Electrical Energy and Capacitance

6 Electrical Energy and Capacitance QUICK QUIZZES. Choice. The field exerts a force on the electron, causing it to accelerate in the direction opposite to that of the field. In this process, electrical potential energy is converted into kinetic energy of the electron. Note that the electron moves to a region of higher potential, but because the electron has negative charge this corresponds to a decrease in the potential energy of the electron.. Choice (a). The electron, a negatively charged particle, will move toward the region of higher electric potential. Because of the electron s negative charge, this corresponds to a decrease in electrical potential energy. 3. Choice. Charged particles always tend to move toward positions of lower potential energy. The electrical potential energy of a charged particle is PE qv and, for positively-charged particles, this decreases as V decreases. Thus, a positively-charged particle located at x A would move toward the left. 4. Choice (d). For a negatively-charged particle, the potential energy (PE qv) decreases as V increases. A negatively charged particle would oscillate around x B which is a position of minimum potential energy for negative charges. 5. Choice (d). If the potential is zero at a point located a finite distance from charges, negative charges must be present in the region to make negative contributions to the potential and cancel positive contributions made by positive charges in the region. 6. Choice. Both the electric potential and the magnitude of the electric field decrease as the distance from the charged particle increases. However, the electric flux through the balloon does not change because it is proportional to the total charge enclosed by the balloon, which does not change as the balloon increases in size. 7. Choice (a). From the conservation of energy, the final kinetic energy of either particle will be given by KE f KE i + ( PE i PE f ) 0 + qv i qv f qv ( f V i ) q( ΔV ) For the electron, q e and ΔV + V giving KE f ( e) ( + V)+ ev. For the proton, q +e and ΔV V, so KE f ( e) ( V)+ ev, the same as that of the electron. 8. Choice. The battery moves negative charge from one plate and puts it on the other. The first plate is left with excess positive charge whose magnitude equals that of the negative charge moved to the other plate. 9. (a) C decreases. Q stays the same. E stays the same. (d) ΔV increases. (e) The energy stored increases. 30 6879_6_ch6_p030-06.indd 30 /7/ :3:36 PM

Electrical Energy and Capacitance 3 Because the capacitor is removed from the battery, charges on the plates have nowhere to go. Thus, the charge on the capacitor plates remains the same as the plates are pulled apart. Because E s 0 (Q A) 0, the electric field is constant as the plates are separated. Because ΔV Ed and E does not change, ΔV increases as d increases. Because the same charge is stored at a higher potential difference, the capacitance (C Q ΔV) has decreased. Because energy stored Q C and Q stays the same while C decreases, the energy stored increases. The extra energy must have been transferred from somewhere, so work was done. This is consistent with the fact that the plates attract one another, and work must be done to pull them apart. 0. (a) C increases. Q increases. E stays the same. (d) ΔV remains the same. (e) The energy stored increases. The presence of a dielectric between the plates increases the capacitance by a factor equal to the dielectric constant. Since the battery holds the potential difference constant while the capacitance increases, the charge stored (Q CΔV) will increase. Because the potential difference and the distance between the plates are both constant, the electric field (E ΔVd) will stay the same. The battery maintains a constant potential difference. With ΔV constant while capacitance increases, the stored energy [energy stored C(ΔV) ] will increase.. Choice (a). Increased random motions associated with an increase in temperature make it more difficult to maintain a high degree of polarization of the dielectric material. This has the effect of decreasing the dielectric constant of the material, and in turn, decreasing the capacitance of the capacitor. ANSWERS TO MULTIPLE CHOICE QUESTIONS. The change in the potential energy of the proton is equal to the negative of the work done on it by the electric field. Thus, 850 N C ΔPE W qe x ( Δx) +.6 0 9 C and is the correct choice for this question..5 m 0 3.4 0 6 J. Because electric forces are conservative, the kinetic energy gained is equal to the decrease in electrical potential energy, or KE PE q( ΔV ) ( e) ( +.00 0 4 V)+.00 0 4 ev so the correct choice is (a). 3. In a uniform electric field, the change in electric potential is ΔV E x ( Δx), giving E x ΔV Δx V f V i x f x i 90 V 0 V 5.0 m 3.0 m and it is seen that the correct choice is (d). 35 V m 35 N C 6879_6_ch6_p030-06.indd 3 /7/ :3:37 PM

3 Chapter 6 4. From conservation of energy, KE f + PE f KE i + PE i, or mv B mv A + qv A qv B, the final speed of the nucleus is v B v A + q(v A V B ) m + (.60 0 9 C).50 4.00 6.0 0 5 ms 0 3 V 6.63 0 7 kg 3.78 0 5 ms Thus, the correct answer is choice. 5. In a series combination of capacitors, the equivalent capacitance is always less than any individual capacitance in the combination, meaning that choice (a) is false. Also, for a series combination of capacitors, the magnitude of the charge is the same on all plates of capacitors in the combination, making both choices (d) and (e) false. The potential difference across the capacitance C i is ΔV i QC i, where Q is the common charge on each capacitor in the combination. Thus, the largest potential difference (voltage) appears across the capacitor with the least capacitance, making choice the correct answer. 6. The total potential at a point due to a set of point charges q i is V kq i r i i where r i is the distance from the point of interest to the location of the charge q i. Note that in this case, the point at the center of the circle is equidistant from the 4 point charges located on the rim of the circle. Note also that q + q 3 + q 4 +.5.0 0.5 mc 0, so we have V center k e q r + k e q r + k e q 3 r + k e q 4 r k e r ( q + q + q 3 + q 4 ) k e r ( q + 0) k q e V r 4.5 0 4 V or the total potential at the center of the circle is just that due to the first charge alone, and the correct answer is choice. 7. With the given specifications, the capacitance of this parallel-plate capacitor will be C k 0 A d (.00 0 )( 8.85 0 C N m )(.00 cm ).00 0 3 m 8.85 0 F 88.5 0 F 88.5 pf and the correct choice is (a). m 0 4 cm 8. Keeping the capacitor connected to the battery means that the potential difference between the plates is kept at a constant value equal to the voltage of the battery. Since the capacitance of a parallel-plate capacitor is C k 0 Ad, doubling the plate separation d, while holding other characteristics of the capacitor constant, means the capacitance will be decreased by a factor of. The energy stored in a capacitor may be expressed as U C(ΔV), so when the potential difference ΔV is held constant while the capacitance is decreased by a factor of, the stored energy decreases by a factor of, making the correct choice for this question. 9. When the battery is disconnected, there is no longer a path for charges to use in moving onto or off of the plates of the capacitor. This means that the charge Q is constant. The capacitance of a 6879_6_ch6_p030-06.indd 3 /7/ :3:39 PM

