4.3. IDENTIFY and SET UP: It is a parallel-plate air capacitor, so we can apply the equations of Section 4.. EXEUTE: (a) (b) = ε 0 A d (c) V ab so Q V = so 0 ab V ab 6 Q 0. 48 0 = = = 604 V. 45 0 F 3 d (45 0 F)(0. 38 0 m) 3 A = = = 9. 08 0 m = 90. 8 cm. ε 8. 854 0 /N m V ab 604 V 6 = Ed so E = = =. 84 0 V/m. d 3 0. 38 0 m (d) E = σ 6 5 so σ = Eε =.. =. ε 0 0 ( 84 0 V/m)(8 854 0 /N m ) 63 0 /m. EVALUATE: We could also calculate σ directly as Q/A. 6 0. 48 0 5 3 63 0 /m, Q σ = = =. A 9. 08 0 m which checks. 4.9. IDENTIFY: Apply the results of Example 4.4. = Q/ V. 0 50 mm, SET UP: r a =. r b = 5. 00 mm. EXEUTE: (a) = Lπε 0 ln(r b /r a ) = (0.80 m)πε 0 = 4.35 0 F. ln(5.00/0.50) (b) V Q = / = (0. 0 0 )/(4. 35 0 F) =. 30 V. EVALUATE: 4 pf. L =. This value is similar to those in Example 4.4. The capacitance is determined entirely by the dimensions of the cylinders. 4.. IDENTIFY: We can use the definition of capacitance to find the capacitance of the capacitor, and then relate the capacitance to geometry to find the inner radius. (a) SET UP: By the definition of capacitance, = Q/ V. EXEUTE: = Q V = 3.30 09.0 0 V =.50 0 F = 5.0 pf. (b) SET UP: The capacitance of a spherical capacitor is = 4πε 0 EXEUTE: Solve for r a and evaluate using r a = 3.09 cm. r a r b r b r a. = 5.0 pf and = 4.00 cm, r giving (c) SET UP: We can treat the inner sphere as a point charge located at its center and use oulomb s law, E = EXEUTE: 4πε 0 q r. E = (8.99 09 N m / )(3.30 0 9 ) (0.0309 m) = 3. 0 4 N/. b
EVALUATE: Outside the capacitor, the electric field is zero because the charges on the spheres are equal in magnitude but opposite in sign. 4.7. 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 4.7a. = = 3 = 4 = 4. 00 µ F. V = 8. 0 V. ab Figure 4.7a EXEUTE: Simplify the circuit by replacing the capacitor combinations by their equivalents: and are in series and are equivalent to (Figure 4.7b). Figure 4.7b = +. 6 6.. 6 6 +. +. (4 00 0 F)(4 00 0 F) 6 = = =. 00 0 F. 4 00 0 F 4 00 0 F and 3 are in parallel and are equivalent to 3 (Figure 4.7c). Figure 4.7c 3 = + 3. 6 6 3 =. 00 0 F + 4. 00 0 F. 6 3 = 6. 00 0 F. 3 and 4 are in series and are equivalent to 34 (Figure 4.7d).
= +. 34 3 4 Figure 4.7d 6 6 34.. 34 6 6 3 + 4. +. (6 00 0 F)(4 00 0 F) 6 = = =. 40 0 F. 6 00 0 F 4 00 0 F The circuit is equivalent to the circuit shown in Figure 4.7e. V34 = V = 8. 0 V. 6 Q34 = 34 V = (. 40 0 F)(8. 0 V) = 67. µ. Figure 4.7e Now build back up the original circuit, step by step. 34 represents 3 and 4 in series (Figure 4.7f). Figure 4.7f Q3 = Q4 = Q 34 = 67. µ (charge same for capacitors in series). Q 67. µ 3 = = =. V. 3 6. 00 µ F Q4 67. µ V4 = = = 6. 8 V. 4. 00 µ F 3 Then V 4 Note that V 4 + V 3 = 6. 8 V +. V = 8. 0 V, as it should. Next consider the circuit as written in Figure 4.7g (next page). Figure 4.7g Q V µ F)( (. 00. V). = = V3 = V = 8. 0 V V4. V 3 =. V. Q3= V 3 3 = (4. 00 µ F)(. V). Q 3 = 44. 8 µ. Q =. 4 µ.
