Fig Solved Problems

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1 CHAP. 25] POTENTIAL; CAPACITANCE 245 Fig. 25- Solved Problems 25. In Fig. 25-2, the potential di erence between the metal plates is 40 V. (a) Which plate is at the higher potential? (b) How much work must be done to carry a 3:0 C charge from B to A? From A to B? (c) How do we know that the electric eld is in the direction indicated? (d ) If the plate separation is 5.0 mm, what is the magnitude of ~ E? Fig (a) (b) A positive test charge between the plates is repelled by A and attracted by B. Left to itself, the positive test charge will move from A to B, and so A is at the higher potential. The magnitude of the work done in carrying a charge q through a potential di erence V is qv. Thus the magnitude of the work done in the present situation is W ˆ 3:0 C 40 V ˆ0:2 kj Because a positive charge between the plates is repelled by A, positive work 20 J) must be done to drag the 3:0 C charge from B to A. To restrain the charge as it moves from A to B, negative work 20 J) is done.

2 246 POTENTIAL; CAPACITANCE [CHAP. 25 (c) A positive test-charge between the plates experiences a force directed from A to B and this is, by de nition, the direction of the eld. (d ) For parallel plates, V ˆ Ed. Therefore, E ˆ V d ˆ 40 V ˆ 8:0 kv=m 0:005 0 m Notice that the SI units for electric eld, V/m and N/C, are identical How much work is required to carry an electron from the positive terminal of a 2-V battery to the negative terminal? Going from the positive to the negative terminal, one passes through a potential drop. In this case it is V ˆ 2 V. Then W ˆ qv ˆ :60 9 C 2 V ˆ:9 0 8 J As a check, we notice that an electron, if left to itself, will move from negative to positive because it is a negative charge. Hence positive work must be done to carry it in the reverse direction as required here How much electrical potential energy does a proton lose as it falls through a potential drop of 5 kv? The proton carries a positive charge. It will therefore move from regions of high potential to regions of low potential if left free to do so. Its change in potential energy as it moves through a potential di erence V is Vq. In our case, V ˆ 5 kv. Therefore, Change in PE E ˆ Vq ˆ 50 3 V :6 0 9 C ˆ 80 6 J 25.4 An electron starts from rest and falls through a potential rise of 80 V. What is its nal speed? Positive charges tend to fall through potential drops; negative charges, such as electrons, tend to fall through potential rises. Change in PE E ˆ Vq ˆ 80 V :6 0 9 C ˆ : J This lost PE E appears as KE of the electron: PE E lost ˆ KE gained : J ˆ 2 mv2 f 2 mv2 i ˆ 2 mv2 f 0 s : J 2 v f ˆ 9: 0 3 ˆ 5:3 0 6 m=s kg 25.5 (a) What is the absolute potential at each of the following distances from a charge of 2:0 C: r ˆ 0 cm and r ˆ 50 cm? (b) How much work is required to carry a 0:05 C charge from the point at r ˆ 50 cm to that at r ˆ 0 cm? b V 0 ˆ k q r ˆ 9:009 Nm 2 =C 2 2:0 0 6 C 0:0 m ˆ :8 05 V V 50 ˆ 0 50 V 0 ˆ 36 kv Work ˆ q V 0 V 50 ˆ 50 8 C : V ˆ7:2 mj 25.6 Suppose, in Problem 25.5(a), that a proton is released at r ˆ 0 cm. How fast will it be moving as it passes a point at r ˆ 50 cm?

