Capacitance and Dielectrics

Transcription

1 6 Capacitance an Dielectrics CHAPTER OUTLINE 6. Definition of Capacitance 6. Calculating Capacitance 6.3 Combinations of Capacitors 6.4 Energy Store in a Charge Capacitor 6.5 Capacitors with Dielectrics 6.6 Electric Dipole in an Electric Fiel 6.7 An Atomic Description of Dielectrics * An asterisk inicates a question or problem new to this eition. ANSWERS TO OBJECTIVE QUESTIONS OQ6. (i) Answer (a). Because C κ 0 A an the ielectric constant κ increases. (ii) Answer (a). Because ΔV is constant, an C increases, so Q CΔV increases. (iii) Answer (c). (iv) Answer (a). Because ΔV is constant, an C increases, U E C( ΔV) increases. OQ6. Answer. The capacitance of a metal sphere is proportional to its raius (C Q/V R/k e ), an its volume is proportional to raius cube; therefore, the capacitance of a metal sphere is proportional to the cube root of the volume: 3 /3. 5

2 5 Capacitance an Dielectrics OQ6.3 Answer (a). C κ 0 A ( C /N m )( m ) m F or 88.5 pf OQ6.4 OQ6.5 OQ6.6 Answer (c). The voltage remains constant, but C ecreases by a factor of because C κ 0 A κ 0 A U E C( ΔV) C. Therefore, C ΔV U E Answer. Choice (a) is not true because /C eq is always larger than /C + /C + /C 3. Choice (c) is not true because capacitors in series carry the same charge Q, an the voltage across capacitance C i is ΔV i Q/C i. Choices () an (e) are not true because capacitors in series carry the same charge. Answer. Let C the capacitance of an iniviual capacitor, an C S represent the equivalent capacitance of the group in series. While being charge in parallel, each capacitor receives charge 800 V Q CΔV charge F While being ischarge in series, ΔV ischarge Q C s Q C 0 (or 0 times the original voltage). OQ6.7 (i) Answer, because Q CΔV C C 8.00 kv F (ii) Answer (a), because U E C( ΔV). OQ6.8 Answer (). Let C be the capacitance of the large capacitor an C that of the small one. The equivalent capacitance is C eq C + C C + C C C C C C + C This is slightly less than C.

3 Chapter 6 53 OQ6.9 OQ6.0 Answer (a). Charge Q remains fixe, but the capacitance oubles: C κ 0 A ( κ ) 0 A C. Therefore, ΔV Q/C Q/(C) ΔV /. (i) Answer (c). For capacitors in parallel, choices (a),, (), an (e) are not true because the potential ifference V is the same, an the charge across capacitance C i is Q i C i ΔV. (ii) Answer (e). Although the charges on capacitors in series are the same, the equivalent capacitance is less than the capacitance of any of the capacitors in the group, because /C eq is always larger than /C + /C + /C 3 ; therefore, choices (a) an (c) are not true. Choices, (c), an () are not true because the charge Q is the same, an choice (c) is also not true because the potential ifference across capacitance C i is ΔV i Q/C i. OQ6. Answer. The charge stays constant but C ecreases by a factor of because C κ 0 A κ 0 A C. Therefore, OQ6. U E Q C Q U E C We fin the capacitance, voltage, charge, an energy for each capacitor. (a) C 0 μf ΔV 4 V Q CΔV 80 μc U E QΔV 60 μj C 30 μf ΔV Q/C 3 V Q 90 μc U E 35 μj (c) C Q/ V 40 μf ΔV V Q 80 μc U E 80 μj () C 0 μf U E 5 μj ΔV (U/C) / 5 V Q 50 μc (e) C U E /( ΔV) 5 µf ΔV 0 V Q 50 μc U E 50 μj (i) The ranking by capacitance is c > b > a > > e. (ii) The ranking by voltage ΔV is e > > a > b > c. (iii) The ranking by charge Q is b > a c > e. (iv) The ranking by energy U E is e > a > b > > c.

4 54 Capacitance an Dielectrics OQ6.3 (a) False. True. The equation C Q/ΔV implies that as charge Q approaches zero, the voltage ΔV also approaches zero so that their ratio remains constant. OQ6.4 (i) Answer. Because C κ 0 A an the plate separation increases. (ii) Answer (c). (iii) Answer ( c). Because E Q/κ 0 A remains the same. (iv) Answer (a). Because ΔV E an increases. ANSWERS TO CONCEPTUAL QUESTIONS CQ6. CQ6. CQ6.3 (a) The capacitor may be charge! Discharge the capacitor by connecting its terminals together. Put a material with higher ielectric strength between the plates, or evacuate the space between the plates. At very high voltages, you may want to cool off the plates or choose to make them of a ifferent chemically stable material, because atoms in the plates themselves can ionize, showing thermionic emission uner high electric fiels. The primary choice woul be the ielectric. You woul want to choose a ielectric that has a large ielectric constant an ielectric strength, such as strontium titanate, where κ 33 (Table 6.). A convenient choice coul be thick plastic or Mylar. Seconly, geometry woul be a factor. To maximize capacitance, one woul want the iniviual plates as close as possible, since the capacitance is proportional to the inverse of the plate separation hence the nee for a ielectric with a high ielectric strength. Also, one woul want to buil, instea of a single parallel plate capacitor, several capacitors in parallel. This coul be achieve through stacking the plates of the capacitor. For example, you can alternately lay own sheets of a conucting material, such as aluminum foil, sanwiche between sheets of insulating ielectric. Making sure that none of the conucting sheets are in contact with their immeiate neighbors, connect every other plate together. ANS. FIG. CQ6.3 illustrates this iea. This technique is often use when home-brewing signal capacitors for raio applications, as they can withstan huge potential ifferences without flashover (without either ischarge between plates aroun the ielectric or ielectric breakown). One variation on this technique is to sanwich together flexible materials such as

5 Chapter 6 55 aluminum roof flashing an thick plastic, so the whole prouct can be rolle up into a capacitor burrito an place in an insulating tube, such as a PVC pipe, an then fille with motor oil (again to prevent flashover). CQ6.4 CQ6.5 ANS. FIG. CQ6.3 The ielectric ecreases the electric fiel between the plates, causing the potential ifference to ecrease for the same amount of charge. More charge may be place on the capacitor before the capacitor experiences ielectric breakown (resulting in charge jumping from one plate to the other, an in a path being burne through the ielectric) because the electric forces between charges on opposite plates are smaller. The capacitor can have a higher maximum operating voltage, allowing it to hol more charge. The work one, W QΔV, is the work one by an external agent, like a battery, to move a charge through a potential ifference, ΔV. To etermine the energy in a charge capacitor, we must a the work one to move bits of charge from one plate to the other. Initially, there is no potential ifference between the plates of an uncharge capacitor. As more charge is transferre from one plate to the other, the potential ifference increases, meaning that more work is neee to transfer each aitional bit of charge. The total work is given by W QΔV. Another explanation is that the charge Q is move through an average potential ifference ΔV, requiring total work W QΔV. *CQ6.6 The potential ifference must ecrease. Since there is no external power supply, the charge on the capacitor, Q, will remain constant that is, assuming that the resistance of the meter is sufficiently large. Aing a ielectric increases the capacitance, which must therefore ecrease the potential ifference between the plates.