Electrical Energy and Capacitance 33 parallel-plate capacitor is C k 0 Ad and the dielectric constant is k when the capacitor is air filled. When a dielectric with dielectric constant k is inserted between the plates, the capacitance is doubled ( C f C i ). Thus, with Q constant, the potential difference between the plates, ΔV QC, is decreased by a factor of, meaning that choice (a) is a true statement. The electric field between the plates of a parallel-plate capacitor is E ΔVd and decreases when ΔV decreases, making choice (e) false and leaving (a) as the only correct choice for this question. 0. Once the capacitor is disconnected from the battery, there is no path for charges to move onto or off of the plates, so the charges on the plates are constant, and choice (e) can be eliminated. The capacitance of a parallel-plate capacitor is C k 0 Ad, so the capacitance decreases when the plate separation d is increased. With Q constant and C decreasing, the energy stored in the capacitor, U Q C, increases, making choice (a) false and choice true. The potential difference between the plates, ΔV QC Q d k 0 A, increases and the electric field between the plates, E ΔVd Q k 0 A, is constant. This means that both choices and (d) are false and leaves choice as the only correct response.. Capacitances connected in parallel all have the same potential difference across them and the equivalent capacitance, + C + C 3 +, is larger than the capacitance of any one of the capacitors in the combination. Thus, choice is a true statement. The charge on a capacitor is Q C(ΔV), so with ΔV constant, but the capacitances different, the capacitors all store different charges that are proportional to the capacitances, making choices (a),, (d), and (e) all false. Therefore, is the only correct answer.. For a series combination of capacitors, the magnitude of the charge is the same on all plates of capacitors in the combination. Also, the equivalent capacitance is always less than any individual capacitance in the combination. Therefore, choice (a) is true while choices and are both false. The potential difference across a capacitor is ΔV QC, so with Q constant, capacitors having different capacitances will have different potential differences across them, with the largest potential difference being across the capacitor with the smallest capacitance. This means that choices (d) and (e) are false, and choice (f) is true. Thus, both choices (a) and (f) are true statements. ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS. The potential energy between a pair of point charges separated by distance R is PE k e q q R. Thus, the potential energy for each of the four systems is: (a) Q( Q) Q PE a k e k e r r Q( Q) PE c k e r k Q e r ( Q) Q PE b k e r ( Q) Q (d) PE d k e r k e Q r k e Q r Therefore, the correct ranking from largest to smallest is (a) > (d) >. 4. To move like charges together from an infinite separation, at which the potential energy of the system of two charges is zero, requires work to be done on the system by an outside agent. Hence energy is stored, and potential energy is positive. As charges with opposite signs move together from an infinite separation, energy is released, and the potential energy of the set of charges becomes negative. 6. A sharp point on a charged conductor would produce a large electric field in the region near the point. An electric discharge could most easily take place at the point. 6879_6_ch6_p030-06.indd 33 /7/ :3:40 PM

34 Chapter 6 8. There are eight different combinations that use all three capacitors in the circuit. These combinations and their equivalent capacitances are: All three capacitors in series: + + C C 3 All three capacitors in parallel: + C + C 3 One capacitor in series with a parallel combination of the other two: + + C C 3, + C 3 + C, + C + C 3 One capacitor in parallel with a series combination of the other two: C C + C + C, C C 3 3 eq C 3 + + C, C C C 3 eq C + C 3 + 0. (a) If the wires are disconnected from the battery and not allowed to touch each other or another object, the charge on the plates is unchanged. If, after being disconnected from the battery, the wires are connected to each other, electrons will rapidly flow from the negatively charged plate to the positively charged plate to leave the capacitor uncharged with both plates neutral.. The primary choice would be the dielectric. You would want to choose a dielectric that has a large dielectric constant and dielectric strength, such as strontium titanate, where k 33 (Table 6.). A convenient choice could be thick plastic or Mylar. Secondly, geometry would be a factor. To maximize capacitance, one would want the individual plates as close as possible, since the capacitance is proportional to the inverse of the plate separation hence the need for a dielectric with a high dielectric strength. Also, one would want to build, instead of a single parallel-plate capacitor, several capacitors in parallel. This could be achieved through stacking the plates of the capacitor. For example, you can alternately lay down sheets of a conducting material, such as aluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none of the conducting sheets are in contact with their nearest neighbors, connect every other plate together as illustrated in the figure below. Dielectric Conductor This technique is often used when home-brewing signal capacitors for radio applications, as they can withstand huge potential differences without flashover (without either discharge between plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich together flexible materials such as aluminum roof flashing and thick plastic, so the whole product can be rolled up into a capacitor burrito and placed in an insulating tube, such as a PVC pipe, and then filled with motor oil (again to prevent flashover). 6879_6_ch6_p030-06.indd 34 /7/ :3:43 PM

Electrical Energy and Capacitance 35 4. The material of the dielectric may be able to withstand a larger electric field than air can withstand before breaking down to pass a spark between the capacitor plates. ANSWERS TO EVEN NUMBERED PROBLEMS. (a) 6.6 0 7 N 3.69 0 0 ms in the direction of the electric field 7.38 cm 4. 6.67 0 electrons 6. (a).0 0 N to the right.98 0 3 J.98 0 3 J (d) 49.5 V 8. (a).3 kv Protons would require a greater potential difference. ΔV p ΔV e m p m e 0. 40. kv. (a) +5.39 kv +0.8 kv 4. 9.08 J 6. (a) See Solution. 3k e qa See Solution. 8. (a) See Solution. V.5 V m.0 m x x 37.5 V (d) x 0.800 m 0. (a) Conservation of energy alone yields one equation with two unknowns. Conservation of linear momentum v p.05 0 7 ms, v α.64 0 6 ms. 5.4 0 5 V 4. (a) V 4 k e Qa W 4 k e qq a 6. (a) 3.00 mf 36.0 mc 8. (a).00 mf 00 V 30. 3.0 Å 3..3 kv 6879_6_ch6_p030-06.indd 35 /7/ 3::6 PM

36 Chapter 6 34. (a) 7.0 mf 9.00 V 45.0 mc on, 08 mc on C 36. 3.00 pf and 6.00 pf 38. (a) 6.00 mf.0 mf 43 mc (d) Q 4 44 mc, Q 7.0 mc, Q rightmost branch (f) 9.00 V (g) 7.0 V 6 mc (e) Q 4 Q 8 6 mc 40. (a) C Q > Q 3 > Q ΔV >ΔV ΔV 3 (d) Q and Q 3 increase, Q decreases 4. (a) 6.04 mf 83.6 mc 44. (a) 5.96 mf 89.4 mc on the 0.0 mf capacitor, 63.0 mc on the 6.00 mf capacitor, 6.3 mc on the 5.0 mf capacitor, and 6.3 mc on the 3.00 mf capacitor 46. (a).0 mf, E stored,total 8.64 0 4 J E stored, 5.76 0 4 J, E stored,.88 0 4 J It will always be true that E stored, + E stored, E stored,total. 5.66 V; C, with the largest capacitance, stores the most energy. 48. 9.79 kg 50. (a) 3.3 nc 7 nc 5..04 m 54. 0.443 mm 56. (a) 3.5 mj E stored, 3.60 mj, E stored,3 5.40 mj, E stored,4.80 mj, E stored,6.70 mj The energy stored in the equivalent capacitance equals the sum of the energies stored in the individual capacitors. 58. C p ± 4 C p C p C s, C C p 4 C p C p C s 60. (a).8 0 4 V 3.6 0 4 V.8 0 4 V (d) 5.4 0 J 6. (a) C k e ab b a See Solution. 6879_6_ch6_p030-06.indd 36 /7/ :3:46 PM