3 CHAP. 25] POTENTIAL; CAPACITANCE 247 As the proton moves from one point to the other, there is a potential drop of Potential drop ˆ : V 0: V ˆ : V The proton acquires KE as it falls through this potential drop: KE gained ˆ PE E lost 2 mv2 f 2 mv2 i ˆ qv 2 : kg v 2 f 0 ˆ :60 9 C : V from which v f ˆ 5:3 0 6 m/s In Fig. 25-2, let E ˆ 2:0 kv/m and d ˆ 5:0 mm. A proton is shot from plate B toward plate A with a speed of 00 km/s. What will be its speed just before it strikes plate A? The proton, being positive, is repelled by A and will therefore be slowed down. We need the potential di erence between the plates, which is V ˆ Ed ˆ 2:0 kv=m 0:005 0 m ˆ0 V Now, from the conservation of energy, for the proton, KE lost ˆ PE E gained 2 mv2 B 2 mv2 A ˆ qv Substituting m ˆ : kg, v B ˆ : m/s, q ˆ : C, and V ˆ 0 V gives v A ˆ 90 km/s. As we see, the proton is indeed slowed A tin nucleus has a charge 50e. (a) Find the absolute potential V at a radius of :0 0 2 m from the nucleus. (b) If a proton is released from this point, how fast will it be moving when it is.0 m from the nucleus? (b) V ˆ k q r ˆ 9:0 09 Nm 2 =C 2 50 :6 0 9 C 0 2 m ˆ 72 kv The proton is repelled by the nucleus and ies out to in nity. The absolute potential at a point is the potential di erence between the point in question and in nity. Hence there is a potential drop of 72 kv as the proton ies to in nity. Usually we would simply assume that.0 m is far enough from the nucleus to consider it to be at in nity. But, as a check, let us compute V at r ˆ :0 m: V mˆ k q r ˆ 9:009 Nm 2 =C 2 50 :6 0 9 C :0 m which is essentially zero in comparison with 72 kv. As the proton falls through 72 kv, KE gained ˆ PE E lost ˆ 7:2 0 8 V 2 mv2 f 2 mv2 i ˆ qv 2 : kg v 2 f 0 ˆ :60 9 C V from which v f ˆ 3:7 0 6 m/s The following point charges are placed on the x-axis: 2:0 C at x ˆ 20 cm, 3:0 C at x ˆ 30 cm, 4:0 C atx ˆ 40 cm. Find the absolute potential on the axis at x ˆ 0.

4 248 POTENTIAL; CAPACITANCE [CHAP. 25 Potential is a scalar, and so V ˆ k X q i r i ˆ 9:0 0 9 Nm 2 =C 2 2:0 0 6 C 0:20 m 3:0 0 6 C 0:30 m 4:0 0 6 C 0:40 m ˆ 9:0 0 9 Nm 2 =C C=m C=m C=m ˆ 90 kv 25.0 Two point charges, q and q, are separated by a distance d. Where, besides at in nity, is the absolute potential zero? At the point (or points) in question, 0 ˆ k q r k q r 2 or r ˆ r 2 This condition holds everywhere on a plane which is the perpendicular bisector of the line joining the two charges. Therefore the absolute potential is zero everywhere on that plane. 25. Four point charges are placed at the four corners of a square that is 30 cm on each side. Find the potential at the center of the square if (a) the four charges are each 2:0 C and (b) two of the four charges are 2:0 C and two are 2:0 C. b V ˆ k X P q i qi ˆ k ˆ 9:00 9 Nm 2 =C 2 4 2:0 0 6 C r i r 0:30 m cos 458 ˆ 3:4 05 V V ˆ 9:0 0 9 Nm 2 =C 2 2:0 2:0 2:0 2:0 0 6 C 0:30 m cos 458 ˆ In Fig. 25-3, the charge at A is 200 pc, while the charge at B is 00 pc. (a) Find the absolute potentials at points C and D. (b) How much work must be done to transfer a charge of 500 C from point C to point D? V C ˆ k X q i r i ˆ 9:0 0 9 Nm 2 =C 2 V D ˆ 9:0 0 9 Nm 2 =C 2 2: C : C 0:80 m 0:20 m 2: C : C 0:20 m 0:80 m ˆ 2:25 V ˆ 2:3 V ˆ 7:88 V ˆ 7:9 V (b) There is a potential rise from C to D of V ˆ V D V C ˆ 7:88 V 2:25 V ˆ0:3 V: So W ˆ Vq ˆ 0:3 V 5: C ˆ5: mj Fig. 25-3