6 56 Capacitance an Dielectrics CQ6.7 CQ6.8 A capacitor stores energy in the electric fiel between the plates. This is most easily seen when using a issectible capacitor. If the capacitor is charge, carefully pull it apart into its component pieces. One will fin that very little resiual charge remains on each plate. When reassemble, the capacitor is suenly recharge by inuction ue to the electric fiel set up an store in the ielectric. This proves to be an instructive classroom emonstration, especially when you ask a stuent to reconstruct the capacitor without supplying him/her with any rubber gloves or other insulating material. (Of course, this is after they sign a liability waiver.) The work you o to pull the plates apart becomes aitional electric potential energy store in the capacitor. The charge is constant an the capacitance ecreases but the potential ifference increases to rive up the potential energy QΔV. The electric fiel between the plates is constant in strength but fills more volume as you pull the plates apart. SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 0. Definition of Capacitance P6. (a) From Equation 6. for the efinition of capacitance, C Q ΔV, we have ΔV Q C Similarly, ΔV Q C 7.0 µc 3.00 µf 9.00 V 36.0 µc 3.00 µf.0 V P6. (a) C Q ΔV C 0.0 V ΔV Q C C F 00 V F.00 µf

7 Chapter 6 57 P6.3 (a) Q CΔV ( F) (.0 V) C 48.0 µc Q CΔV ( F) (.50 V) C 6.00 µc Section 6. Calculating Capacitance P6.4 (a) For a spherical capacitor with inner raius a an outer raius b, C k e ( 0.40 m) ( 0.40 m m) ab ( b a) m N m /C 5.6 pf ΔV Q C C.56 0 F V 57 kv P6.5 (a) The capacitance of a cylinrical capacitor is C k e ln b/ a 50.0 m ( N m /C )ln 7.7 mm/.58 mm.68 nf Metho : ΔV k e λ ln b a P6.6 (a) C κ 0 A λ Q C 50.0 m C m 7.7 mm ln.58 mm ΔV ( N m /C ) C/m 3.0 kv Metho : ΔV Q C C F. nf kv ( m) C / N m 800 m

8 58 Capacitance an Dielectrics The potential between groun an clou is ( 800 m) V ( V) 6.6 C ΔV E N/C Q C( ΔV). 0 9 C/ V P6.7 We have Q CΔV an C 0 A/. Thus, Q 0 AΔV/ The surface charge ensity on each plate has the same magnitue, given by Thus, σ Q A 0 ΔV 0 ΔV Q/A P6.8 (a) C κ 0 A ( C /N m )(50 V) C/cm V C cm N m.00 ( m ) ( 0 4 cm ) C N m.36 0 F.36 pf J N m 4.43 µm V C J ( m ) m Q CΔV (.36 pf) (.0 V) 6.3 pc (c) E ΔV.0 V m V/m P6.9 (a) The potential ifference between two points in a uniform electric fiel is ΔV E, so E ΔV 0.0 V m. 04 V/m The electric fiel between capacitor plates is E σ 0, so σ 0 E: σ ( C /N m )(. 0 4 V/m ) C/m 98.3 nc/m

9 Chapter 6 59 (c) For a parallel-plate capacitor, C 0 A : ( m ) C C /N m m F 3.74 pf () The charge on each plate is Q CΔV: Q ( F) ( 0.0 V) 74.7 pc P6.0 With θ π, the plates are out of mesh an the overlap area is zero. With θ 0, the overlap area is that of a semi-circle, π R. By proportion, the effective area of a single sheet of charge is ( π θ)r When there are two plates in each comb, the number of ajoining sheets of positive an negative charge is 3, as shown in the sketch. When there are N plates on each comb, the number of parallel capacitors is N an the total capacitance is C ( N ) 0 A effective istance N ( N ) 0 ( π θ)r 0 ( π θ)r P6. (a) The electric fiel outsie a spherical charge istribution of raius R is E k e q/r. Therefore, Then q Er k e ( N/C) ( 0.0 m) 0.40 µc N m / C σ q C.33 µc/m A 4π 0.0 m For an isolate charge sphere of raius R, ANS. FIG. P6.0 C 4π 0 r 4π ( C N m )( 0.0 m) 3.3 pf

10 60 Capacitance an Dielectrics P6. F y 0: T cosθ mg 0 F x 0: T sinθ Eq 0 Diviing, tanθ Eq mg, so E mg q tanθ an ΔV E mgtanθ q. Section 6.3 Combinations of Capacitors P6.3 (a) Capacitors in parallel a. Thus, the equivalent capacitor has a value of C eq C + C 5.00 µf +.0 µf 7.0 µf The potential ifference across each branch is the same an equal to the voltage of the battery. ΔV 9.00 V (c) Q 5 CΔV ( 5.00 µf) ( 9.00 V) 45.0 µc Q CΔV (.0 µf) ( 9.00 V) 08 µc P6.4 (a) In series capacitors a as C eq C + C C eq 3.53 µf 5.00 µf +.0 µf (c) We must answer part (c) first before we can answer part. The charge on the equivalent capacitor is Q eq C eq ΔV ( 3.53 µf) ( 9.00 V) 3.8 µc Each of the series capacitors has this same charge on it. So Q Q 3.8 µc.

11 Chapter 6 6 The potential ifference across each is ΔV Q 3.8 µc C 5.00 µf 6.35 V ΔV Q 3.8 µc C.0 µf.65 V P6.5 (a) When connecte in series, the equivalent capacitance is +, or C eq C C C eq C + C 4.0 µf µf C eq.8 µf When connecte in parallel, the equivalent capacitance is C eq C + C 4.0 µf µf.70 µf P6.6 (a) When connecte in series, the equivalent capacitance is +, or C eq C C C eq C + C.50 µf µf C eq.79 µf Capacitors in series carry the same charge as their equivalent capacitance: Q C eq ( ΔV).79 µf ( 6.00 V) 0.7 µc on each capacitor When connecte in parallel, each capacitor has the same potential ifference across it. The charge store on each capacitor is then For C.50 µf : Q C ( ΔV).50 µf For C 6.5 µf : Q C ( ΔV) 6.5 µf P6.7 (a) In series, to reuce the effective capacitance: 3.0 µf 34.8 µf + C s C s C s 398 µf ( 6.00 V) 5.0 µc ( 6.00 V) 37.5 µc 3.0 µf 34.8 µf