Electrical Energy and Capacitance 37 64. k.33 66. (a) k e q d 5 4k e q d 5 4k e q d 5 (d) 8k e q md 5 68. (a) 0. mm 4.4 mm PROBLEM SOLUTIONS 6. (a) Because the electron has a negative charge, it experiences a force in the direction opposite to the field and, when released from rest, will move in the negative x-direction. The work done on the electron by the field is W F x ( Δx) ( qe x )Δx.60 0 9 C.9 0 8 J ( 375 N C) ( 3.0 0 m) The change in the electric potential energy is the negative of the work done on the particle by the field. Thus, ΔPE W.9 0 8 J Since the Coulomb force is a conservative force, conservation of energy gives ΔKE +ΔPE 0, or KE f m e v f KE i ΔPE 0 ΔPE, and v f ( ΔPE ) m e (.9 0 8 J).05 0 6 m s in the x-direction 9. 0 3 kg 6. (a) F qe (.60 0 9 C) ( 385 N C) 6.6 0 7 N a F m p 6.6 0 7 N.67 0 7 kg 3.69 00 ms in the direction of the electric field Δx v 0 t + at 0 + ( 3.69 00 ms )(.00 0 6 s) 7.38 0 m 7.38 cm 6.3 The work done by the agent moving the charge out of the cell is W input W field ( ΔPE e )+q( ΔV ) (.60 0 9 C) ( + 90 0 3 JC).4 0 0 J 6.4 Assuming the sphere is isolated, the excess charge on it is uniformly distributed over its surface. Under this spherical symmetry, the electric field outside the sphere is the same as if all the excess charge on the sphere were concentrated as a point charge located at the center of the sphere. continued on next page 6879_6_ch6_p030-06.indd 37 /7/ :3:50 PM

38 Chapter 6 Thus, at r 8.00 cm > R sphere 5.00 cm, the electric field is E k e Qr. The required charge then has magnitude Q Er k e, and the number of electrons needed is ( 8.00 0 m) (.60 0 9 C) 6.67 0 electrons n Q e Er k e e.50 0 5 NC 8.99 0 9 N m C 6.5 E ΔV d 5 03 JC.5 0 m.7 06 NC 6.6 (a) F q E +40.0 0 6 C +75 N C.0 0 N directed toward the right (d) W AB F ( Δx)cosq (.0 0 N) ( 0.80 m)cos0.98 0 3 J ΔPE W AB.98 0 3 J ΔV V B V A ΔPE q.98 0 3 J +40.0 0 6 C 49.5 V 6.7 (a) E ΔV d 600 J C 5.33 0 3 m.3 05 NC F q E q ΔV d W F s cosq 600 J C.60 0 9 C 5.33 0 3 m.80 0 4 N 5.33.90.80 0 4 N 0 3 m cos 0 4.37 0 7 J 6.8 (a) Using conservation of energy, ΔKE +ΔPE 0, with KE f 0 since the particle is stopped, we have ΔPE ΔKE 0 m v e i + 9. 0 3 kg +3.70 0 6 J The required stopping potential is then ΔV ΔPE q (.85 0 7 ms) +3.70 0 6 J.60 0 9 C.3 03 V.3 kv Being more massive than electrons, protons traveling at the same initial speed will have more initial kinetic energy and require a greater magnitude stopping potential. q mv Since ΔV stopping ΔPE q ΔKE proton to that for an electron having the same initial speed is q, the ratio of the stopping potential for a ΔV p ΔV e m v p i (+e) m e v i ( e) m p m e 6879_6_ch6_p030-06.indd 38 /7/ :3:5 PM

Electrical Energy and Capacitance 39 6.9 (a) We use conservation of energy, Δ( KE )+Δ( PE s )+Δ( PE e ) 0, recognizing that Δ(KE) 0 since the block is at rest at both the beginning and end of the motion. The change in the elastic potential energy is given by Δ ( PE s ) kx max 0, where x max is the maximum stretch of the spring. The change in the electrical potential energy is the negative of the work the electric field does, Δ ( PE e ) W F e (Δx) ( QE )x max. Thus, 0 + kx max ( QE )x max 0, which yields x max QE k ( 4.86 0 4 Vm) 35.0 0 6 C 78.0 N m 4.36 0 m 4.36 cm At equilibrium, ΣF F s + F e 0, or kx eq + QE 0. Therefore, x eq QE k x max.8 cm The amplitude is the distance from the equilibrium position to each of the turning points at x 0 and x 4.36 cm, so A.8 cm x max. From conservation of energy, Δ( KE )+Δ( PE s )+Δ( PE e ) 0, we have 0 + kx max + QΔV 0. Since x max A, this gives ΔV kx max Q A k Q or ΔV ka Q 6.0 Using Δy v 0 y t + a y t for the full flight gives 0 v 0 y t f + a y t f, or a y v 0 y t f, where t f is the full time of the flight. Then, using v y v 0 y + a y (Δy) for the upward part of the flight gives v 0 y ( Δy) max 0 v 0 y a y v 0 y t f From Newton s second law, v t 0 y f 4 ( 4.0 s) 0. m s 4 0.6 m a y ΣF y m mg qe m g + qe m Equating this to the earlier result gives a y ( g + qe m) v 0 y t f, so the electric field strength is E m q v 0 y t f 4.0 s g.00 kg 0. m s 5.00 0 6 C 9.80 m s.95 0 3 NC Thus, ( ΔV ) max ( Δy max )E ( 0.6 m) (.95 0 3 NC) 4.0 0 4 V 40. kv 6. (a) V A k eq r A V B k eq r B ( 8.99 0 9 N m C ).60 0 9 C 0.50 0 m ( 8.99 0 9 N m C ).60 0 9 C 0.750 0 m 5.75 0 7 V.9 0 7 V continued on next page 6879_6_ch6_p030-06.indd 39 /7/ :3:54 PM

40 Chapter 6 ΔV V B V A.9 0 7 V ( 5.75 0 7 V)+3.83 0 7 V No. The original electron will be repelled by the negatively charged particle which suddenly appears at point A. Unless the electron is fixed in place, it will move in the opposite direction, away from points A and B, thereby lowering the potential difference between these points. 6. (a) V A V B k e q i 8.99 0 9 N m C i r i k e q i 8.99 0 9 N m C i r i 5.0 0 9 C.00 0 m + 7.0 0 9 C.00 0 m +5.39 kv 5.0 0 9 C.00 0 m + 7.0 0 9 C.00 0 m +0.8 kv 6.3 (a) Calling the.00 mc charge q 3, V k e q i q k e + q + i r i r r 8.99 0 9 N m C q 3 r + r 8.00 0 6 C 0.060 0 m + 4.00 0 6 C + ( 0.030 0) m 0.030 0 m +.00 0 6 C 0.060 0 V.67 0 6 V Replacing.00 0 6 C by.00 0 6 C in part (a) yields V.3 0 6 V 6.4 W q( ΔV ) qv ( f V i ), and V f 0 since the final location of the 8.00 mc is an infinite distance from other charges. The potential, due to the other charges, at the initial location of the 8.00 mc is V i k e ( q r + q r ). Thus, the required energy for the move is 6.5 (a) V q W q 0 k e + q r r 8.99 0 9 N m 8.00 0 6 C W 9.08 J k e q i 8.99 0 9 N m i r i C PE k q q e i 8.99 0 9 N m r C C.00 0 6 C 0.030 0 m + 4.00 0 6 C 0.030 0 + ( 0.060 0) m 5.00 0 9 C 0.75 m 3.00 0 9 C 0.75 m 03 V ( 5.00 0 9 C) 3.00 0 9 C 0.350 m 3.85 0 7 J The negative sign means that positive work must be done to separate the charges by an infinite distance (that is, bring them up to a state of zero potential energy). 6879_6_ch6_p030-06.indd 40 /7/ :3:57 PM