5 CHAP. 25] POTENTIAL; CAPACITANCE Find the electrical potential energy of three point charges placed as follows on the x-axis: 2:0 C at x ˆ 0, 3:0 C atx ˆ 20 cm, and 6:0 C atx ˆ 50 cm. Take the PE E to be zero when the charges are far separated. Let us compute how much work must be done to bring the charges from in nity to their places on the axis. We bring in the 2:0 C charge rst; this requires no work because there are no other charges in the vicinity. Next we bring in the 3:0 C charge, which is repelled by the 2:0 C charge. The potential di erence between in nity and the position to which we bring it is due to the 2:0 C charge and is 2:0 C V x ˆ 0:2 ˆ k 0:20 m ˆ 9:009 Nm 2 =C 2 Therefore the work required to bring in the 3 C charge is C 0:20 m ˆ 9:0 0 4 V W 3 C ˆ qv x ˆ 0:2 ˆ 3:00 6 C 9:0 0 4 V ˆ0:270 J Finally we bring the 6:0 C charge in to x ˆ 0:50 m. The potential there due to the two charges already present is V x ˆ 0:5 ˆ k 2:0 0 6 C 0:50 m 3:0 0 6 C ˆ 2:6 0 4 V 0:30 m Therefore the work required to bring in the 6:0 C charge is W 6 C ˆ qv x ˆ 0:5 ˆ 6:00 6 C 2:6 0 4 V ˆ0:756 J Adding the amounts of work required to assemble the charges gives the energy stored in the system: PE E ˆ 0:270 J 0:756 J ˆ :0 J Can you show that the order in which the charges are brought in from in nity does not a ect this result? 25.4 Two protons are held at rest, 5:0 0 2 m apart. When released, they y apart. How fast will each be moving when they are far from each other? Their original PE E will be changed to KE. We proceed as in Problem The potential at 5:0 0 2 m from the rst charge due to that charge alone is V ˆ 9:00 9 Nm 2 =C 2 : C ˆ 288 V m The work needed to bring in the second proton is then W ˆ qv ˆ : C 288 V ˆ4:6 0 7 J and this is the PE E of the original system. From the conservation of energy, Original PE E ˆ final KE 4:6 0 7 J ˆ 2 m v 2 2 m 2v 2 2 Since the particles are identical, v ˆ v 2 ˆ v. Solving, we nd that v ˆ :7 0 5 m=s when the particles are far apart In Fig we show two large metal plates connected to a 20-V battery. Assume the plates to be in vacuum and to be much larger than shown. Find (a) E between the plates, (b) the force experienced by an electron between the plates, (c) the PE E lost by an electron as it moves from plate B to plate A, and (d ) the speed of the electron released from plate B just before striking plate A: (a) E is directed from the positive plate A to the negative plate B. It is uniform between large parallel plates and is given by

6 250 POTENTIAL; CAPACITANCE [CHAP. 25 E ˆ V d ˆ 20 V ˆ 6000 V=m ˆ 6:0 kv=m 0:020 m directed from left to right. b F E ˆ qe ˆ :6 0 9 C 6000 V=m ˆ 9:6 0 6 N The minus sign tells us that ~ FE is directed oppositely to ~ E. Since plate A is positive, the electron is attracted by it. The force on the electron is toward the left. c Change in PE E ˆ Vq ˆ 20 V :6 0 9 C ˆ : J ˆ :9 0 7 J Notice that V is a potential rise from B to A: d from which v f ˆ 6:5 0 6 m/s. PE E lost ˆ KE gained : J ˆ 2 mv2 f 2 mv2 i : J ˆ 2 9: 0 3 kg v 2 f 0 Fig As shown in Fig. 25-5, a charged particle remains stationary between the two horizontal charged plates. The plate separation is 2.0 cm, and m ˆ 4:0 0 3 kg and q ˆ 2:4 0 8 C for the particle. Find the potential di erence between the plates. Fig. 25-5