12 6 Capacitance an Dielectrics In parallel, to increase the total capacitance: 9.8 µf + C p 3.0 µf C p.0 µf P6.8 The capacitance of the combination of extra capacitors must be 7 3 C C 4 C. The possible combinations are: one capacitor: C; two 3 capacitors: C or C; three capacitors: 3C, 3 C, 3 C or 3 C. None of these is 4 C, so the esire capacitance cannot be achieve. 3 P6.9 (a) The equivalent capacitance of the series combination in the upper branch is C upper C + C 3.00 µf µf C upper.00 µf Likewise, the equivalent capacitance of the series combination in the lower branch is C lower C + C.00 µf µf C lower.33 µf These two equivalent capacitances are connecte in parallel with each other, so the equivalent capacitance for the entire circuit is C eq C upper + C lower.00 µf +.33 µf 3.33 µf Note that the same potential ifference, equal to the potential ifference of the battery, exists across both the upper an lower branches. Each of the capacitors in series combination hols the same charge as that on the equivalent capacitor. For the upper branch: Q 3 Q 6 Q upper C upper ( ΔV).00 µf ( 90.0 V) 80 µc s so, 80 µc on the 3.00-µF an the 6.00-µF capacitors For the lower branch: Q Q 4 Q lower C lower ( ΔV).33 µf ( 90.0 V) 0 µc so, 0 µc on the.00-µf an 4.00-µF capacitors

13 Chapter 6 63 (c) The potential ifference across each of the capacitors in the circuit is: ΔV Q C 0 µc.00 µf 60.0 V ΔV 3 Q 3 C 3 80 µc 3.00 µf 60.0 V 60.0 V across the 3.00-µF an the.00-µf capacitors ΔV 4 Q 4 C 4 0 µc 4.00 µf 30.0 V ΔV 6 Q 6 80 µc C µf 30.0 V 30.0 V across the 6.00-µF an the 4.00-µF capacitors P6.0 (a) Capacitors an 3 are in parallel an present equivalent capacitance 6C. This is in series with capacitor, so the battery sees capacitance 3C + 6C C. If they were initially uncharge, C stores the same charge as C an C 3 together. With greater capacitance, C 3 stores more charge than C. Then Q > Q 3 > Q. (c) The ( C C 3 ) equivalent capacitor stores the same charge as C. Since it has greater capacitance, ΔV Q C implies that it has smaller potential ifference across it than C. In parallel with each other, C an C 3 have equal voltages: ΔV > ΔV ΔV 3. () If C 3 is increase, the overall equivalent capacitance increases. More charge moves through the battery an Q increases. As V increases, V must ecrease so Q ecreases. Then Q 3 must increase even more: Q 3 an Q increase; Q ecreases. P6. Call C the capacitance of one capacitor an n the number of capacitors. The equivalent capacitance for n capacitors in parallel is C p C + C C n nc

14 64 Capacitance an Dielectrics The relationship for n capacitors in series is Therefore, C s C + C C n n C C p C s nc C/n n or n C p /C s 00 0 P6. (a) In the upper section, each C -C pair, on either sie of C 3, are in series: C s µf an both C -C pairs are in parallel to C 3 : C upper ( 3.33) µf In the lower section, the C -C pair are in parallel: C lower ( 0.0) 0.0 µf The upper section is in series with the lower section: ANS. FIG. P6. C eq µf Capacitors in series carry the same charge as their equivalent capacitor; therefore, the upper section, equivalent to a 8.67-µF capacitor, an the lower section, equivalent to a 0.0-µF capacitor, carry the same charge as a 6.05-µF capacitor: Q upper Q eq C eq ΔV ( 6.05 µf) ( 60.0 V) 363 µc The upper section is equivalent to capacitor C 3 an two 3.33-µF capacitors in parallel, an the voltage across each is the same as that across a 8.67-µF capacitor: ΔV upper Q eq C eq 363 µc 8.67 µf 4.9 V Therefore, the charge on C 3 is Q 3 C 3 ΔV 3 (.00 µf) ( 4.9 V) 83.7 µc

15 Chapter 6 65 P6.3 (a) We simplify the circuit of Figure P6.3 in three steps as shown in ANS. FIG. P6.3 panels (a),, an (c). First, the 5.0-µF an 3.00-µF capacitors in series are equivalent to (/5.0 µf) + (/3.00 µf).50 µf Next, the.50-µf capacitor combines in parallel with the 6.00-µF capacitor, creating an equivalent capacitance of 8.50 µf. At last, this 8.50-µF equivalent capacitor an the 0.0-µF capacitor are in series, equivalent to 5.96 µf (/8.50 µf) + (/0.00 µf) We fin the charge on each capacitor an the voltage across each by working backwars through solution figures (c) (a), alternately applying Q CΔV an ΔV Q/C to every capacitor, real or equivalent. For the 5.96-µF capacitor, we have Q CΔV ( 5.96 µf) ( 5.0 V) 89.5 µc (a) (c) ANS. FIG. P6.3 Thus, if a is higher in potential than b, just 89.5 µc flows between the wires an the plates to charge the capacitors in each picture. In we have, for the 8.5-µF capacitor, ΔV ac Q C 89.5 µc 8.50 µf 0.5 V an for the 0.0-µF capacitor in, (a), an the original circuit, we have Q µc. Then ΔV cb Q C 89.5 µc 0.0 µf 4.47 V Next, the circuit in iagram (a) is equivalent to that in, so ΔV cb 4.47 V an ΔV ac 0.5 V. For the.50-µf capacitor, ΔV 0.5 V an Q CΔV (.50 µf) ( 0.5 V) 6.3 µc

16 66 Capacitance an Dielectrics For the 6.00-µF capacitor, V 0.5 V an Q 6 C V (6.00 µf)(0.5 V) 63. µc Now, 6.3 µc having flowe in the upper parallel branch in (a), back in the original circuit we have Q µc an Q µc. P6.4 (a) C Q ΔV. When S is close, the charge on C will be Q CΔV ( 6.00 µf) ( 0.0 V) 0 µc When S is opene an S is close, the total ANS. FIG. P6.4 charge will remain constant an be share by the two capacitors. We let prime symbols represent the new charges on the capacitors, in Q 0 µc Q. The potential ifferences across the two capacitors will be equal. Δ V Q Q C C or 0 µc Q 6.00 µf Q 3.00 µf Then we o the algebra to fin Q µc 40.0 µc an Q 0 µc 40.0 µc 80.0 µc. P6.5 C s µf C p µf P6.6 (a) First, we replace the parallel combination between points b an c by its equivalent capacitance, C bc.00 µf µf 8.00 µf. Then, we have three capacitors in series between points a an. The equivalent capacitance for this circuit is therefore + + C eq C ab C bc C c µf ANS. FIG. P6.5 ANS. FIG. P6.6

17 Chapter 6 67 giving C eq 8.00 µf 3.67 µf The charge on each capacitor in the series is the same as the charge on the equivalent capacitor: Q ab Q bc Q c C eq (.67 µf) 9.00 V ΔV a 4.0 µc Then, note that ΔV bc Q bc 4.0 µc 3.00 V. The charge on each C bc 8.00 µf capacitor in the original circuit is: On the 8.00 µf between a an b: Q 8 Q ab 4.0 µc On the 8.00 µf between c an : Q 8 Q c 4.0 µc On the.00 µf between b an c: Q C (.00 µf) 3.00 V ΔV bc On the 6.00 µf between b an c: 6.00 µc (c) Q 6 C 6 ( 6.00 µf) 3.00 V ΔV bc 8.0 µc We earlier foun that ΔV bc 3.00 V. The two 8.00 µf capacitors have the same voltage: ΔV 8 ΔV 8 Q 4.0 µc 3.00 V, so we C 8.00 µf conclue that the potential ifference across each capacitor is the same: ΔV 8 ΔV ΔV 6 ΔV V. P6.7 C p C + C an C s C + C. Substitute C C p C : Simplifying, C s C + C p C + C C p C C C p C C C C p + C p C s 0 C C p ± C p 4C p C s C ± p 4 C p C p C s