Electrical Energy and Capacitance 4 6.6 (a) At the center of the triangle, each of the identical charges produce a field contribution of magnitude E k e q a. The three contributions are oriented as shown at the right and the components of the resultant field are: E 30 y E 30 x E x ΣE x +E cos30 E cos30 0 E y ΣE y +E sin 30 E + E sin 30 0 Thus, the resultant field has magnitude E E E x + E y 0 The total potential at the center of the triangle is V ΣV i Σ k e q i r i k q e a + k q e a + k q e a 3k q e a Imagine a test charge placed at the center of the triangle. Since the field is zero at the center, the test charge will experience no electrical force at that point. The fact that the potential is not zero at the center means that work would have to be done by an external agent to move a test charge from infinity to the center. 6.7 The Pythagorean theorem gives the distance from the midpoint of the base to the charge at the apex of the triangle as r 3 ( 4.00 cm) (.00 cm) 5 cm 5 0 m Then, the potential at the midpoint of the base is V k e q i r i, or i V 8.99 0 9 N m ( 7.00 0 9 C) C + 7.00 0 9 C 0.00 0 m 0.00 0 m + +7.00 0 9 C 5 0 m.0 0 4 V.0 kv 6.8 (a) See the sketch below: 000 y 500 x. 0.8 0.4 0 0.4 0.8..6 500 000 continued on next page 6879_6_ch6_p030-06.indd 4 /7/ :3:59 PM

4 Chapter 6 At the point (x, 0), where 0 < x <.0 m, the potential is V k e q i k e q i r i x k + e q.0 m x k q e.0 m x x or V 8.99 0 9 N m C.50 0 9 C.0 m x x (.5 V m).0 m x x At x +0.600 m, the potential is V.5 V m.0 m 0.600 m 0.600 m.5 V m 0.600 m 37.5 V (d) When 0 < x <.0 m and V 0, we have (.0 m x) x 0, or x.40 m x. This yields x.40 m 3 0.800 m. 6.9 (a) When the charge configuration consists of only the two protons ( q and q in the sketch), the potential energy of the configuration is PE a k e q q r or PE a 3.84 0 4 J ( 8.99 0 9 N m C ).60 0 9 C 6.00 0 5 m When the alpha particle ( q 3 in the sketch) is added to the configuration, there are three distinct pairs of particles, each of which possesses potential energy. The total potential energy of the configuration is now PE b k q q e + k q q e 3 + k q q e 3 PE a + k e e r r 3 r 3 r 3 where use has been made of the facts that q q 3 q q 3 e( e) e and + ( 3.00 fm) 8.0 fm 8.0 0 5 m. Also, note that the first r 3 r 3 3.00 fm term in this computation is just the potential energy computed in part (a). Thus, PE b PE a + 4k ee r 3 (.60 0 9 C) 3.84 0 4 J + 4 8.99 09 N m C 8.0 0 5 m.55 0 3 J If we start with the three-particle system of part and allow the alpha particle to escape to infinity [thereby returning us to the two-particle system of part (a)], the change in electric potential energy will be ΔPE PE a PE b 3.84 0 4 J.55 0 3 J.7 0 3 J (d) Conservation of energy, ΔKE +ΔPE 0, gives the speed of the alpha particle at infinity in the situation of part as m a v a 0 ΔPE, or continued on next page 6879_6_ch6_p030-06.indd 4 /7/ :33:00 PM

Electrical Energy and Capacitance 43 v a ( ΔPE ) m a (.7 0 3 J) 8.08 0 6 ms 6.64 0 7 kg (e) When, starting with the three-particle system, the two protons are both allowed to escape to infinity, there will be no remaining pairs of particles and hence no remaining potential energy. Thus, ΔPE 0 PE b PE b, and conservation of energy gives the change in kinetic energy as ΔKE ΔPE +PE b. Since the protons are identical particles, this increase in kinetic energy is split equally between them giving KE proton m p v p ( PE b ), or v p PE b m p.55 0 3 J.67 0 7 kg.4 07 ms 6.0 (a) If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and allowed to recede to infinity, the final speeds of the two particles will differ because of the difference in the masses of the particles. Thus, attempting to solve for the final speeds by use of conservation of energy alone leads to a situation of having one equation with two unknowns, and does not permit a solution. In the situation described in part (a) above, one can obtain a second equation with the two unknown final speeds by using conservation of linear momentum. Then, one would have two equations which could be solved simultaneously for both unknowns. From conservation of energy: m a v a + ( m p v p ) 0 + 0 k q q e a p m a v a + m p v p k e q a q p r i ( 8.99 0 9 N m C ) 3.0 0 9 C r i 0, or (.60 0 9 C) 4.00 0 5 m yielding m a v a + m p v p.30 0 3 J [] From conservation of linear momentum, m a v a + m p v p 0 or v a m p v p [] m a Substituting Equation [] into Equation [] gives m a m p m a v p + m p v p.30 0 3 J or m p + m a m pv p.30 0 3 J and v p.30 0 3 J ( m p m a +)m p.30 0 3 J (.67 0 7 6.64 0 7 +).67 0 7 kg.05 07 ms Then, Equation [] gives the final speed of the alpha particle as v a m p v.67 0 7 kg p 6.64 0 7 kg (.05 07 ms).64 0 6 ms m a 6879_6_ch6_p030-06.indd 43 /7/ :33:03 PM

44 Chapter 6 6. (a) Conservation of energy gives KE f KE i + ( PE i PE f ) 0 + k e q q r i r f With q +8.50 nc, q.80 nc, r i.60 mm, and r f 0.500 mm, this becomes KE f 8.99 0 9 N m C 8.50 0 9 C yielding KE f 0.94 J (.80 0 9 C).60 0 6 m 0.500 0 6 m When r r f 0.500 mm and KE KE f 0.94 J, the speed of the sphere having mass m 8.00 mg 8.00 0 6 kg is v f ( KE f ) m ( 0.94 J) 8.00 0 6 kg 7 m s 6. The excess charge on the metal sphere will be uniformly distributed over its surface. In this spherically symmetric situation, the electric field and the electric potential outside the sphere is the same as if all the excess charge were concentrated as a point charge at the center of the sphere. Thus, for points outside the sphere, Q E k e r and V k Q e r E r If the sphere has a radius of r 8 cm 0.8 m and the air breaks down when E 3.0 0 6 the electric potential at the surface of the sphere when breakdown occurs is Vm, V ( 3.0 0 6 Vm) ( 0.8 m) 5.4 0 5 V 6.3 From conservation of energy, KE + PE e 0 + k e Qq r f m a v i + 0, or ( e) r f k Qq e k e 79e m a v i m a v i r f 8.99 09 N m C 6.64 0 7 kg f ( KE + PE e ) i, which gives ( 58) (.60 0 9 C).74 0 (.00 0 7 ms) 4 m 6.4 (a) The distance from any one of the corners of the square to the point at the center is one half the length of the diagonal of the square, or r diagonal a + a a a Since the charges have equal magnitudes and are all the same distance from the center of the square, they make equal contributions to the total potential. Thus, V total 4V single 4 k Q e charge r 4 k Q e a 4 k Q e a continued on next page 6879_6_ch6_p030-06.indd 44 /7/ :33:05 PM