7 CHAP. 25] POTENTIAL; CAPACITANCE 25 Since the particle is in equilibrium, the weight of the particle is equal to the upward electrical force. That is, mg ˆ qe E ˆ mg q ˆ 4:0 0 3 kg 9:8 m=s 2 or 2:4 0 8 ˆ : V=m C But for a parallel-plate system, V ˆ Ed ˆ : V=m 0:020 m ˆ33 kv 25.7 An alpha particle q ˆ 2e, m ˆ 6: kg) falls from rest through a potential drop of 3:0 0 6 V (3.0 MV). (a) What is its KE in electron volts? (b) What is its speed? b Energy in ev ˆ qv e ˆ 2e 3:0 06 ˆ 6:0 0 6 ev ˆ 6:0 MeV e PE E lost ˆ KE gained qv ˆ 2 mv2 f 2 mv2 i 2 :6 0 9 C 3:0 0 6 V ˆ 2 6: kg v 2 f 0 from which v f ˆ :7 0 7 m/s What is the speed of a 400 ev (a) electron, (b) proton, and (c) alpha particle? In each case we know that the particle's kinetic energy is : J 2 mv2 ˆ 400 ev ˆ 6: J :00 ev Substituting m e ˆ 9: 0 3 kg for the electron, m p ˆ : kg for the proton, and m ˆ 4 : kg) for the alpha particle gives their speeds as (a) : m/s, (b) 2: m/s, and (c) : m/s A capacitor has a capacitance of 8:0 F with air between its plates. Determine its capacitance when a dielectric with dielectric constant 6.0 is placed between its plates. C with dielectric ˆ K C with air ˆ 6:0 8:0 F ˆ48 F What is the charge on a 300 pf capacitor when it is charged to a voltage of.0 kv? q ˆ CV ˆ F 000 V ˆ3:0 0 7 C ˆ 0:30 C 25.2 A metal sphere mounted on an insulating rod carries a charge of 6.0 nc when its potential is 200 V higher than its surroundings. What is the capacitance of the capacitor formed by the sphere and its surroundings? C ˆ q V ˆ 6:0 0 9 C ˆ 30 pf 200 V

8 252 POTENTIAL; CAPACITANCE [CHAP A :2 F capacitor is charged to 3.0 kv. Compute the energy stored in the capacitor. Energy ˆ 2 qv ˆ 2 CV 2 ˆ 2 :2 0 6 F 3000 V 2 ˆ 5:4 J The series combination of two capacitors shown in Fig is connected across 000 V. Compute (a) the equivalent capacitance C eq of the combination, (b) the magnitudes of the charges on the capacitors, (c) the potential di erences across the capacitors, and (d ) the energy stored in the capacitors. (b) from which C ˆ 2:0 pf: ˆ ˆ C eq C C 2 3:0 pf 6:0 pfˆ 2:0 pf In a series combination, each capacitor carries the same charge, which is the charge on the combination. Thus, using the result of (a), we have q ˆ q 2 ˆ q ˆ C eq V ˆ 2:0 0 2 F 000 V ˆ2:0 nc c d V ˆ q ˆ 2:0 0 9 C C 3:0 0 2 ˆ 667 V ˆ 0:67 kv F V 2 ˆ q2 ˆ 2:0 0 9 C C 2 6:0 0 2 ˆ 333 V ˆ 0:33 kv F Energy in C ˆ 2 q V ˆ 2 2:0 0 9 C 667 V ˆ6:7 0 7 J ˆ 0:67 J Energy in C 2 ˆ 2 q 2V 2 ˆ 2 2:0 0 9 C 333 V ˆ3:3 0 7 J ˆ 0:33 J Energy in combination ˆ 6:7 3:3 0 7 J ˆ J ˆ :0 J The last result is also directly given by 2 qv or 2 C eqv 2. Fig Fig The parallel capacitor combination shown in Fig is connected across a 20 V source. Determine the equivalent capacitance C eq, the charge on each capacitor, and the charge on the combination. For a parallel combination, C eq ˆ C C 2 ˆ 2:0 pf 6:0 pfˆ 8:0 pf