18 68 Capacitance an Dielectrics We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus sign, we woul get the same two answers with their names interchange.) C C p + 4 C p C p C s ( 9.00 pf ) + ( 9.00 pf 4 ) 9.00 pf 6.00 pf (.00 pf) C C p C C p 4 C p C p C s P6.8 C p C + C an Substitute Simplifying, ( 9.00 pf).50 pf 3.00 pf C C p C : C s C + C. C s C + C C C p + C p C s 0 C C + C p C p C C C p C an C C ± C p p 4C p C s C + p 4 C p C p C s where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values for the capacitances, with the names reverse). Then, from C C p C, we obtain C C p 4 C p C p C s P6.9 For C connecte by itself, C ΔV 30.8 µc where ΔV is the battery 30.8 µc voltage: ΔV. C For C an C in series: C + C ΔV 3. µc

19 Chapter 6 69 substituting, 30.8 µc 3. µc 3. µc + which gives C C C C 0.333C For C an C 3 in series: For all three: C + C 3 ΔV 5. µc 30.8 µc 5. µc 5. µc + which gives C C C C 0.C 3 3 Q C + C + C 3 ΔV C ΔV + C C + C C µc 9.8 µc This is the charge on each one of the three. Section 6.4 Energy Store in a Charge Capacitor P6.30 From U E CΔV, we have ΔV U E C 300 J F V P6.3 The energy store in the capacitor is given by U E Q C QΔV ( C) (.0 V) J P6.3 (a) U E C( ΔV) ( 3.00 µf ) (.0 V) 6 µj U E C( ΔV) ( 3.00 µf ) ( 6.00 V) 54.0 µj P6.33 (a) Q CΔV ( 50 0 F) ( V) C.50 µc From U E C( ΔV), ΔV U E C J 50 0 F V.83 kv

20 70 Capacitance an Dielectrics P6.34 (a) The equivalent capacitance of a series combination of C an C is + C eq C C 8.0 µf µf C eq.0 µf This series combination is connecte to a.0-v battery, the total store energy is.0 V U E, eq C ( ΔV eq ) F J (c) Capacitors in series carry the same charge as their equivalent capacitor. The charge store on each of the two capacitors in the series combination is Q Q Q total C eq ( ΔV).0 µf 44 µc C (.0 V) an the energy store in each of the iniviual capacitors is: 8.0 µf capacitor: J U E Q C C F 36.0 µf capacitor: J U E Q C C F () U E +U E J J J U E, eq, which is one reason why the.0 µf capacitor is consiere to be equivalent to the two capacitors. (e) (f) The total energy of the equivalent capacitance will always equal the sum of the energies store in the iniviual capacitors. If C an C were connecte in parallel rather than in series, the equivalent capacitance woul be C eq C + C 8.0 µf µf 54.0 µf. If the total energy store in this parallel combination is to be the same as store in the original series combination, it is necessary that C eq ΔV U E, eq

21 Chapter 6 7 From which we obtain ΔV U E, eq J C eq F 5.66 V (g) Because the potential ifference is the same across the two capacitors when connecte in parallel, an U E C( ΔV), the larger capacitor C stores more energy. P6.35 (a) Because the capacitors are connecte in parallel, their voltage remains the same: U E C( ΔV) + C( ΔV) C( ΔV) ( µf) ( 50.0 V).50 0 J Because C κ 0 A an, the altere capacitor has new capacitance to C C. The total charge is the same as before: Q initial Q final C( ΔV) + C( ΔV) C( Δ V ) + C ( Δ V ) 3 C( Δ V ) Δ C ΔV (c) New U E C( ΔV ) + C ΔV V 4 3 ΔV V 3 4 C ΔV 66.7 V 3 4 C 4ΔV 3 U E 4 3 C( ΔV) 4 3 U E 4 ( J) J () Positive work is one by the agent pulling the plates apart. P6.36 Before the capacitors are connecte, each has voltage V an charge Q. (a) Connecting plates of like sign places the capacitors in parallel, so the voltage on each capacitor remains the same. U E, total C( ΔV) + C( ΔV) C( ΔV)

22 7 Capacitance an Dielectrics Because C 0 A, the altere capacitor has new capacitance C 0 A C, an this change in capacitance results in a new potential ifference Δ V across the parallel capacitors. We can solve for the new potential ifference because the total charge remains the same: Q C( ΔV)+ C ΔV C Δ V + C Δ V Δ V 4ΔV 3 (c) Each capacitor has potential ifference V : U E, total C( ΔV ) + C ΔV C 4ΔV 3 9 C( ΔV) 4C ΔV 3 + C 4ΔV 3 () Positive work is one by the agent pulling the plates apart. P6.37 (a) The circuit iagram for capacitors connecte in parallel is shown in ANS. FIG. P6.37(a). U E C( ΔV), an C p C + C 5.0 µf µf 30.0 µf U E ( ) J ANS. FIG. P6.37(a) (c) C s + C C U E C( ΔV) ΔV 5.0 µf µf U E C 0.50 J F 68 V () The circuit iagram for capacitors connecte in series is shown in ANS. FIG. P6.37(). 4.7 µf ANS. FIG. P6.37()

23 Chapter 6 73 P6.38 To prove this, we follow the hint, an calculate the work one in separating the plates, which equals the potential energy store in the charge capacitor: U E Q C Fx Now from the funamental theorem of calculus, U E Fx an F x U E x Q C Q x A 0 /x. Performing the ifferentiation, F Q x x A 0 Q 0 A P6.39 The energy transferre is T ET QΔV ( 50.0 C) ( V) J an % of this (or ΔE int J ) is absorbe by the tree. If m is the amount of water boile away, then ( 00 C 30.0 C) + m( J kg) ΔE int m 4 86 J kg C J giving m 9.79 kg. P6.40 (a) Accoring to Equation 6., we may think of a sphere of raius R that hols charge Q as having a capacitance C R k e. The energy store is U E C( ΔV) R k e k e Q R k eq R The total energy is U E U E +U E q + q q + C C R k e k q e + k e Q q R R Q q R k e

24 74 Capacitance an Dielectrics (c) For a minimum we set U E q 0: k e q + k e Q q R R which gives ( ) 0 R q R Q R q q R Q R + R () q Q q R Q R + R (e) V k e q R k e R Q V R R + R k e Q R + R, an V k e q R (f) V V 0 k e R Q V k e Q R R + R R + R P6.4 Originally, the capacitance of each pair of plates is C 0 A, but after the switch is close an the istance is change to 0.500, the plates have new capacitance C A 0 A 0 A 0 C The capacitors are ientical an in series, so each has half the total voltage (ΔV) 00 V. (a) The plates are in series, so each collects the same charge: Q C ( ΔV) C ΔV (.00 µc) ( 00 V) 400 µc Each plate contributes half of the total electric fiel between the plates, E σ Q, where Q C(ΔV) is the magnitue of 0 0 A the charge on a plate, from (a) above. The electric force that each plate exerts on the charge of its neighboring plate is [ ] F Q E Q C ΔV 0 A 0 A C ΔV ( 0 A ) C ΔV an this force is balance by the spring force F kx on each plate.

