CHAPTER 6 ELECTROSTATIC ENERGY AND CAPACITORS. Three point charges, each of +q, are moved from infinity to the vertices of an equilateral triangle of side l. How much work is required? The sentence preceding Example 6- allows us to rewrite Equation 6- (for the electrostatic energy of a distribution of point charges) as W pairs kq q r. For three equal charges (three different pairs) at the corners of an equilateral triangle (r ij l for each pair) W 3kq i j ij l. 3. Four 50-µ C charges are brought from far apart onto a line where they are spaced at.0-cm intervals. How much work does it take to assemble this charge distribution? Number the charges qi 50 µ C, i,, 3, 4, as they are spaced along the line at a cm intervals. There are six pairs, so W pairs kqiq j rij k( qq a + qq3 a + qq4 3a + qq3 a + qq4 a + q3q4 a) ( kq a)( + + 3 + + + ) 3kq 3a 3( 9 0 9 m/ F )( 50 µ C ) ( 3 cm ) 4. 88 kj. (See solution to.) 3. 6. To a very crude approximation, a water molecule consists of a negatively charged oxygen atom and two bare protons, as shown in Fig. 6-5. Calculate the electrostatic energy of this configuration, which is therefore the magnitude of the energy released in forming this molecule from widely separated atoms. Your answer is an overestimate because electrons are actually shared among the three atoms, spending more time near the oxygen. The electrostatic potential energy of the water molecule (in this approximation) is U W pairs kq q r (generalization of Equation 6-). The two oxygen-hydrogen pairs have separation a 0 0 m, while the hydrogen-hydrogen pair has separation a cos 37. 5. 59a. Therefore, U k( e)( e) a + ke. 59a 3. 37ke a 3. 37( 9 0 9 ) 9 0 8 (. 6 0 ) J 0 7. 76 0 J 48. 5 ev. ( U is called the ionic separation energy.) i j ij a 0 0m e 05 +e a cos 37 +e FIGURE 6-5 6.
68 CHAPTER 6 7. Four identical charges q, initially widely separated, are brought to the vertices of a tetrahedron of side a (Fig. 6-6). Find the electrostatic energy of this configuration. q q q q FIGURE 6-6 7. There are six different pairs of equal charges and the separation of any pair is a. Thus, W pairs kqiq j a 6kq a. (See.) 8. A charge Q 0 is at the origin. A second charge, Qx Q 0, is brought to the point x a, y 0. Then a third charge Q y is brought to the point x 0, y a. If it takes twice as much work to bring in Q y as it did Q x,what is Q y in terms of Q 0? The work necessary to bring up Q x is Wx kq0qx a kq0 a, while the work necessary to subsequently bring up Q y is W kq Q a + kq Q a kq Q ( + ) a. If W W, then Q ( + ) 4Q, or Q 4Q ( + ). 66Q. y 0 y x y 0 y (Note: ( + ).) y x y 0 y 0 0. Two parallel, circular metal plates of 5 cm radius are initially uncharged. It takes 6.3 J to transfer 45 µ C from one plate to the other. How far apart are the plates? Equation 6- can be rearranged to give d ε 0 AU Q ( 8. 85 pf/ m ) π ( 5 cm ) ( 6. 3 J ) ( 45 µ C ) 3. 89 mm. (The fact that d 5 cm is justification for using the approximation of closely spaced plates.) 3. A conducting sphere of radius a is surrounded by a concentric spherical shell of radius b. Both are initially uncharged. How much work does it take to transfer charge from one to the other until they carry charges ±Q? When a charge q (assumed positive) is on the inner sphere, the potential difference between the spheres is V kq( a b ). (See the solution to 5-63(a).) To transfer an additional charge dq from the outer sphere requires work dw V dq, so the Q z total work required to transfer charge Q (leaving the spheres oppositely charged) is W 0 V dq z Q 0 kq dq( a b ) kq ( a b ). (Incidentally, this shows that the capacitance of this spherical capacitor is k( a b ) ab k( b a) ; see Equation 6-8a.)