Electrical Energy and Capacitance 45 The work required to carry charge q from infinity to the point at the center of the square is equal to the increase in the electric potential energy of the charge, or Q qq W PE center PE qv total 0 q 4 k e a 4 k e a A 6.5 (a) C 0 d 8.85 C.0 0 6 m 0 N m 800 m. 0 8 F A Q max C( ΔV ) max C( E max d) 0 d ( E d max ) 0 AE max ( 8.85 0 C N m )(.0 0 6 m )( 3.0 0 6 NC) 7 C 6.6 (a) C Q ΔV 7.0 mc 9.00 V 3.00 mf Q C( ΔV ) ( 3.00 mf) (.0 V) 36.0 mc 6.7 (a) The capacitance of this air-filled ( dielectric constant, k.00) parallel-plate capacitor is C k 0 A d 8.85 0 C N m.00 (.30 0 4 m ).50 0 3 m.36 0 F.36 pf Q C ( ΔV ) (.36 0 F) (.0 V).63 0 C 6.3 0 C 6.3 pc E ΔV d.0 V.50 0 3 m 8.00 03 Vm 8.00 0 3 NC 6.8 (a) C Q V 0.0 mc 0.0 V.00 mf V Q C 00 mc.00 mf 00 V 6.9 (a) E ΔV d 0.0 V.80 0 3 m. 04 Vm. kv m toward the negative plate C 0 A d ( 7.60 0 4 m ) 8.85 0 C N m.80 0 3 m 3.74 0 F 3.74 pf 0.0 V Q C ( ΔV ) 3.74 0 F 7.48 0 C 74.8 pc on one plate and 74.8 pc on the other plate. 6.30 C 0 Ad, so d 0 A C d 3.0 0 9 m (.0 0 m ) 8.85 0 C N m 60.0 0 5 F Å 0 0 m 3.0 Å 3.0 0 9 m 6879_6_ch6_p030-06.indd 45 /7/ :33:08 PM

46 Chapter 6 6.3 (a) Assuming the capacitor is air-filled ( k ), the capacitance is ( 0.00 m ) C A 0 8.85 0 C N m d 3.00 0 3 m 5.90 0 0 F Q C ( ΔV ) ( 5.90 0 0 F) ( 6.00 V) 3.54 0 9 C E ΔV d 6.00 V 3.00 0 3 m.00 03 Vm.00 0 3 NC (d) (e) s Q A 3.54 0 9 C 0.00 m.77 0 8 Cm Increasing the distance separating the plates decreases the capacitance, the charge stored, and the electric field strength between the plates. This means that all of the previous answers will be decreased. 6.3 ΣF y 0 T cos5.0 mg or T mg cos5.0 ΣF x 0 qe T sin5.0 mg tan5.0 or E mg tan5.0 q ΔV Ed mgd tan5.0 q ( 9.80 m s ) 0.040 0 m ΔV 350 0 6 kg tan5.0.3 0 3 V.3 kv 30.0 0 9 C 6.33 (a) Capacitors in a series combination store the same charge, Q (ΔV), where is the equivalent capacitance and ΔV is the potential difference maintained across the series combination. The equivalent capacitance for the given series combination is + C, or C ( + C ), giving (.50 mf) 6.5 mf.50 mf + 6.5 mf.79 mf and the charge stored on each capacitor in the series combination is 6.00 V Q ( ΔV ).79 mf 0.7 mc When connected in parallel, each capacitor has the same potential difference, ΔV 6.00 V, maintained across it. The charge stored on each capacitor is then For.50 mf: Q ( ΔV ) (.50 mf) ( 6.00 V) 5.0 mc For C 6.5 mf: Q C ( ΔV ) ( 6.5 mf) ( 6.00 V) 37.5 mc 6879_6_ch6_p030-06.indd 46 /7/ :33:0 PM

Electrical Energy and Capacitance 47 6.34 (a) + C 5.00 mf +.0 mf 7.0 mf In a parallel combination, the full potential difference maintained between the terminals of the battery exists across each capacitor. Thus, ΔV ΔV ΔV battery 9.00 V Q ( ΔV ) ( 5.00 mf) ( 9.00 V) 45.0 mc Q C ( ΔV ) (.0 mf) ( 9.00 V) 08 mc 6.35 (a) First, we replace the parallel combination between points b and c by its equivalent capacitance, C bc.00 mf + 6.00 mf 8.00 mf. Then, we have three capacitors in series between points a and d. The equivalent capacitance for this circuit is therefore + + 3 C ab C bc C cd 8.00 mf giving 8.00 mf 3.67 mf The charge stored on each capacitor in the series combination is Q ab Q bc Q cd ( ΔV ad ) (.67 mf) ( 9.00 V) 4.0 mc Then, note that ΔV bc Q bc in the original circuit is: C bc 4.0 mc 8.00 mf 3.00 V. The charge on each capacitor On the 8.00 mf between a and b: On the 8.00 mf between c and d: Q 8 Q ab 4.0 mc Q 8 Q cd 4.0 mc On the.00 mf between b and c: Q C ( ΔV bc ) (.00 mf) ( 3.00 V) 6.00 mc On the 6.00 mf between b and c: Q 6 C 6 ( ΔV bc ) ( 6.00 mf) ( 3.00 V) 8.0 mc Note that ΔV ab Q ab C ab 4.0 mc 8.00 mf 3.00 V, and that ΔV cd Q cd C cd 4.0 mc 8.00 mf 3.00 V. We earlier found that ΔV bc 3.00 V, so we conclude that the potential difference across each capacitor in the circuit is ΔV 8 ΔV ΔV 6 ΔV 8 3.00 V 6.36 C parallel + C 9.00 pf 9.00 pf C [] C series + C C series C + C.00 pf ( 9.00 pf C Thus, using Equation [], C series )C.00 pf which reduces to ( 9.00 pf C )+ C continued on next page 6879_6_ch6_p030-06.indd 47 /7/ :33:3 PM

48 Chapter 6 C ( 9.00 pf)c + 8.0 ( pf) 0, or ( C 6.00 pf) ( C 3.00 pf) 0 Therefore, either C 6.00 pf and, from Equation [], 3.00 pf or C 3.00 pf and 6.00 pf. We conclude that the two capacitances are 3.00 pf and 6.00 pf. 6.37 (a) The equivalent capacitance of the series combination in the upper branch is 3.00 mf 6.00 mf C upper 3.00 mf + 6.00 mf + 6.00 mf or C upper.00 mf.00 mf 4.00 mf Likewise, the equivalent capacitance of the series combination in the lower branch is C lower.00 mf + 4.00 mf + 4.00 mf or C lower.33 mf 90.0 V These two equivalent capacitances are connected in parallel with each other, so the equivalent capacitance for the entire circuit is C upper + C lower.00 mf +.33 mf 3.33 mf Note that the same potential difference, equal to the potential difference of the battery, exists across both the upper and lower branches. The charge stored on each capacitor in the series combination in the upper branch is Q 3 Q 6 Q upper C upper ( ΔV ) (.00 mf) ( 90.0 V) 80 mc and the charge stored on each capacitor in the series combination in the lower branch is Q Q 4 Q lower C lower ( ΔV ) (.33 mf) ( 90.0 V) 0 mc The potential difference across each of the capacitors in the circuit is: ΔV Q C ΔV 3 Q 3 C 3 0 mc.00 mf 60.0 V ΔV 4 Q 4 C 4 80 mc 3.00 mf 60.0 V ΔV 6 Q 6 C 6 6.38 (a) The equivalent capacitance of the series combination in the rightmost branch of the circuit is 0 mc 4.00 mf 30.0 V 80 mc 6.00 mf 30.0 V or C right 4.0 mf + 8.00 mf + 3 4.0 mf C right 6.00 mf Figure P6.38 continued on next page 6879_6_ch6_p030-06.indd 48 /7/ :33:6 PM