25 Chapter 6 75 Each spring stretches by istance x, so we obtain 4 C( ΔV) k 4 an solving for the force constant gives k 8C ( ΔV ) ( 00 V).50 kn m F m Section 6.5 Capacitors with Dielectrics P6.4 (a) Consier two sheets of aluminum foil, each 40 cm by 00 cm, with one sheet of plastic between them. (c) Suppose the plastic has κ 3, E max ~ 0 7 V/m, an thickness.54 cm mil 000. Then, C κ 0 A ( m ) ~ C / N m m ΔV max E max ~ ( 0 7 V/m) ( m) ~ 0 V P6.43 Q max CΔV max, but ΔV max E max. Also, Thus, (a) C κ 0 A. Q max κ 0 A ( E max ) κ 0 AE max. With air between the plates, from Table 6., the ielectric constant is κ.00, an the ielectric strength is E max V m. Therefore, Q max κ 0 AE max F/m ~ 0 6 F ( m )( V/m) 3.3 nc

26 76 Capacitance an Dielectrics With polystyrene between the plates, from Table 6., κ.56 an E max V/m. Q max κ 0 AE max ( m ) ( V/m) F/m 7 nc P6.44 (a) Note that the charge on the plates remains constant at the original value, Q 0, as the ielectric is inserte. Thus, the change in the potential ifference, ΔV Q/C, is ue to a change in capacitance alone. The ratio of the final an initial capacitances is C f C i κ 0 A 0 A κ an C f Q 0 ΔV C i ΔV Q 0 f ( ΔV) i 85.0 V i ( ΔV) f 5.0 V 3.40 Thus, the ielectric constant of the inserte material is κ The material is probably nylon (see Table 6.). (c) P6.45 (a) C κ 0 A The presence of a ielectric weakens the fiel between plates, an the weaker fiel, for the same charge on the plates, results in a smaller potential ifference. If the ielectric only partially fille the space between the plates, the fiel is weakene only within the ielectric an not in the remaining air-fille space, so the potential ifference woul not be as small. The voltage woul lie somewhere between 5.0 V an 85.0 V. 8.3 pf ( m ) F m m F ΔV max E max ( V/m) ( m).40 kv

27 Chapter 6 77 P6.46 ANS. FIG. P6.46 exaggerates how the strips can be offset to avoi contact between the two foils. It shows how a secon paper strip can be use to roll the capacitor into a convenient cylinrical shape with electrical contacts at the two ens. We suppose that the overlapping with of the two metallic strips is still w 7.00 cm. Then for the area of the plates we have A w in C κ 0 A/ κ 0 w/. Solving the equation gives P6.47 Originally, C i 0 A (a) C κ 0 w ( F)( m) 3.70( C /N m )( m).04 m Q. ( ΔV) i The charge is the same before an after immersion, with value Finally, an i Q C i ( ΔV) i 0 A ΔV ( m ) 50 V Q C / N m 369 pc C f κ 0 A C f 80 Q : ( ΔV) f.50 0 m C / N m ( m ) F i.50 0 m ΔV ( ΔV) f Q C i ΔV A / 0 C f C f κ 0 A / 3.0 V (c) Originally, U i C i ΔV Finally, U f C f ΔV i 0 A( ΔV) i f κ 0 A. ( ΔV) i i ( ΔV) i κ ANS. FIG. P6.46 κ 50 V 80 0 A( ΔV) i, κ

28 78 Capacitance an Dielectrics where, from Table 6., κ 80 for istille water. So, ΔU U f U i i A ΔV 0 κ i A ΔV 0 i A ΔV 0 κ A ΔV 0 κ ΔU C N m κ i ( m ) 50 V (.50 0 m) ( 80) J 45.5 nj ( 80) P6.48 The given combination of capacitors is equivalent to the circuit iagram shown in ANS. FIG. P6.48. ANS. FIG. P6.48 Put charge Q on point A. Then, Q ( 40.0 µf)δv AB ( 0.0 µf)δv BC ( 40.0 µf)δv CD So, ΔV BC 4ΔV AB 4ΔV CD, an the center capacitor will break own first, at V BC 5.0 V. When this occurs, ΔV AB ΔV CD ( 4 ΔV BC ) 3.75 V an V AD V AB + V BC + V CD 3.75 V V V.5 V. P6.49 (a) We use the equation U E Q /C to fin the potential energy of the capacitor. As we will see, the potential ifference ΔV changes as the ielectric is withrawn. The initial an final energies are U E,i Q C i an U E, f Q C f. But the initial capacitance (with the ielectric) is C i κc f. Therefore, U E, f κ Q C i κu E,i. Since the work one by the external force in removing the ielectric equals the change in potential energy, we have W U f U i κu i U i κ U i κ Q C i

29 Chapter 6 79 To express this relation in terms of potential ifference ΔV i, we substitute Q C i ( ΔV i ), an evaluate: W C i κ ΔV i J 40.0 µj 00 V F ( ) The positive result confirms that the final energy of the capacitor is greater than the initial energy. The extra energy comes from the work one on the system by the external force that pulle out the ielectric. The final potential ifference across the capacitor is ΔV f Q C f. Substituting C f C i κ an Q C ΔV i ( i ) gives ΔV f kδv i 5.00(00 V) 500 V Even though the capacitor is isolate an its charge remains constant, the potential ifference across the plates oes increase in this case. Section 6.6 Electric Dipole in an Electric Fiel P6.50 (a) The isplacement from negative to positive charge is mm (.40î.30ĵ ) mm 0 3 m a.0î +.0ĵ.60î +.40ĵ The electric ipole moment is p aq 0 3 m p C.60î +.40ĵ 0 C m 9.0î ĵ The torque exerte by the fiel on the ipole is τ p E 0 C m 9.0î ĵ 0 3 N C 7.80î 4.90ĵ 0 9 N m ˆk N m ˆk 65.5 ˆk

