Sol Ass 4 250 ap

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					Chap. 24, Exercises
1, 2, 3,a, 3, b, 9, a. 32, a, b, c 38, a, 15
24.9. A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The
inner cylinder is negatively charged and the outer is positively charged; the magnitude of the
charge on each is 10.0 pC. The inner cylinder has radius 0.50 mm, the outer one has radius 5.00
mm, and the length of each cylinder is 18.0 cm. (a) What is the capacitance?
Exercises
24.1. A capacitor has a capacitance of 728 F What amount of charge must be placed on each
of its plates to make the potential difference between its plates equal to 25.0 V?
24.2. The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of
122 cm2  Each plate carries a charge of magnitude 435 102 8 C The plates are in vacuum. (a)
What is the capacitance? (b) What is the potential difference between the plates? (c) What is the
magnitude of the electric field between the plates?
24.3. A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0148 C
on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the
plates? (b) What is the area of each plate?
24.32. For the capacitor network shown in Fig. 24.28, the potential difference across ab is 36 V.
Find (a) the total charge stored in this network; (b) the charge on each capacitor; (c) the total
energy stored in the network; (d) the energy stored in each capacitor; (e) the potential differences
across each capacitor.
24.38. A parallel-plate capacitor has capacitance C0  500 pF when there is air between the
plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of
charge Q that can be placed on each plate if the electric field in the region between the plates is
not to exceed 300 104 V/m? (b) A dielectric with K  270 is inserted between the plates of
the capacitor, completely filling the volume between the plates. Now what is the maximum
magnitude of charge on each plate if the electric field between the plates is not to exceed
300 104 V/m?
24.15. In Fig. 24.25, each capacitor has C  400 F and Vab  1 280 V Calculate (a) the
charge on each capacitor; (b) the potential difference across each capacitor; (c) the potential
difference between points a and d.
Figure 24.25 Exercise 24.15.
                              Q
24.1.    IDENTIFY:       C
                              Vab
         SET UP: 1  F  106 F
         EXECUTE: Q  CVab  (7.28 106 F)(25.0 V)  1.82 104 C  182 C
         EVALUATE: One plate has charge Q and the other has charge Q .
                                       PA        Q
24.2.    IDENTIFY and SET UP: C  0 , C           and V  Ed .
                                        d        V
                    A     0.00122 m 2
         (a) C  P  P
                  0     0              3.29 pF
                    d     0.00328 m
                 Q 4.35  108 C
         (b) V                    13.2 kV
                 C 3.29  1012 F
                 V 13.2  103 V
         (c) E                   4.02  106 V/m
                 d    0.00328 m
         EVALUATE: The electric field is uniform between the plates, at points that aren't close to the edges.
24.3.    IDENTIFY and SET UP: It is a parallel-plate air capacitor, so we can apply the equations of Sections 24.1.
                               Q            Q 0.148 106 C
         EXECUTE: (a) C          so Vab                    604 V
                              Vab           C 245 1012 F

         (b) C 
                  PA
                   0
                      so A 
                             Cd
                                 
                                                     
                                    245  1012 F 0.328 103 m        
                                                                    9.08  103 m2  90.8 cm2
                   d          P0       8.854  1012 C2 / N  m 2
                             V         604 V
         (c) Vab  Ed so E  ab                   1.84  106 V/m
                              d    0.328  103 m
                   
         (d) E        so   EP  1.84  106 V/m  8.854  1012 C2 / N  m 2   1.63  105 C/m 2
                                0
                   P
                   0
                                                                              Q 0.148  106 C
         EVALUATE: We could also calculate  directly as Q/A.                                 1.63  105 C/m 2 , which
                                                                              A 9.08  103 m 2
         checks.
24.9.    IDENTIFY: Apply the results of Example 24.4. C  Q / V .
         SET UP:       ra  0.50 mm , rb  5.00 mm
                                     L2 P      (0.180 m)2 P
         EXECUTE: (a) C                   0
                                                           0
                                                               4.35 1012 F .
                                    ln(rb ra ) ln(5.00 0.50)
         (b) V  Q / C  (10.0 1012 C) /(4.35 1012 F)  2.30 V
                         C
         EVALUATE:          24.2 pF . This value is similar to those in Example 24.4. The capacitance is determined
                         L
         entirely by the dimensions of the cylinders.
                                                          1      1   1          Q
24.32.   IDENTIFY: The two capacitors are in series.                   C  . U  1 CV 2 .
                                                                                         2
                                                        Ceq C1 C2               V
         SET UP: For capacitors in series the voltages add and the charges are the same.
                        1     1     1              CC      (150 nF)(120 nF)
         EXECUTE: (a)                so Ceq  1 2                          66.7 nF .
                       Ceq C1 C2                 C1  C2 150 nF  120 nF
         Q  CV  (66.7 nF)(36 V)  2.4 106 C  2.4 C
         (b) Q  2.4 C for each capacitor.
24.38.   IDENTIFY: V  Ed and C  Q / V . With the dielectric present, C  KC0 .
         SET UP: V  Ed holds both with and without the dielectric.
         EXECUTE: (a) V  Ed  (3.00 104 V/m)(1.50  103 m)  45.0 V .
         Q  C0V  (5.00  1012 F)(45.0 V)  2.25 1010 C .
         (b) With the dielectric, C  KC0  (2.70)(5.00 pF)  13.5 pF . V is still 45.0 V, so
         Q  CV  (13.5 1012 F)(45.0 V)  6.08 1010 C .
         EVALUATE: The presence of the dielectric increases the amount of charge that can be stored for a given
         potential difference and electric field between the plates. Q increases by a factor of K.