CHAPTER 6 69 5. Two conducting spheres of radius a are separated by a distance l À a; since the distance is large, neither sphere affects the other s electric field significantly, and the fields remain spherically symmetric. (a) If the spheres carry equal but opposite charges ±q, show that the potential difference between them is kq a. (b) Write an expression for the work dw involved in moving an infinitesimal charge dq from the negative to the positive sphere. (c) Integrate your expression to find the work involved in transferring a charge Q from one sphere to the other, assuming both are initially uncharged. (a) The potential difference between the two (essentially isolated) spheres is V kq a k( q) a kq a (see Equation 5- ). (b) V is the work per unit positive charge transferred between the spheres, so dw dq V kq dq a. (c) The Q z z 0 integration yields W dw kq dq a kq a. 9. Find the electric field energy density at the surface of a proton, taken to be a uniformly charged sphere fm in radius. For this model of the proton, the field strength at the surface is E ke R (from spherical symmetry and Gauss s law). Thus, 4 9 9 the energy density in the surface electric field is u ε E ke 8π R ' ( 9 0 m/ F )(. 6 0 C ) 8π 4 30 3 3 ( fm ) 9. 7 0 J/ m 57. 3 kev/ fm. 0 0. A pair of closely spaced square conducting plates measure 0 cm on a side. The electric field energy density between the 3 plates is 4. 5 kj/ m. What is the charge on the plates? Combining Equation 6-3 with Equation 4- (see last paragraph of Section 4-6), one finds E q σ A A uε ( 0 cm ) ( 4. 5 kj/ m )( 8. 85 pf/ m ). 8 µ C. 0 3 σ ε u ε, or 0 0 3. A sphere of radius R carries a total charge Q distributed over its surface. Show that the total energy stored in its electric field is U kq R. The calculation of the electrostatic energy for a sphere with uniform surface charge density is, in fact, given in Example 6-3. We simply set R R, the radius of the sphere, and R (so the integral covers all the space where the field is nonzero). 34. A parallel-plate capacitor with.-mm plate spacing has ±. 3 µ C on its plates when charged to 50 V. What is the plate area? From Equation 6-6, A Qd ε V (. 3 µ C )(. mm ) ( 8. 85 pf/ m )( 50 V ). 9 m. 0 35. Find the capacitance of a.0-m-long piece of coaxial cable whose inner conductor radius is 0.80 mm and whose outer conductor radius is. mm, with air in between.
60 CHAPTER 6 The capacitance of air-filled ( κ ) cylindrical capacitor was found in Example 6-5: C πε 0l ln( b a) π( 8. 85 pf/ m ) ( m ) ln(. 0. 8) 55. 0 pf. 36. A capacitor consists of a conducting sphere of radius a surrounded by a concentric conducting shell of radius b. Show ab that its capacitance is C k( b a). This result, mentioned in the solution to 3, also follows from Equation 6-5 and the potential difference between two concentric conducting spheres, V kq( a b ) Q( b a) 4πε ab Q C. 37. Figure 6-8 shows a capacitor consisting of two electrically connected plates with a third plate between them, spaced so its surfaces are a distance d from the other plates. The plates have area A. Neglecting edge effects, show that the capacitance is ε 0 A d. 0 d d FIGURE 6-8 37. When the third (middle) plate is positively charged, the electric field (not near an edge) is approximately uniform and away from the plate, with magnitude E σ ε 0. Since half of the total charge Q is on either side (by symmetry), σ Q A. The potential difference between the third plate and the outer two plates (which are both at the same potential and carry charges of Q on their inner surfaces) is V Ed σ d ε 0 Qd ε 0 A. Therefore the capacitance is C Q V ε 0 A d. (The arrangement is like two capacitors in parallel.) 