Electrical Energy and Capacitance 49 The equivalent capacitance of the three capacitors now connected in parallel with each other and with the battery is 4.00 mf +.00 mf + 6.00 mf.0 mf The total charge stored in this circuit is Q total ( ΔV ).0 mf ( 36.0 V) Diagram or Q total 43 mc (d) The charges on the three capacitors shown in Diagram are: Q 4 C 4 ( ΔV ) ( 4.00 mf) ( 36.0 V) 44 mc 36.0 V Q C ( ΔV ).00 mf 7 mc 36.0 V Q right C right ( ΔV ) 6.00 mf 6 mc Yes. Q 4 + Q + Q right Q total as it should. Diagram (e) The charge on each capacitor in the series combination in the rightmost branch of the original circuit (Figure P6.38) is Q 4 Q 8 Q right 6 mc (f) ΔV 4 Q 4 C 4 (g) ΔV 8 Q 8 C 8 6 mc 4.0 mf 9.00 V 6 mc 8.00 mf 7.0 V Note that ΔV 8 +ΔV 4 ΔV 36.0 V as it should. 6.39 The circuit may be reduced in steps as shown above. Using Figure 3, Q ac ( 4.00 mf) ( 4.0 V) 96.0 mc Then, in Figure, ( ΔV ) ab Q ac C ab 96.0 mc 6.00 mf 6.0 V continued on next page 6879_6_ch6_p030-06.indd 49 /7/ :33:9 PM

50 Chapter 6 and ( ΔV ) bc ( ΔV ) ac ( ΔV ) ab 4.0 V 6.0 V 8.00 V Finally, using Figure, 6.0 V Q ( ΔV ) ab.00 mf 6.0 mc, Q 5 5.00 mf ( ΔV ) ab 80.0 mc Q 8 ( 8.00 mf) ( ΔV ) bc 64.0 mc, and Q 4 ( 4.00 mf) ( ΔV ) bc 3.0 mc 6.40 (a) Consider the simplification of the circuit as shown below: a a ΔV + C C 3 ΔV + C ab ΔV + b b Since C and C 3 are connected in parallel, C ab C + C 3 C + 5C 6C. Now observe that and C ab are connected in series, giving + or C ab C ab + C ab ( 6C ) 3C 3C + 6C C Since capacitors and C ab are connected in series, Q Q ab Q eq ( ΔV ) C ( ΔV ) Then, ΔV ab Q ab C ab C ( ΔV ) ΔV 6C 3, giving Q C ΔV ab C ΔV 3 3 5C ΔV Also, Q 3 C 3 ( ΔV ab ). Therefore, Q > Q 3 > Q Since capacitors and C ab are in series with the battery, ΔV ΔV ΔV ab ΔV ΔV 3 3 ΔV Also, with capacitors C and C 3 in parallel between points a and b, ΔV ΔV 3 ΔV ab ΔV 3 Thus, ΔV >ΔV ΔV 3 (d) Consider the following steps: (i) Increasing C 3 while and C remain constant will increase C ab C + C 3. C Therefore, the equivalent capacitance, C ab + C ab, will increase. continued on next page 6879_6_ch6_p030-06.indd 50 /7/ :33: PM

Electrical Energy and Capacitance 5 (ii) Since and C ab are in series, Q Q ab ( ΔV ). Thus, Q will increase as increases. Also, Q ab experiences the same increase. (iii) Because ΔV Q, an increase in Q causes ΔV to increase and causes ΔV ab ΔV ΔV to decrease. Thus, since Q C ( ΔV ab ), Q will decrease. (iv) With capacitors C and C 3 in parallel between points a and b, we have Q ab Q + Q 3 or Q 3 Q ab Q. Thus, with Q ab increasing [see Step (ii)] while Q is decreasing [see Step (iii)], we see that Q 3 will increase. 6.4 (a) From Q C( ΔV ), Q 5 ( 5.0 mf) ( 50.0 V).5 0 3 mc.5 mc and Q 40 ( 40.0 mf) ( 50.0 V).00 0 3 mc.00 mc Since the negative plate of one capacitor was connected to the positive plate of the other, the net charge stored in the new parallel combination is Q Q 40 Q 5.00 0 3 mc.5 0 3 mc 750 mc The two capacitors, now in parallel, have a common potential difference ΔV across them. The new charges on each of the capacitors are Q 5 ( ΔV ) and Q 40 C ( ΔV ). Thus, Q 5 Q 40 C 5 mf 40 mf Q 40 5 8 and the total change now stored in the combination may be written as Q 40 Q Q 40 + Q 5 Q 40 + 5 8 Q 3 40 Q 40 750 mc 8 giving Q 40 8 ( 750 mc) 46 mc and 3 Q Q 5 Q 40 ( 750 46) mc 88 mc The potential difference across each capacitor in the new parallel combination is ΔV Q Q + C 750 mc 65.0 mf.5 V 6.4 (a) The original circuit reduces to a single equivalent capacitor in the steps shown below. eq continued on next page 6879_6_ch6_p030-06.indd 5 /7/ :33:5 PM

5 Chapter 6 C s + C C p C s + C 3 + C s 5.00 mf + 0.0 mf 3.33 mf ( 3.33 mf)+.00 mf 8.66 mf C p C + C ( 0.0 mf) 0.0 mf + C p C p 8.66 mf + 0.0 mf 6.04 mf The total charge stored between points a and b is 60.0 V Q total ( ΔV ) ab 6.04 mf 36 mc Then, looking at the third figure, observe that the charges of the series capacitors of that figure are Q p Q p Q total 36 mc. Thus, the potential difference across the upper parallel combination shown in the second figure is ( ΔV ) p Q p 36 mc C p 8.66 mf 4.8 V Finally, the charge on C 3 is 4.8 V Q 3 C 3 ( ΔV ) p.00 mf 83.6 mc 6.43 From Q C( ΔV ), the initial charge of each capacitor is Q (.00 mf) ( 0.0 V) 0.0 mc and Q (.00 mf) ( 0) 0 After the capacitors are connected in parallel, the potential difference across one is the same as that across the other. This gives Q ΔV.00 mf Q.00 mf or Q Q [] From conservation of charge, Q + Q Q + Q 0.0 mc. Then, substituting from Equation [], this becomes Q + Q 0.0 mc, giving Q 0 mc 3 3.33 mc Finally, from Equation [], Q 0 mc 3 6.67 mc 6.44 (a) We simplify the circuit in stages as shown below: 5.0 mf 3.00 mf C s a C p b a b a b a b 0.0 mf 0.0 mf 0.0 mf 6.00 mf 6.00 mf continued on next page 6879_6_ch6_p030-06.indd 5 /7/ :33:8 PM