30 80 Capacitance an Dielectrics (c) Relative to zero energy when it is perpenicular to the fiel, the ipole has potential energy U p E 0 C m 0 3 N C 9.0î ĵ 7.80î 4.90ĵ ( ) 0 9 J nj () For convenience we compute the magnitues an p C m.4 0 C m E ( 7.80) + ( 4.90) 0 3 N C N C The maximum potential energy occurs when the ipole moment is opposite in irection to the fiel, an is U max p E p E ( ) p E 4 nj The minimum potential energy configuration is the stable equilibrium position with the ipole aligne with the fiel. The value is U min 4 nj Then the ifference, representing the range of potential energies available to the ipole, is U max U min 8 nj. P6.5 (a) The electric fiel prouce by the line of charge has raial symmetry about the y axis. Accoring to Equation 4.7 in Example 4.4, the electric fiel to the right of the y axis is E E r î k e λ î r Let x 5.0 cm represent the ANS. FIG. P6.5 coorinate of the center of the ipole charge, an let a.00 cm represent the istance between the charges. Then r x acosθ is the coorinate of the negative charge an r + x + acosθ is the coorinate of the positive charge. The force on the positive charge is F + qe( r + )î q k e λ î r + k e qλ î x + acosθ

31 Chapter 6 8 an the force on the negative charge is F qe( r )î q k e The force on the ipole is λ î r k e qλ î x acosθ F F + + F qλ k e x + acosθ k qλ e x acosθ î k e qλ x + acosθ x acosθ î ( k e qλ x acosθ ) ( x + acosθ ) x + ( acosθ ) î 4k eaqλ cosθ x + ( acosθ ) î Substituting numerical values an suppressing units, 4( )( ) ( )( )cos35.0 F î N [ ] î + ( ) ( cos35.0 ) P6.5 Let x represent the coorinate of the negative charge. Then x + acosθ is the coorinate of the positive charge. The force on the negative charge is F qe( x)î. The force on the positive charge is F + +qe( x + acosθ )î q E ( x E )+ x ( a cosθ ) î ANS. FIG. P6.5 The force on the ipole is altogether F F + F + q E ( x acosθ E )î p x cosθ î

32 8 Capacitance an Dielectrics Section 6.7 An Atomic Description of Dielectrics P6.53 (a) Consier a gaussian surface in the form of a cylinrical pillbox with ens of area A << A parallel to the sheet. The sie wall of the cyliner passes no flux of electric fiel since this surface is everywhere parallel to the fiel. Gauss s law becomes E A + E A Q A, A irecte away from the positive sheet. Q In the space between the sheets, each creates fiel A away from the positive an towar the negative sheet. Together, they create a fiel of E Q A (c) Assume that the fiel is in the positive x-irection. Then, the potential of the positive plate relative to the negative plate is ΔV + plate plate E s + plate Q î A îx + Q A plate () Capacitance is efine by: C Q ΔV Q Q A A κ 0 A. Aitional Problems P6.54 The stages for the reuction of this circuit are shown in ANS. FIG. P6.54 below. Thus, C eq 6.5 µf ANS. FIG. P6.54

33 Chapter 6 83 P6.55 (a) Each face of P carries charge, so the three-plate system is equivalent to what is shown in ANS. FIG. P6.55 below. ANS. FIG. P6.55 Each capacitor by itself has capacitance C κ 0 A 5.58 pf ( m ) C /N m m Then equivalent capacitance 5.58 pf pf. pf. (c) Q CΔV + CΔV (. 0 F) ( V) 34 pc Now P 3 has charge on two surfaces an in effect three capacitors are in parallel: C 3( 5.58 pf) 6.7 pf () Only one face of P 4 carries charge: Q CΔV ( F) ( V) 66.9 pc P6.56 The upper pair of capacitors, 3-µF an 6-µF, are in series. Their equivalent capacitance is µf The lower pair of capacitors, -µf an 4-µF, are in series. Their equivalent capacitance is µf The upper pair are in parallel to the lower pair, so the total capacitance is C eq.00µf +.33µF 3.33µF (a) The total energy store in the full circuit is then ( Energy store ) total C ΔV eq ( F) ( 90.0 V).35 0 J J 3.5 mj

34 84 Capacitance an Dielectrics Refer to P6.9 for the calculation of the charges use below. The energy store in each iniviual capacitor is For.00 µf: ( Energy store) Q C C F 3.60 mj J For 3.00 µf: ( Energy store) 3 Q C C F For 4.00 µf: 5.40 mj J ( Energy store) 4 Q C C F For 6.00 µf:.80 mj ( Energy store) 6 Q C C F.70 mj J J mj 3.5 mj (c) Energy store ( Energy store) total *P6.57 From Equation 6.3, The total energy store by the system equals the sum of the energies store in the iniviual capacitors. u E U E V 0 E Solving for the volume gives V U E 0 E J ( C /N m )( V/m) 000 L m m 3.5 L m 3

35 Chapter 6 85 P6.58 Imagine the center plate is split along its miplane an pulle apart. We have two capacitors in parallel, supporting the same V an carrying total charge Q. The upper capacitor has capacitance C 0 A an the lower C 0 A. Charge flows from groun onto each of the outsie plates so that ANS. FIG. P6.58 Q + Q Q an ΔV ΔV ΔV. Then Q C Q C Q 0 A Q 0 A Q Q Q + Q Q. (a) Q Q 3. On the lower plate the charge is Q 3. Q Q 3. On the upper plate the charge is Q 3 ΔV Q Q C 3 0 A P6.59 The ielectric strength is E max V/m ΔV max so we have for the istance between plates ΔV max E max. Now to also satisfy C κ A F with κ 3.00, we combine by substitution to solve for the plate area: A C κ m CΔV max κ 0 E max ( F)(4 000 V) (3.00)( F/m)( V/m) P6.60 We can use the energy U C store in the capacitor to fin the potential ifference across the plates:, U C C( ΔV) ΔV U C C

36 86 Capacitance an Dielectrics When the particle moves between the plates, the change in potential energy of the charge-fiel system is ΔU system qδv q U C C where we have note that the potential ifference is negative from the positive plate to the negative plate. Apply the isolate system (energy) moel to the charge-fiel system: ΔK + ΔU system 0 ΔK ΔU system q Substitute numerical values: ΔK ( C) U C C ( J) F J This ecrease in kinetic energy of the particle is more than the energy with which it began. Therefore, the particle oes not arrive at the negative plate but rather turns aroun an moves back to the positive plate. *P6.6 (a) V m ρ.00 0 kg 00 kg m m 3 (c) Since V 4π r 3, the raius is r 3V 3 4π A 4π r 4π 3V 4π C κ 0 A m F 3 3 4π m 3 4π ( 5.00) C N m m, an the surface area is ( m ) Q C( ΔV) ( F) ( V) C, an the number of electronic charges is n Q e C C 3

37 Chapter 6 87 P6.6 (a) With the liqui filling the space between the plates to height f, the top of the flui at the air-flui interface evelops an inuce ipole layer of charge so that it acts as a thin plate with opposite charge on its upper an lower sies; thus, the partially fille capacitor behaves as two capacitors in series connecte at the interface. The upper an lower capacitors have separate capacitances: C up 0 A ( f ) an C own A f The equivalent series capacitance is C f A ( f ) 0 A + f f + f A 0 A f 5.0 µf( f ) For f 0, the capacitor is empty so we can expect capacitance 5.0 µf. For f 0, C f 5.0 µf( f ) 5.0 µf( 0) 5.0 µf an the general expression agrees. (c) For f, we expect 6.5(5.0 µf) 6 µf. For f, C f 5.0 µf( f ) 5.0 µf( 0.846) 6 µf an the general expression agrees. P6.63 The initial charge on the larger capacitor is Q CΔV ( 0.0 µf) ( 5.0 V) 50 µc An aitional charge q is pushe through the 50.0-V battery, giving the smaller capacitor charge q an the larger charge 50 µc + q. Then 50.0 V q 50 µc + q µf 0.0 µf. 500 µc q + 50 µc + q q 7 µc