14:
                                                                            1  1  1                  1   1   1
(b) C1  15 pF is in series with C23  20 pF . For capacitors in series,                    so             and
                                                                           Ceq C1 C2                C123 C1 C23
                   24.14. IDENTIFY: The capacitors between b and c are in parallel. This combination is in series with
         the 15 pF capacitor.
                 CC       (15 pF)(20 pF)
          C123  1 23                    8.6 pF .
                C1  C23 15 pF  20 pF
         EVALUATE: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors.
         For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors.
24.15.   IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents. In each equivalent
         network apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work
         back to the original circuit.
         SET UP: Do parts (a) and (b) together. The capacitor network is drawn in Figure 24.15a.




                                                                                C1  C2  C3  C4  400 F
                                                                                Vab  28.0 V



                                     Figure 24.15a
         EXECUTE: Simplify the circuit by replacing the capacitor combinations by their equivalents: C1 and C2 are in
         series and are equivalent to C12 (Figure 24.15b).
                                                                                    1  1  1
                                                                                       
                                                                                   C12 C1 C2
                                           Figure 24.15b

          C12 
                 C1C2
                        
                           4.00 106 F 4.00 106 F  2.00 106 F
                C1  C2     4.00  106 F  4.00  106 F
          C12 and C3 are in parallel and are equivalent to C123 (Figure 24.15c).
                                                                      C123  C12  C3
                                                                      C123  2.00 106 F  4.00 106 F
                                                                      C123  6.00  106 F
                                 Figure 24.15c
          C123 and C4 are in series and are equivalent to C1234 (Figure 24.15d).

                                                                                 1     1    1
                                                                                         
                                                                                C1234 C123 C4

                                          Figure 24.15d

          C1234 
                     C123C4
                              
                                 6.00 106 F 4.00 106 F  2.40 106 F
                    C123  C4    6.00  106 F  4.00  106 F
The circuit is equivalent to the circuit shown in Figure 24.15e.
                                            V1234  V  28.0 V
                                            Q1234  C1234V   2.40  106 F   28.0 V   67.2  C

              Figure 24.15e
Now build back up the original circuit, step by step. C1234 represents C123 and C4 in series (Figure 24.15f).

                                                    Q123  Q4  Q1234  67.2 C
                                                   (charge same for capacitors in series)

                    Figure 24.15f
            Q123 67.2 C
Then V123                   11.2 V
            C123 6.00 F
     Q 67.2 C
V4  4               16.8 V
     C4 4.00 F
Note that V4  V123  16.8 V  11.2 V  28.0 V, as it should.
Next consider the circuit as written in Figure 24.15g.
                                                                V3  V12  28.0 V  V4
                                                                V3  11.2 V
                                                                Q3  C3V3   4.00  F11.2 V 
                                                                Q3  44.8 C
                                                                Q12  C12V12   2.00 F11.2 V 
                                                                Q12  22.4 C
                    Figure 24.15g
Finally, consider the original circuit, as shown in Figure 24.15h.

                                                              Q1  Q2  Q12  22.4 C
                                                              (charge same for capacitors in series)
                                                                   Q 22.4 C
                                                              V1  1             5.6 V
                                                                   C1 4.00 F
                                                                   Q 22.4 C
                                                              V2  2              5.6 V
                                                                   C2 4.00  F

                    Figure 24.15h
Note that V1  V2  11.2 V, which equals V3 as it should.
Summary: Q1  22.4 C, V1  5.6 V
Q2  22.4 C, V2  5.6 V
Q3  44.8 C, V3  11.2 V
Q4  67.2 C, V4  16.8 V
(c) Vad  V3  11.2 V
EVALUATE:      V1  V2  V4  V , or V3  V4  V . Q1  Q2 , Q1  Q3  Q4 and Q4  Q1234 .

				
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