-5.
V
k  3Q 
L

kQ
2L

k  2Q 
L

kQ 
1 
1 

L 
2
2kQ
2L


2 1
*29. (II) Calculate the electric potential due to a dipole whose dipole moment is
4.8 1030 C  m at a point 1.110 9 m away if this point is (a) along the axis of the
dipole nearer the positive charge; (b) 45° above the axis but nearer the positive
charge; (c) 45° above the axis but nearer the negative charge.
Solution
The potential due to the dipole is given by Eq. 17-6b.
8.99  109 N m 2 C 2 4.8  10 30 C m cos 0
kp cos 

(a) V 
2
r2
1.1  109 m



 3.6  102 V
(b) V 
kp cos 
r
2


8.99  109 N m 2 C 2
(c) V 
r
2
8.99 10


9
9
r
Q
Q
r

4.8  1030 C m cos 45o
1.110 m 
 2.5  102 V
kp cos 


2
Q
 4.8 10
1.110 m 
N m 2 C2
9

2
30
Q

C m cos135o
r

2
 2.5  10 V
Q
Q
37.
(II) An electric field of 8.50 105 V m is desired between two parallel plates, each
of area 35.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?
Solution
The desired electric field is the value of E= V d for the capacitor. Combine Eq. 17-7 and
Eq. 17-8 to find the charge.
 AV
Q  CV  0
  0 AE  8.85  1012 C2 N  m 2 35.0  104 m 2 8.50  105 V m
d




 2.63  108 C
*45. (II) The electric field between the plates of a paper-separated ( K  3.75) capacitor is
8.24 104 V m . The plates are 1.95 mm apart, and the charge on each plate is
0.775 C. Determine the capacitance of this capacitor and the area of each plate.
Solution
The capacitance is found from Eq. 17-7, with the voltage given by Eq. 17-4 (ignoring the sign).
Q
0.775 106 C
Q  CV  C  Ed   C 

 4.82  109 F
Ed
8.24 104 V m 1.95 103 m



The plate area is found from Eq. 17-9.
4.82  109 F 1.95  10 3 m
A
Cd
C  K 0
 A

 0.283 m 2
d
K  0  3.75  8.85  1012 C 2 N  m 2





51. (II) How does the energy stored in a capacitor change if (a) the potential difference
is doubled, and (b) the charge on each plate is doubled, as the capacitor remains
connected to a battery?
Solution
(a) The energy stored in the capacitor is given by Eq. 17-10, PE  12 CV 2 . Assuming the
capacitance is constant, then if the potential difference is doubled, the stored energy is
multiplied by 4 .
(b) Now we assume the potential difference is constant, since the capacitor remains connected to
a battery. Then the energy stored in the capacitor is given by PE  12 QV , and so the stored
energy is multiplied by 2 . Note, in this situation, the capacitance of the capacitor is not a
constant (it is a capacitor with changeable geometry).