44. A medical defibrillator stores 950 J of energy in a 00-µ F capacitor. (a) What is the voltage across the capacitor? (b) If the capacitor discharges 300 J of its stored energy in.5 ms, what is the power delivered during this time? (a) From Equation 6-8b, V U C 950 J 00 µ F 4. 36 kv. (b) P av U t 300 J. 5 ms 0 kw. 47. A solid conducting slab is inserted between the plates of a charged capacitor, as shown in Fig. 6-9. The slab thickness is 60% of the plate spacing, and its area is the same as the plates. (a) What happens to the capacitance? (b) What happens to the stored energy, assuming the capacitor is not connected to anything? (a) The charge on the plates remains the same, and so does the electric field ( E σ ε 0 ) in the gaps between either plate and the slab. However, the separation (i.e., the thickness of the field region) between the plates is reduced to 40% of its original value d d + d 0.4 d, therefore the capacitance is increased, C ε 0 A d ε 0 A 0.4 d. 5 C. (The equations V El and C Q V lead to the same result.) In fact, the configuration behaves like a series combination of two parallel
CHAPTER 6 6 plate capacitors, C C + C ( d ε 0 A) + ( d ε 0 A) ( d + d ) ε 0 A 0.4 d ε 0 A. 5 C. (b) When the charge is constant (no connections to anything isolates the system), the energy stored is inversely proportional to the capacitance, U Q C. Thus U Q C Q (. 5C) 0.4 U, or the energy decreases to 40% of its original value. (With the slab inserted, there is less field region and less energy stored. While the slab is being inserted, work is done by electrical forces to conserve energy.) FIGURE 6-9 47. 5. (a) What is the equivalent capacitance of the combination shown in Fig. 6-30? (b) If a 00-V battery is connected across the combination, what is the charge on each capacitor? (c) What is the voltage across each? (a) C is in series with the parallel combination of C and C 3. Thus, C C ( C + C3 ) ( C + C + C3 ) ( 0. 0 µ F) ( + ) ( + + ) 0. 0 µ F. (b) The net charge on the entire combination is Q CV ( 0. 0 µ F )( 00 V ). µ C. Since C is in series with the capacitors in parallel, Q. µ C Q Q + Q3. Moreover, for the parallel capacitors, V Q C V3 Q3 C3, so Q3 Q C3 C. Thus, Q ( 3) Q 0.4 µ C and Q3 ( 3) Q 0. 8 µ C. (In general, for two capacitors in parallel, Q CQ( C + C3 ) etc.) (c) Equation 6-5, applied to each capacitor, gives V Q C. µ C 0. 0 µ F 60 V, and V V3 40 V. (Alternatively, one can first use the general result in the solution to 5 (with C replaced by C + C3 ) to obtain the voltages, V ( C + C3) V ( C + C + C3) ( 3 5)( 00 V ), V V CV ( C + C + C3 ) ( 5)( 00 V ), and then use Equation 6-5 to find the charges.) 3 FIGURE 6-30 5. 54. What is the equivalent capacitance of the four identical capacitors in Fig. 6-3, measured between A and B? Relative to points A and B, the combination of capacitors, 3, and 4 is in parallel with (see numbering added to Fig. 6-8), so Ctot C + C34. However, C 34 consists of in series with the parallel combination of 3 and 4, so
6 CHAPTER 6 C34 CC34 ( C + C34 ) C ( C3 + C4 ) ( C + C3 + C4 ). Since each individual capacitance is equal to C, C34 C and Ctot 5 3 C. 3 FIGURE 6-3 54. 57. What is the equivalent capacitance in Fig. 6-3? Number the capacitors as shown. Relative to points A and B, C, C 4, and the combination of C and C 3 are in series, so the capacitance is given by CAB C + C4 + C 3. C 3 is a parallel combination, hence C3 C + C3, therefore C AB 6 ( 3 µ F) + ( µ F ) + ( µ F + µ F ), or µ F 0. 857 µ F. C AB 7 FIGURE 6-3 57. 58. In Fig. 6-3, find the energy stored in the -µ Fcapacitor when a 50-V battery is connected between points A and B. The energy of the µ F capacitor is U3 C3V3, where V3 V3, since C and C 3 are in parallel (see the solution to the previous problem for a figure and the numbering). But C, C 4, and C 3 are in series, so V V + V4 + V3 and 3 3 Q CV C4V4 C3V3. Therefore V V3 ( + C3 C + C3 C4 ) V3 ( + 3 + ), or V 3 ( 7) V. Finally, U 3 ( µ F )( 50 V 7) 0 µ J. 59. Two capacitors C and C are in series, with a voltage V across the combination. Show that the voltages across the individual capacitors are This is shown in the solution to 5. V CV C + C CV and V. C + C