9 CHAP. 25] POTENTIAL; CAPACITANCE 253 Each capacitor has a 20 V potential di erence impressed on it. Therefore, q ˆ C V ˆ 2:00 2 F 20 V ˆ0:24 nc q 2 ˆ C 2 V 2 ˆ 6:00 2 F 20 V ˆ0:72 nc The charge on the combination is q q 2 ˆ 960 pc. Or, we could write q ˆ C eq V ˆ 8:00 2 F 20 V ˆ0:96 nc A certain parallel-plate capacitor consists of two plates, each with area 200 cm 2, separated by a 0.40-cm air gap. (a) Compute its capacitance. (b) If the capacitor is connected across a 500 V source, nd the charge on it, the energy stored in it, and the value of E between the plates. (c) Ifa liquid with K ˆ 2:60 is poured between the plates so as to ll the air gap, how much additional charge will ow onto the capacitor from the 500 V source? (a) b (c) For a parallel-plate capacitor with air gap, C ˆ K 0 A d ˆ 8: F=m m 2 4:0 0 3 m ˆ 4:4 0 F ˆ 44 pf q ˆ CV ˆ 4:4 0 F 500 V ˆ2:2 0 8 C ˆ 22 nc Energy ˆ 2 qv ˆ 2 2:2 0 8 C 500 V ˆ5:5 0 6 J ˆ 5:5 J E ˆ V d ˆ 500 V 4:0 0 3 m ˆ :3 05 V=m The capacitor will now have a capacitance K ˆ 2:60 times larger than before. Therefore, q ˆ CV ˆ 2:60 4:4 0 F 500 V ˆ5:7 0 8 C ˆ 57 nc The capacitor already had a charge of 22 nc and so 57 nc 22 nc or 35 nc must have been added to it Two capacitors, 3:0 F and 4:0 F, are individually charged across a 6.0-V battery. After being disconnected from the battery, they are connected together with a negative plate of one attached to the positive plate of the other. What is the nal charge on each capacitor? The situation is shown in Fig Before being connected, their charges are q 3 ˆ CV ˆ 3:00 6 F 6:0 V ˆ8 C q 4 ˆ CV ˆ 4:00 6 F 6:0 V ˆ24 C These charges partly cancel when the capacitors are connected together. Their nal charges are given by q3 0 q4 0 ˆ q 4 q 3 ˆ 6:0 C Fig. 25-8

10 254 POTENTIAL; CAPACITANCE [CHAP. 25 Also, the potentials across them are now the same, so that V ˆ q=c gives q3 0 3:0 0 6 F ˆ q4 0 4:0 0 6 F or q 0 3 ˆ 0:75q 0 4 Substitution in the previous equation gives 0:75q4 0 q4 0 ˆ 6:0 C or q4 0 ˆ 3:4 C Then q3 0 ˆ 0:75q4 0 ˆ 2:6 C: Supplementary Problems Two metal plates are attached to the two terminals of a.50-v battery. How much work is required to carry a 5:0-C charge (a) from the negative to the positive plate, (b) from the positive to the negative plate? Ans. (a) 7:5 J, (b) 7:5 J The plates described in Problem are in vacuum. An electron q ˆ e, m e ˆ 9: 0 3 kg) is released at the negative plate and falls freely to the positive plate. How fast is it going just before it strikes the plate? Ans. 7:3 0 5 m/s A proton q ˆ e, m p ˆ : kg) is accelerated from rest through a potential di erence of.0 MV. What is its nal speed? Ans. :4 0 7 m/s An electron gun shoots electrons q ˆ e, m e ˆ 9: 0 3 kg) at a metal plate that is 4.0 mm away in vacuum. The plate is 5.0 V lower in potential than the gun. How fast must the electrons be moving as they leave the gun if they are to reach the plate? Ans. :3 0 6 m/s 25.3 The potential di erence between two large parallel metal plates is 20 V. The plate separation is 3.0 mm. Find the electric eld between the plates. Ans. 40 kv/m toward negative plate An electron q ˆ e, m e ˆ 9: 0 3 kg) is shot with speed 5:0 0 6 m/s parallel to a uniform electric eld of strength 3.0 kv/m. How far will the electron go before it stops? Ans. 2.4 cm A potential di erence of 24 kv maintains a downward-directed electric eld between two horizontal parallel plates separated by.8 cm. Find the charge on an oil droplet of mass 2:2 0 3 kg that remains stationary in the eld between the plates. Ans. :6 0 8 C ˆ 0e Determine the absolute potential in air at a distance of 3.0 cm from a point charge of 500 C. Ans. 5 kv Compute the magnitude of the electric eld and the absolute potential at a distance of.0 nm from a helium nucleus of charge 2e. What is the potential energy (relative to in nity) of a proton at this position? Ans. 2:9 0 9 N=C, 2.9 V, 4:6 0 9 J A charge of 0:20 C is 30 cm from a point charge of 3:0 C in vacuum. What work is required to bring the 0:20-C charge 8 cm closer to the 3:0-C charge? Ans J A point charge of 2:0 C is placed at the origin of coordinates. A second, of 3:0 C, is placed on the x-axis at x ˆ 00 cm. At what point (or points) on the x-axis will the absolute potential be zero? Ans. x ˆ 40 cm and x ˆ 0:20 m In Problem 25.37, what is the di erence in potential between the following two points on the x-axis: point A at x ˆ 0: m and point B at x ˆ 0:9 m? Which point is at the higher potential? Ans V, point A