38 88 Capacitance an Dielectrics *P6.64 So across the 5.00-µF capacitor, ΔV q C 7 µc 5.00 µf 3.3 V Across the 0.0-µF capacitor, ΔV 50 µc + 7 µc 0.0 µf 6.7 V From Gauss's Law, for the electric fiel insie the cyliner, π r E q in 0. so E λ π r 0. r ΔV E r r λ λ r ln r r r π r 0 π 0 r ANS. FIG. P6.64 Recognizing that λ max π 0 E max r inner, we obtain ΔV ( V m) ( m)ln 5.0 m 0.00 m ΔV max 579 V P6.65 Where the metal block an the plates overlap, the electric fiel between the plates is zero. The plates o not lose charge in the overlapping region, but opposite charge inuce on the surfaces of the inserte portion of the block cancels the fiel from charge on the plates. The unfille portion of the capacitor has capacitance C 0 A 0 ( x) The effective charge on this portion (the charge proucing the remaining electric fiel between the plates) is proportional to the unblocke area: (a) ( Q x )Q 0 The store energy is U Q C [ ( x)q 0 ] 0 x Q 0 x 0 3

39 Chapter 6 89 F U x x Q 0 ( x) Q F Q 0 to the right (into the capacitor: the block is pulle in) 3 0 (c) Stress F Q () The energy ensity is u E 0 E σ 0 Q Q A 0 ( x)q 0 ( x) (e) They are precisely the same. P6.66 (a) Put charge Q on the sphere of raius a an Q on the other sphere. Relative to V 0 at infinity, because is larger compare to a an to b. The potential at the surface of a is approximately V a k e Q a k e Q an the potential of b is approximately V b k e Q b + k e Q. The ifference in potential is V a V b k e Q a + k e Q b k e Q k e Q Q 4π an C 0 V a V b a + b As, becomes negligible compare to a an b. Then, C 4π 0 a + b an as for two spheres in series. C 4π 0 a + 4π 0 b

40 90 Capacitance an Dielectrics P6.67 Call the unknown capacitance C u. The charge remains the same: Q C u C u ( ΔV i ) ( C u + C) ΔV f C ΔV f ( ΔV i ) ΔV f ( 30.0 V) 0.0 µf 00 V 30.0 V 4.9 µf P6.68 (a) C 0 A 0 Q 0 for a capacitor with air or vacuum beteen its ΔV 0 plates. When the ielectric is inserte at constant voltage, C κc 0 Q ΔV 0 The original energy is U E0 C ΔV 0 ( 0 ) an the final energy is therefore, U E C ( ΔV 0 ) U E U E0 κ κc ΔV 0 ( 0 ) (c) The electric fiel between the plates polarizes molecules within the ielectric; therefore the fiel oes work on charge within the molecules to create electric ipoles. The extra energy comes from (part of the) electrical work one by the battery in separating that charge. The charge on the plates increases because the voltage remains the same: Q 0 C 0 ΔV 0 an Q CΔV 0 κc 0 ΔV 0 so the charge increases accoring to Q Q 0 κ.

41 Chapter 6 9 P6.69 Initially (capacitors charge in parallel), q C ( ΔV) ( 6.00 µf) 50 V q C ( ΔV) (.00 µf) 50 V 500 µc 500 µc After reconnection (positive plate to negative plate), q total q q 000 µc q total an Δ V C total Therefore, q C Δ V q C Δ V 000 µc 8.00 µf 5 V ( 6.00 µf) ( 5 V) 750 µc (.00 µf) ( 5 V) 50 µc P6.70 The conition that we are testing is that the capacitance increases by less than 0%, or, C C <.0 Substituting the expressions for C an C from Example 6., we have C C This becomes k e ln b.0a k e ln b a <.0 ln b a ln b.0a ln b b a <.0ln.0a.0ln b a +.0ln.0 We can rewrite this as 0.0ln b a ln b a.0ln b a.0ln.0 <.0ln (.0) >.0ln (.0) ln (.0).0 where we have reverse the irection of the inequality because we multiplie the whole expression by to remove the negative signs.

42 9 Capacitance an Dielectrics Comparing the arguments of the logarithms on both sies of the inequality, we see that b a > Thus, if b >.85a, the increase in capacitance is less than 0% an it is more effective to increase. P6.7 Placing two ientical capacitor is series will split the voltage evenly between them, giving each a voltage of 45 V, but the total capacitance will be half of what is neee. To ouble the capacitance, another pair of series capacitors must be place in parallel with the first pair, as shown in ANS. FIG. P6.7A. The equivalent capacitance is 00 µf + 00 µf + 00 µf + 00 µf 00 µf Another possibility shown in ANS. FIG. P6.7B: two capacitors in parallel, connecte in series to another pair of capacitors in parallel; the voltage across each parallel section is then 45 V. The equivalent capacitance is 00 µf ( 00 µf + 00 µf) + 00 µf + 00 µf ANS. FIG. P6.7A ANS. FIG. P6.7B (a) One capacitor cannot be use by itself it woul burn out. She can use two capacitors in series, connecte in parallel to another two capacitors in series. Another possibility is two capacitors in parallel, connecte in series to another two capacitors in parallel. In either case, one capacitor will be left over. Each of the four capacitors will be expose to a maximum voltage of 45 V.

43 Chapter 6 93 Challenge Problems P6.7 From Example 6., when there is a vacuum between the conuctors, the voltage between them is ΔV V b V a k e λ ln b a λ ln b π 0 a With a ielectric, a factor /κ must be inclue, an the equation becomes ΔV The electric fiel is λ ln πκ b 0 ( a ) E λ πκ 0 r So when E E max at r a, Thus, λ max E πκ max a an ΔV max λ max ln b 0 πκ 0 a E a ln b max a ΔV max ( V/m) m 9.0 kv P6.73 Accoring to the suggestion, the combination of capacitors shown is equivalent to 3.00 mm ln mm Then, from ANS. FIG. P6.73, C C 0 + C + C 0 + C 0 C + C + C + C + C C 0 ( C + C 0 ) C 0 C + C 0 C + 3C 0 C C + C 0 C C 0 0 C C ± 4C + 4 C Only the positive root is physical: C C 0 ( 3 ) ANS. FIG. P6.73

44 94 Capacitance an Dielectrics P6.74 Let charge λ per length be on one wire an λ be on the other. The electric fiel ue to the charge on the positive wire is perpenicular to the wire, raial, an of magnitue E + λ π 0 r The potential ifference between the surfaces of the wires ue to the presence of this charge is ΔV E r λ π 0 + wire wire r r λ ln D r r π 0 r The presence of the linear charge ensity λ on the negative wire makes an ientical contribution to the potential ifference between the wires. Therefore, the total potential ifference is ΔV ( ΔV ) λ D r ln D r π 0 r With D much larger than r we have nearly ΔV λ ln D π 0 r an the capacitance of this system of two wires, each of length, is C Q ΔV λ ΔV λ ( λ π 0 )ln D r The capacitance per unit length is [ ] π 0 ln[ D r] C π 0 ln D r [ ]. P6.75 By symmetry, the potential ifference across 3C is zero, so the circuit reuces to (see ANS. FIG. P6.75): C eq C + 4C 8 6 C 4 3 C ANS. FIG. P6.75 P6.76 (a) Consier a strip of with x an length W at position x from the front left corner. The capacitance of the lower portion of this strip is κ 0 W x t x L. The capacitance of the upper portion is κ 0 W x t ( x L).