Electrical Energy and Capacitance 53 C s 5.0 mf + 3.00 mf ( 5.0 mf) 3.00 mf or C s 5.0 mf + 3.00 mf.50 mf C p C s + 6.00 mf.50 mf + 6.00 mf 8.50 mf + C p 0.0 mf ( 8.50 mf) 0.0 mf or 8.50 mf + 0.0 mf 5.96 mf Q 0 Q C p Q eq ( ΔV ab ) ( 5.96 mf) ( 5.0 V) 89.4 mc ΔV C p Q C p C p 89.4 mc 8.50 mf 0.5 V so Q 6 C 6 ( ΔV C p ) 6.00 mf and Q 5 Q 3 Q s C s ( ΔV C p ).50 mf ( 0.5 V) 63.0 mc ( 0.5 V) 6.3 mc The charges are 89.4 mc on the 0 mf capacitor, 63.0 mc on the 6 mf capacitor, and 6.3 mc on both the 5 mf and 3 mf capacitors. 6.45 Energy stored Q C C( ΔV 4.50 0 6 F) (.0 V) 3.4 0 4 J 6.46 (a) The equivalent capacitance of a series combination of and C is 8.0 mf + 36.0 mf + 36.0 mf or.0 mf When this series combination is connected to a.0-v battery, the total stored energy is.0 V Total energy stored ( ΔV ).0 0 6 F 8.64 0 4 J The charge stored on each of the two capacitors in the series combination is.0 V Q Q Q total ( ΔV ).0 mf 44 mc.44 0 4 C and the energy stored in each of the individual capacitors is Energy stored in Q and Energy stored in C Q C (.44 0 4 C) 8.0 0 6 F 5.76 0 4 J (.44 0 4 C) 36.0 0 6 F.88 0 4 J Energy stored in + Energy stored in C 5.76 0 4 J +.88 0 4 J 8.64 0 4 J, which is the same as the total stored energy found in part (a). This must be true if the computed equivalent capacitance is truly equivalent to the original combination. continued on next page 6879_6_ch6_p030-06.indd 53 /7/ :33:3 PM

54 Chapter 6 If and C had been connected in parallel rather than in series, the equivalent capacitance would have been + C 8.0 mf + 36.0 mf 54.0 mf. If the total energy stored ( ΔV ) in this parallel combination is to be the same as was stored in the original series combination, it is necessary that ΔV ( Total energy stored ) 4 8.64 0 J 54.0 0 6 F 5.66 V Since the two capacitors in parallel have the same potential difference across them, the energy stored in the individual capacitors C( ΔV ) is directly proportional to their capacitances. The larger capacitor, C, stores the most energy in this case. 6.47 (a) The energy initially stored in the capacitor is ( Energy stored) Q i C i C i ΔV i ( 3.00 mf ) ( 6.00 V 54.0 mj ) When the capacitor is disconnected from the battery, the stored charge becomes isolated with no way off the plates. Thus, the charge remains constant at the value Q i as long as the capacitor remains disconnected. Since the capacitance of a parallel-plate capacitor is C k 0 Ad, when the distance d separating the plates is doubled, the capacitance is decreased by a factor of ( C f C i.50 mf). The stored energy (with Q unchanged) becomes ( Energy stored) Q i Q i C f C i C i Q i ( Energy stored ) 08 mj When the capacitor is reconnected to the battery, the potential difference between the plates is reestablished at the original value of ΔV ( ΔV ) i 6.00 V, while the capacitance remains at C f C i.50 mf. The energy stored under these conditions is ( Energy stored) 3 C f ( ΔV ) i (.50 mf ) ( 6.00 V 7.0 mj ) 6.48 The energy transferred to the water is W 00 Q( ΔV ) 50.0 C 00.00 08 V.50 0 7 J Thus, if m is the mass of water boiled away, W m c( ΔT )+ L v becomes.50 0 7 J m 486 J kg C ( 00 C 30.0 C )+.6 06 J kg.50 0 7 J giving m 9.79 kg.93 0 5 J kg +.6 0 6 J kg 6.49 (a) Note that the charge on the plates remains constant at the original value, Q 0, as the dielectric is inserted. Thus, the change in the potential difference, ΔV QC, is due to a change in capacitance alone. The ratio of the final and initial capacitances is C f k Ad 0 C i 0 Ad k and C f Q 0 ΔV C i ΔV Q 0 f ( ΔV ) i 85.0 V i ( ΔV ) f 5.0 V 3.40 continued on next page 6879_6_ch6_p030-06.indd 54 /7/ :33:34 PM

Electrical Energy and Capacitance 55 Thus, the dielectric constant of the inserted material is k 3.40, and the material is probably nylon (see Table 6.). If the dielectric only partially filled the space between the plates, leaving the remaining space air-filled, the equivalent dielectric constant would be somewhere between k.00 (air) and k 3.40. The resulting potential difference would then lie somewhere between ΔV i 85.0 V and ( ΔV ) f 5.0 V. 6.50 (a) If the maximum electric field that can exist between the plates before breakdown (i.e., the dielectric strength) is E max, the maximum potential difference across the plates is ΔV max E max d, where d is the plate separation. The maximum change on either plate then has magnitude Q max C ( ΔV max ) C( E max d) Since the capacitance of a parallel-plate capacitor is C k 0 Ad, the maximum charge is Q max k A 0 d ( E d max ) k 0 AE max The area of each plate is A 5.00 cm 5.00 0 4 m, and when air is the dielectric, k.00 and E max 3.00 0 6 V m (see Table 6.). Thus, Q max (.00) ( 8.85 0 CN m )( 5.00 0 4 m )( 3.00 0 6 Vm).33 0 8 C 3.3 nc If the dielectric is now polystyrene ( k.56 and E max 4.0 0 6 Vm), then Q max (.56) ( 8.85 0 CN m )( 5.00 0 4 m )( 4.0 0 6 Vm).7 0 7 C 7 nc 6.5 (a) The dielectric constant for Teflon is k., so the capacitance is C k 0 A d 8.85 0 C N m. ( 75 0 4 m ) 0.040 0 0 3 m C 8. 0 9 F 8. nf For Teflon, the dielectric strength is E max 60 0 6 Vm, so the maximum voltage is ΔV max E max d ( 60 0 6 Vm) ( 0.040 0 0 3 m ) ΔV max.4 0 3 V.4 kv 6.5 Before the capacitor is rolled, the capacitance of this parallel-plate capacitor is C k 0 A d k 0 ( w L) d where A is the surface area of one side of a foil strip. Thus, the required length is L C d 8 k 0 w 9.50 0 F 3.70 ( 0.05 0 0 3 m) ( 7.00 0 m).04 m 8.85 0 C N m 6879_6_ch6_p030-06.indd 55 /7/ :33:36 PM

56 Chapter 6 6.53 (a) V m r.00 0 kg 00 kg m 3 9.09 0 6 m 3 Since V 4pr 3 3, the radius is r [ 3V 4p ] 3, and the surface area is A 4pr 4p 3V 3 4p 4p 3 9.09 0 6 m 3 4p C k 0 A d 8.85 0 C N m 5.00 ( 4.54 0 0 m ) 00 0 9 m Q C( ΔV ) (.0 0 3 F) ( 00 0 3 V).0 0 4 C 3 4.54 0 0 m.0 0 3 F and the number of electronic charges is n Q e.0 0 4 C.6.60 0 9 05 C ΔV 6.54 For a parallel-plate capacitor, C k 0 Ad and Q s A C( ΔV ). Thus, s A k 0 A d, and d ( k 0 s )( ΔV ). With air as the dielectric material ( k.00), the separation of the plates must be ( 50 V) ( 0 4 cm m ) 4.43 0 4 m 0.443 mm 8.85 0 C N m d.00 3.00 0 0 C cm 6.55 Since the capacitors are in series, the equivalent capacitance is given by + + d C C 3 0 A + d 0 A + d 3 0 A d + d + d 3 A 0 or 0 A d where d d + d + d 3 6.56 (a) Please refer to the solution of Problem 6.37 where the following results were obtained: 3.00 mf 6.00 mf 3.33 mf Q 3 Q 6 80 mc Q Q 4 0 mc The total energy stored in the full circuit is then ( Energy stored) total ( ΔV ) ( 3.33 0 6 F) ( 90.0 V).00 mf 4.00 mf.35 0 J 3.5 0 3 J 3.5 mj 90.0 V The energy stored in each individual capacitor is For.00 mf: For 3.00 mf: ( Energy stored) Q 0 0 6 C C.00 0 6 F ( Energy stored) 3 Q 80 0 6 C 3 C 3 3.00 0 6 F 3.60 0 3 J 3.60 mj 5.40 0 3 J 5.40 mj continued on next page 6879_6_ch6_p030-06.indd 56 /7/ :33:39 PM