Finally, consider the original circuit, as shown in Figure 4.7h. Figure 4.7h Q = Q = Q =. 4 µ (charge same for capacitors in series). Q. 4 µ V = = = 5. 6 V. 4. 00 µ F Q. 4 µ V = = = 5. 6 V. 4. 00 µ F Note that V + V =. V, which equals V 3 as it should. Summary: Q =. 4 µ, V = 5. 6 V. Q =. 4 µ, V = 5. 6 V. Q = 44. 8 µ, V =. V. 3 3 Q = 67. µ, V = 6. 8 V. 4 4 (c) V = V3 =. V. ad EVALUATE: V + V + V4 = V V3 + V4 = V. Q = Q Q + Q3 = Q4 Q4 = Q34, or, and. 4.7. IDENTIFY: Use the rules for series and for parallel capacitors to express the voltage for each capacitor in terms of the applied voltage. Express U, Q, and E in terms of the capacitor voltage. SET UP: Let the applied voltage be V. Let each capacitor have capacitance. U = V for a single capacitor with voltage V. EXEUTE: (a) Series: The voltage across each capacitor is V /. The total energy stored is U s = ( ( V /) ) = V. 4 Parallel: The voltage across each capacitor is V. The total energy stored is p U = ( V ) = V Up = 4 Us. V (b) Q = for a single capacitor with voltage V. Qs = [ ( V/)] = V; Qp = ( V ) = V; Qp = Qs. E V d (c) = / for a capacitor with voltage V. E s = V/ d; E p = V/ d; E p = E s. EVALUATE: The parallel combination stores more energy and more charge since the voltage for each capacitor is larger for parallel. More energy stored and larger voltage for parallel means larger electric field in the parallel case.
4.3. IDENTIFY: U = QV. Solve for Q. = Q/ V. SET UP: Example 4.4 shows that for a cylindrical capacitor, L = πε 0 ln(r b /r a ). U = EXEUTE: (a) (b) a QV gives L = πε 0 ln(r b /r a ). Solving for r b /r a gives 9 U (3. 0 0 J) 9 Q = = =. 60 0. V 4. 00 V rb 9 = exp( πε0l/ ) = exp( πε0 LV / Q) = exp[ πε0(5. 0 m)(4. 00 V)/(. 60 0 )] = 8. 05. r The radius of the outer conductor is 8.05 times the radius of the inner conductor. EVALUATE: When the ratio rb / r a increases, / L decreases and less charge is stored for a given potential difference. 4.37. IDENTIFY and SET UP: For a parallel-plate capacitor with a dielectric we can use the equation = Kε 0 A/d. Minimum A means smallest possible d. d is 7 limited by the requirement that E be less than. 60 0 V/m when V is as large as 5500 V. EXEUTE: V 5500 V 4 V = Ed so d = = = 3. 44 0 m. E 7. 60 0 V/m Then A = d = (.5 09 F)(3.44 0 4 m) Kε 0 (3.60)(8.854 0 /N m ) = 0.035 m. EVALUATE: The relation V = Ed applies with or without a dielectric present. A would have to be larger if there were no dielectric. Q/ V. 4.39. IDENTIFY: = = K0. V = Ed. SET UP: Table 4. gives K = 3. for mylar. 7 6 EXEUTE: (a) Q = Q Q = ( K ) Q = ( K ) V = (. )(. 5 0 F)( V) = 6. 3 0. 0 0 0 0 6 6 (b) σi = σ ( / K) so Qi = Q( / K) = (9. 3 0 )( /3. ) = 6. 3 0. (c) The addition of the mylar doesn t affect the electric field since the induced charge cancels the additional charge drawn to the plates. EVALUATE: E = V/ d and V is constant so E doesn t change when the dielectric is inserted. 4.45. IDENTIFY: P E/ t, in the capacitor is U = V = where E is the total light energy output. The energy stored.
SET UP: E = 0. 95 U. EXEUTE: (a) The power output is 5.70 0 W, 5 3 energy is converted, so E Pt (b) U = V so and 95% of the original = = (. 70 0 W)(. 48 0 s) = 400 J. U (4 J) = = = 0. 054 F. V (5 V) 400 J U = = 4 J. 0. 95 EVALUATE: For a given V, the stored energy increases linearly with. 4.53. (a) IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents. SET UP: The network is sketched in Figure 4.53a. = 5 = 8. 4 µ F. = 3 = 4 = 4. µ F. Figure 4.53a EXEUTE: Simplify the circuit by replacing the capacitor combinations by their equivalents: 3 and 4 are in series and can be replaced by 34 (Figure 4.53b): Figure 4.53b = +. 34 3 4 + = 3 4 34 3 4. 34 (4. µ F)(4. µ F) = = =. µ F. + 4. µ F+ 4. µ F 3 4 3 4 and 34 are in parallel and can be replaced by their equivalent (Figure 4.53c): 34 = + 34. 34 = 4. µ F+. µ F. 34 = 6. 3 µ F.