45 Chapter 6 95 (c) The series combination of the two elements has capacitance tx κ 0 WL x + t(l x) κ 0 W Lx κ κ 0 W Lx κ tx +κ tl κ tx The whole capacitance is a combination of elements in parallel: C 0 L κ κ 0 W Lx ( κ κ )tx +κ tl ( κ κ )t 0 L κ κ 0 W L ( κ κ )t κ κ 0 W L( κ κ )tx ( κ κ )tx +κ tl ln ( κ κ )tx +κ tl κ κ 0 WL ( κ κ )t ln ( κ κ )tl +κ tl 0 +κ tl κ κ 0 WL ( κ κ )t ln κ κ κ κ 0 WL ( κ κ )t ln κ κ 0 L κ κ 0 WL ( )( κ κ )t ln κ κ The capacitor physically has the same capacitance if it is turne upsie own, so the answer shoul be the same with κ an κ interchange. We have proven that it has this property in the solution to part (a). Let κ κ ( + x). Then C κ (+ x)κ 0 WL ln[ + x]. κ xt As x approaches zero we have C κ (+ 0) 0 WL xt was to be shown. x κ 0 WL t P6.77 Assume a potential ifference across a an b, an notice that the potential ifference across 8.00 µf the capacitor must be zero by symmetry. Then the equivalent capacitance can be etermine from the circuit shown in ANS. FIG. P6.77, an is C ab 3.00 µf. as ANS. FIG. P6.77

46 96 Capacitance an Dielectrics P6.78 (a) The portion of the evice containing the ielectric has plate area x an capacitance C κ 0 x. The unfille part has area an capacitance C 0 x x C + C 0 [ + x( κ ) ].. The total capacitance is The store energy is U Q C Q 0 + x κ [ ]. (c) F U x î Q κ 0 + x κ [ ] î. When x 0, the original value of the force is Q ( κ ) î. As the ielectric slies in, the 0 3 charges on the plates reistribute themselves. The force ecreases to its final value, when x, of Q κ î. 0 3 κ î. () At x, F Q κ 0 3 κ + For the constant charge on the capacitor an the initial voltage we have the relationship Q C 0 ΔV 0 ΔV Then the force is F 0 ΔV F C /N m κ î. κ m 05î µn V î ( m) ( )

47 Chapter 6 97 ANSWERS TO EVEN-NUMBERED PROBLEMS P6. (a).00 µf; 00 V P6.4 (a) 5.6 pf; 57 kv P6.6 (a). nf; 6.6 C P6.8 (a).36 pf; 6.3 pc; (c) V/m P6.0 P6. ( N ) 0 ( π θ )R mg tanθ q P6.4 (a) 3.53 µf; 6.35 V an.65 V; (c) 3.8 µc P6.6 (a) 0.7 µc; 5.0 µc an 37.5 µc P6.8 None of the possible combinations of the extra capacitors is 4 C, so the 3 esire capacitance cannot be achieve. P6.0 (a) C; Q > Q 3 > Q ; (c) ΔV > ΔV > ΔV 3 ; () Q 3 an Q increase; Q ecreases P6. (a) 6.05 µf ; 83.7 µc P6.4 0 µc ; 40.0 µc an 80.0 µc P6.6 (a).67 µf ; 4.0 µc, 4.0 µc, 6.00 µc, 8.0 µc ; (c) 3.00 V P6.8 C C + p 4 C C p p C s an C C p 4 C C p p C s P V P6.3 (a) 6 µj ; 54.0 µj P6.34 (a).0 µf ; J; (c) U J an U J; () U + U J J J U eq, which is one reason why the.0 µf capacitor is consiere to be equivalent to the two capacitors; (e) The total energy of the equivalent capacitance will always equal the sum of the energies store in the iniviual capacitors; (f) 5.66 V; (g) The larger capacitor C stores more energy. P6.36 (a) C( ΔV) ; Δ V 4ΔV 3 the agent pulling the plates apart. ΔV ; (c) 4C 3 ; () Positive work is one by

48 98 Capacitance an Dielectrics P6.38 Q 0 A P6.40 (a) k e Q R ; k eq + k e Q q R R (e) V k e Q R + R an V ; (c) k e Q R + R ; (f) 0 R Q R + R ; () R Q R + R ; P6.4 (a) Consier two sheets of aluminum foil, each 40 cm by 00 cm, with one sheet of plastic between them; 0 6 F; (c) 0 V P6.44 (a) κ 3.40 ; nylon; (c) The voltage woul lie somewhere between 5.0 V an 85.0 V. P m P V 0 C m; ˆk N m; (c) nj; () 8 nj P6.50 (a) 9.0î ĵ P6.5 p E x cosθî P µf P6.56 (a) 3.5 mj; 3.60 mj, 5.40 mj,.80 mj,.70 mj; (c) The total energy store by the system equals the sum of the energies store in the iniviual capacitors. P6.58 (a) On the lower plate the charge is Q 3, an on the upper plate the charge is Q 3 ; Q 3 0 A P6.60 The ecrease in kinetic energy of the particle is more than the energy with which it began. Therefore, the particle oes not arrive at the negative plate but rather turns aroun an moves back to the positive plate. P6.6 (a).50µf( f ) ; 5.0µF, the general expression agrees; (c) 6 µf; The general expression agrees. P V

49 Chapter 6 99 P6.66 (a) See P6.66(a) for full explanation; 4π 0 a + 4π 0 b P6.68 (a) See P6.68(a) for full explanation; The electric fiel between the plates polarizes molecules within the ielectric; therefore the fiel oes work on charge within the molecules to create electric ipoles. The extra energy comes from (part of the) electrical work one by the battery in separating that charge; (c) Q Q 0 κ P6.70 See P6.70 for full mathematical verification. P kv P6.74 C π 0 n D r [ ] P6.76 (a) κ κ 0 WL ( κ κ )t ln κ κ ; The capacitor physically has the same capacitance if it is turne upsie own, so the answer shoul be the same with κ an κ interchange. We have proven that it has this property in the solution to part (a); (c) See P6.76(c) for full explanation. P6.78 (a) 0 () 05î µn [ + x( κ ) ]; [ ] ; (c) Q ( κ ) 0 + x( κ ) Q 0 + x κ [ ] î;