CHAPTER 6 63 68. An air-insulated parallel-plate capacitor has plate area 76 cm and spacing. mm. It is charged to 900 V and then disconnected from the charging battery. A plexiglass sheet is then inserted to fill the space between the plates. What are (a) the capacitance, (b) the potential difference between the plates, and (c) the stored energy both before and after the plexiglass is inserted? Before the plexiglass is inserted, (a) the capacitance is C ε A d ( 8. 85 pf/ m )( 76 cm ) (. mm ) 56. pf, 0 0 (b) the voltage is V 0 900 V, and (c) the stored energy is U0 C0V0. 7 µ J. With the plexiglass insulation inserted, (a) the capacitance is C κ C 0 ( 3.4)( 56. pf ) 9 pf. Since the capacitor was disconnected before the process of insertion, i.e., the plates are isolated and their charge Q is constant, (b) the voltage is reduced by a factor of κ, V V0 κ 900 V 3.4 65 V (see the discussion in the text preceding Equation 6-), and (c) so is the stored energy, U U0 κ. 7 µ J 3.4 6. 68 µ J (see Equation 6-). 69. The capacitor of the preceding problem is connected to its 900-V charging battery and left connected as the plexiglass sheet is inserted, so the potential difference remains at 900 V. What are (a) the charge on the plates and (b) the stored energy both before and after the plexiglass is inserted? (a) The capacitances before and after the insertion of the plexiglass insulation are C0 ε 0 A d ( 8. 85 pf/m )( 76 cm ) (. mm ) 56. pf, and C κ C 0 ( 3.4)( 56. pf ) 9 pf, as found previously. Therefore, since the voltage stays at 900 V in this case (due to the battery), Q C ( 900 V ) 50.4 nc, and Q C ( 900 V ) κ Q 7 nc, before and after 0 0 insertion, respectively. (b) The stored energy is U0 C0( 900 V ). 7 µ J before, and U C( 900 V ) κ U 0 77. µ J after. (The difference between this situation and the one in the previous problem is that the battery does additional work moving more charge to the capacitor plates, while maintaining the constant voltage. Equation 6- applies to an isolated capacitor only.) 80. Show that the result of 36 reduces to that of a parallel-plate capacitor when the separation b a is much less than the radius a. Let b a d a for the spherical capacitor in 36. Then C 4πε a( a + d) d ¼ 4πε a d, which is the result of Equation 6-6, with A 4π a being the area of the spherical plates. 0 0 86. A solid sphere contains a uniform volume charge density. What fraction of the total electrostatic energy of this configuration is contained within the sphere? The results of s and 4 (with the aid of the argument in the solution to 3) show that the fraction is just ( kq 0R) ( 3kQ 5R) 6. 90. A classical view of the electron pictures it as a purely electrical entity, whose rest mass energy mc (see Section 8-7) is the energy stored in its electric field. If the electron were a sphere with charge distributed uniformly over its surface, what 0
64 CHAPTER 6 radius would it have to satisfy this condition? (Your answer for the electron s size is not consistent with modern quantum mechanics nor with experiments that suggest the electron is a true point particle.) The electrostatic energy stored in the field of a classical electron, whose charge e is distributed uniformly over the surface of a sphere of radius R, is U ke R (see 3). If we set this equal to the electron s mass energy, U m c, then R ke m e c.4 fm, where constants from the inside front cover were used. (The classical radius of the electron, based on a consideration of the scattering of electromagnetic waves from free electrons, called Thomson scattering, is actually R r ke m c. ) e e e