Figure 4.53c, 5, and 34 are in series and can be replaced by eq (Figure 4.53d): Figure 4.53d = + +. eq 5 34 = +. 84. µ F 63. µ F eq eq =. 5 µ F. EVALUATE: For capacitors in series the equivalent capacitor is smaller than any of those in series. For capacitors in parallel the equivalent capacitance is larger than any of those in parallel. (b) IDENTIFY and SET UP: 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. EXEUTE: The equivalent circuit is drawn in Figure 4.53e. Q eq = eq V. Q eq = (. 5 µ F)(0 V) = 550 µ. Figure 4.53e Q Q5 Q 34 550 µ Q 550 µ V = = = 65 V. 84. µ F Q5 550 µ V5 = = = 65 V. 5 84. µ F Q34 550 µ V34 = = = 87 V. 63. µ F = = = (capacitors in series have same charge). 34 Now draw the network as in Figure 4.53f.
V = V34 = V34 = 87 V capacitors in parallel have the same potential. Figure 4.53f Q = V = (4. µ F)(87 V) = 370 µ. Q34 = 34V 34 = (. µ F)(87 V) = 80 µ. Finally, consider the original circuit (Figure 4.53g). Q3 = Q4 = Q 34 = 80 µ capacitors in series have the same charge. Figure 4.53g Q3 80 µ V3 = = = 43 V. 4. µ F 3 Q4 80 µ V4 = = = 43 V. 4. µ F 4 Summary: Q µ V Q = 370 µ, V = 87 V. Q = 80 µ, V = 43 V. 3 3 Q = 80 µ, V = 43 V. 4 4 Q = 550 µ, V = 65 V. 5 5 = 550, = 65 V. V EVALUATE: 3 + V4 = V and V + V + V5 = 0 V (apart from some small rounding error) Q = Q + Q and Q = Q + Q. 3 5 4 4.59. 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. (a) SET UP: The network is sketched in Figure 4.59a.
= 6. 9 µ F. = 4. 6 µ F. Figure 4.59a EXEUTE: Simplify the network by replacing the capacitor combinations by their equivalents. Make the replacement shown in Figure 4.59b. 3 =. eq Figure 4.59b eq 69. µ F = = =. 3 µ F. 3 3 Next make the replacement shown in Figure 4.59c. = 3. µ F +. eq eq = 3. µ F+ 46. µ F= 69. µ F. Figure 4.59c Make the replacement shown in Figure 4.59d. Figure 4.59d Make the replacement shown in Figure 4.59e. 3 = + =. 69. µ F 69. µ F eq eq =. 3 µ F. eq = + 3. µ F= 46. µ F+ 3. µ F. Figure 4.59e eq = 6. 9 µ F. Make the replacement shown in Figure 4.59f.
3 = + =. 69. µ F 69. µ F eq eq =. 3 µ F. Figure 4.59f (b) SET UP and EXEUTE: onsider the network as drawn in Figure 4.59g. Figure 4.59g The equivalent network is shown in Figure 4.59h. Figure 4.59h From part (a). 3 µ F is the equivalent capacitance of the rest of the network. The capacitors are in series, so all three capacitors have the same Q. But here all three have the same, so by V = Q/ all three must have the same V. The three voltages must add to 40 V, so each capacitor hasv = 40 V. The 6. 9 µ F to the right is the equivalent of and the. 3 µ F capacitor in parallel, so V = 40 V. (apacitors in parallel have the same potential 4 difference.) Hence Q V µ = = (6. 9 F)(40 V) = 9. 7 0 and 4 Q = V = (4. 6 µ F)(40 V) = 6. 4 0. (c) From the potentials deduced in part (b) we have the situation shown in Figure 4.59i.
From part (a) 6. 9 µ F is the equivalent capacitance of the rest of the network. Figure 4.59i The three right-most capacitors are in series and therefore have the same charge. But their capacitances are also equal, so by V = Q/ they each have the same potential difference. Their potentials must sum to40 V, so the potential across each is 47 V and V cd = 47 V. EVALUATE: In each capacitor network the rules for combining V for capacitors in series and parallel are obeyed. Note that V < V, in fact V (40 V) (47 V) = V cd. cd