11 CHAP. 25] POTENTIAL; CAPACITANCE An electron is moving in the x-direction with a speed of 5:0 0 6 m/s. There is an electric eld of 3.0 kv/m in the x-direction. What will be the electron's speed after it has moved.00 cm? Ans. 3:8 0 6 m/s An electron has a speed of 6:0 0 5 m/s as it passes point A on its way to point B. Its speed at B is m/s. What is the potential di erence between A and B, and which is at the higher potential? Ans. 3. V, B 25.4 A capacitor with air between its plates has capacitance 3:0 F. What is its capacitance when wax of dielectric constant 2.8 is placed between the plates? Ans. 8:4 F Determine the charge on each plate of a F capacitor when the potential di erence between the plates is 200 V. Ans. 0 C A capacitor is charged with 9.6 nc and has a 20 V potential di erence between its terminals. Compute its capacitance and the energy stored in it. Ans. 80 pf, 0:58 J Compute the energy stored in a 60-pF capacitor (a) when it is charged to a potential di erence of 2.0 kv and (b) when the charge on each plate is 30 nc. Ans. (a) 2 mj; (b) 7:5 J Three capacitors, each of capacitance 20 pf, are each charged to 0.50 kv and then connected in series. Determine (a) the potential di erence between the end plates, (b) the charge on each capacitor, and (c) the energy stored in the system. Ans. (a).5 kv; (b) 60 nc; (c) 45J Three capacitors 2:00 F, 5.00 F, and 7:00 F are connected in series. What is their equivalent capacitance? Ans. :9 F Three capacitors 2:00 F, 5.00 F, and 7:00 F are connected in parallel. What is their equivalent capacitance? Ans. 4:00 F The capacitor combination in Problem is connected in series with the combination in Problem What is the capacitance of this new combination? Ans. :09 F Two capacitors (0.30 and 0:50 F are connected in parallel. (a) What is their equivalent capacitance? A charge of 200 C is now placed on the parallel combination. (b) What is the potential di erence across it? (c) What are the charges on the capacitors? Ans. (a) 0:80 F; (b) 0:25 kv; (c) 75 C, 0.3 mc A 2:0-F capacitor is charged to 50 V and then connected in parallel (positive plate to positive plate) with a 4:0-F capacitor charged to 00 V. (a) What are the nal charges on the capacitors? (b) What is the potential di erence across each? Ans. (a) 0.7 mc, 0.33 mc, (b) 83 V 25.5 Repeat Problem if the positive plate of one capacitor is connected to the negative plate of the other. Ans. (a) 0.0 mc, 0.20 mc; (b) 50 V (a) Calculate the capacitance of a capacitor consisting of two parallel plates separated by a layer of para n wax 0.50 cm thick, the area of each plate being 80 cm 2. The dielectric constant for the wax is 2.0. (b) If the capacitor is connected to a 00-V source, calculate the charge on the capacitor and the energy stored in the capacitor. Ans. (a) 28 pf; (b) 2.8 nc, 0:4 J