Electrical Energy and Capacitance 57 For 4.00 mf: For 6.00 mf: ( Energy stored) 4 Q 0 0 6 C 4 C 4 4.00 0 6 F ( Energy stored) 6 Q 80 0 6 C 6 C 6 6.00 0 6 F.80 0 3 J.80 mj.70 0 3 J.70 mj The total energy stored in the individual capacitors is Energy stored ( 3.60 + 5.40 +.80 +.70) mj 3.5 mj ( Energy stored) total Thus, the sums of the energies stored in the individual capacitors equals the total energy stored by the system. 6.57 In the absence of a dielectric, the capacitance of the parallel-plate capacitor is C 0 0 Ad. With the dielectric inserted, it fills one-third of the gap between the plates as shown in sketch (a) at the right. We model this situation as consisting of a pair of capacitors, and C, connected in series as shown in sketch at the right. In reality, the lower plate of and the upper plate of C are one 3 d 3 d d k 3 d k c and the same, consisting of the lower surface of the dielectric shown in sketch (a). The capacitances in the model of sketch are given by (a) 3 d c k A 0 d 3 3k A 0 d and C A 0 d 3 3 A 0 d The equivalent capacitance of the series combination is d 3k 0 A + d 3 0 A k + d k + d k + d k + 3 0 A k 3 0 A 3k 0 A 3k C 0 and [ 3k (k +) ]C 0. 6.58 For the parallel combination: C p + C which gives C C p [] For the series combination: C s + C or C C s C s C s Thus, we have C C s C s and equating this to Equation [] above gives C p C C s or C p C p C s C + C s C s C s We write this result as: C p + C p C s 0 continued on next page 6879_6_ch6_p030-06.indd 57 /7/ :33:4 PM

58 Chapter 6 and use the quadratic formula to obtain C p ± 4 C p C p C s Then, Equation [] gives C C p 4 C p C p C s 6.59 For a parallel-plate capacitor with plate separation d, ΔV max E max d or d ΔV max E max The capacitance is then C k 0 d A k 0 A E max ΔV max and the needed area of the plates is A C ΔV max k 0 E max, or ( 0.50 0 6 F) ( 4.00 0 3 V) (.00 0 8 Vm) A ( 3.00) 8.85 0 C N m 0.88 m 6.60 (a) The.0-mC is located 0.50 m from point P, so its contribution to the potential at P is.0 0 6 C V k e q r 8.99 0 9 N m C 0.50 m.8 04 V The potential at P due to the.0-mc charge located 0.50 m away is V k e q r 8.99 0 9 N m C.0 0 6 0.50 m C 3.6 04 V The total potential at point P is V P V + V ( +.8 3.6) 0 4 V.8 0 4 V (d) The work required to move a charge q 3.0 mc to point P from infinity is W qδv qv ( P V ) ( 3.0 0 6 C) (.8 0 4 V 0) 5.4 0 J 6.6 The stages for the reduction of this circuit are shown below. Thus, 6.5 mf 6879_6_ch6_p030-06.indd 58 /7/ :33:45 PM

Electrical Energy and Capacitance 59 6.6 (a) Due to spherical symmetry, the charge on each of the concentric spherical shells will be uniformly distributed over that shell. Inside a spherical surface having a uniform charge distribution, the electric field due to the charge on that surface is zero. Thus, in this region, the potential due to the charge on that surface is constant and equal to the potential at the surface. Outside a spherical surface having a uniform charge distribution, the potential due to the charge on that surface is given by V k e qr, where r is the distance from the center of that surface and q is the charge on that surface. In the region between a pair of concentric spherical shells, with the inner shell having charge +Q and the outer shell having radius b and charge Q, the total electric potential at distance r from the center is given by V V due to + V due to k Q e inner shell outer shell r + k e ( Q) b k Q e r b The potential difference between the two shells is therefore ΔV V r a V r b k e Q a b k e Q b b k e Q The capacitance of this device is given by b a ab C Q ΔV k e ab b a When b >> a, then b a b. Thus, in the limit as b, the capacitance found above becomes C k e ab b a k e 4p 0 a 6.63 The energy stored in a charged capacitor is E stored C ΔV. Hence, ΔV E stored C ( 300 J) 30.0 0 6 F 4.47 V 4.47 kv 03 6.64 From Q C( ΔV ), the capacitance of the capacitor with air between the plates is C 0 Q 0 50 mc ΔV ΔV After the dielectric is inserted, the potential difference is held to the original value, but the charge changes to Q Q 0 + 00 mc 350 mc. Thus, the capacitance with the dielectric slab in place is C Q ΔV 350 mc ΔV The dielectric constant of the dielectric slab is therefore k C C 0 350 mc ΔV ΔV 50 mc 350 50.33 6879_6_ch6_p030-06.indd 59 /7/ :33:47 PM

60 Chapter 6 6.65 The charges initially stored on the capacitors are 50 V Q ( ΔV ) i 6.0 mf.5 0 3 mc and Q C ( ΔV ) i (.0 mf) ( 50 V) 5.0 0 mc When the capacitors are connected in parallel, with the negative plate of one connected to the positive pl ate of the other, the net stored charge is Q Q Q.5 0 3 mc 5.0 0 mc.0 0 3 mc The equivalent capacitance of the parallel combination is + C 8.0 mf. Thus, the final potential difference across each of the capacitors is ( ΔV ) Q.0 03 mc 5 V 8.0 mf and the final charge on each capacitor is Q ΔV and Q C ΔV ( 6.0 mf) 5 V (.0 mf) 5 V 750 mc 0.75 mc 50 mc 0.5 mc 6.66 (a) The distance from the charge q to either of the charges on the y-axis is r d + ( d) 5 d. Thus, V k e q i i r i PE q k e q q 3 r + k e q q 3 r k q e 5 d + k q e 5 d k q e 5 d k q( q) e 5 d + k q q e 5 d 4k e q 5 d From conservation of energy with PE 0 at r, KE f KE i + PE i PE f 0 + 4k e q 5 d 0 4k e q 5 d (d) v f ( KE f ) m 4k e q m 5 d 8k e q 5 md 6.67 When excess charge resides on a spherical surface that is far removed from any other charge, this excess charge is uniformly distributed over the spherical surface, and the electric potential at the surface is the same as if all the excess charge were concentrated at the center of the spherical surface. In the given situation, we have two charged spheres, initially isolated from each other, with charges and potentials of Q A +6.00 mc and V A k e Q A R A, where R A.0 cm, Q B 4.00 mc, and V B k e Q B R B, with R B 8.0 cm. When these spheres are then connected by a long conducting thread, the charges are redistributed ( yielding charges of Q A and Q B, respectively) until the two surfaces come to a common potential ( V A k Q A R A V B k Q B R B ). When equilibrium is established, we have: From conservation of charge: Q A + Q B Q A + Q B Q A + Q B +.00 mc [] continued on next page 6879_6_ch6_p030-06.indd 60 /7/ :33:49 PM