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CHAPTER 24 ELECTROSTATIC ENERGY and CAPACITANCE • Capacitance and Capacitors • Storage of Electrical Energy • Energy

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CHAPTER 24 ELECTROSTATIC ENERGY and CAPACITANCE • Capacitance and Capacitors • Storage of Electrical Energy • Energy CAPACITANCE: In the previous chapter, we saw that an object with charge Q, will have a potential V. Conversely, if an CHAPTER 24 object has a potential V it will have a charge Q. The Q ELECTROSTATIC ENERGY capacitance (C) of the object is the ratio . and CAPACITANCE V • Capacitance and capacitors + Example: A charged spherical • Storage of electrical energy + + R conductor with a charge Q. The • Energy density of an electric field + + potential of the sphere is + Q • Combinations of capacitors V = k . • In parallel R • In series Therefore, the capacitance of the charged sphere is Q Q R • Dielectrics C = = = = 4πε!R V kQ k • Effects of dielectrics R UNITS: Capacitance Coulombs/Volts • Examples of capacitors ⇒ ⇒ Farad (F). Example: A sphere with R = 5.0 cm (= 0.05m) −12 C = 5.55 ×10 F (⇒ 5.55pF). Capacitance is a measure of the “capacity” that an object has for “holding” charge, i.e., given two objects at From before, the capacitance of a sphere is the same potential, the one with the greater capacitance −12 3 C = 4πε R = 4π × 8.85×10 × 6400 ×10 will have more charge. As we have seen, a charged ! −4 object has potential energy (U); a device that is = 7.1×10 F. specifically designed to hold or store charge is called a capacitor. Earlier, question 22.4, we found the charge on the Earth was 5 Q = −9.11×10 C. So, what is the corresponding potential? By definition Q 9.11×105 V = = =1.28 ×109V Question 24.1: The Earth is a conductor of radius C 7.1×10−4 6400 km. If it were an isolated sphere what would be its which is what we found in chapter 23 (question 23.7). capacitance? Capacitance of two parallel plates: Two parallel plates (Practical considerations): + Q − Q Example: A = 5.0 cm × 5.0 cm with d = 0.5 cm. A −12 C = ε! = 4.4 ×10 F = 4.4pF. + − d Area = A The maximum possible value of E in air (from earlier) d 6 We can use a battery or a ≈ 3×10 V/m. Therefore, the maximum potential generator to move an amount difference (voltage) we can get between this pair of of charge Q from one plate to plates (in air) is: 6 the other. The electric field between the plates is: V max = Emax.d ≈ 3 ×10 × 0.005 ≈ 15,000 V. σ Q E = = . (From ch. 22) Note: V max depends only on the spacing. ε! ε!A Also, the maximum charge we can achieve is Also, the potential difference (voltage) between the two −12 3 Q max = CVmax = 4.4 ×10 ×15 ×10 = 66 nC. plates is: A pair of parallel plates is a useful capacitor. Later, we σ V = E.d = d. (From ch. 23) will find an expression for the amount of energy stored. ε ! _________________________________ So the capacitance of this pair of plates is To have a 1F capacitance the area would have to be ~ 5.6 × 108 m2, Q σA A C = = = ε . i.e., the length of each side of the plates would be ~ 23.8 km (i.e., V V ! d about 14 miles!) with a spacing of 0.50 cm. Two parallel plates: A C = ε! d A cylindrical (coaxial) capacitor: How can we increase the capacitance, i.e, get more charge per unit of potential difference? − Q • increase A - - - - - - • decrease d r a + + Metal foil - - - + + + + r - - - + Q b Insulating spacer (dielectric) - - - L The capacitance of an air-filled coaxial capacitor of length L is: The capacitor is rolled up into a cylindrical shape. 2πε L C = ! rb ln r ( a ) • increase ε ! by changing the medium between the plates, i.e., ε ! ⇒ ε = κε! (later). A cylindrical (coaxial) capacitor (continued): Example: coaxial (antenna) wire. copper wire insulation (dielectric) copper braid Assume an outer conductor (braid) radius r b ≈ 2.5mm, and an inner conductor (wire) radius r a ≈ 0.5mm, with Question 24.2: What is the relationship between the neoprene insulation (dielectric) ( ε ! ⇒ κε! = 6.9ε!). charge density on the inner and outer plates of a The capacitance per meter is: cylindrical capacitor? Is it C 2πκε! A: larger on the outer plate? = r L ln b B: larger on the inner plate? ( ra ) C: the same on both plates. 2 × π × 6.9 × 8.85×10−12 = = 2.384 ×10−10 F/m 1.609 = 238.4 pF/m Storing energy in a capacitor + Q − Q When an uncharged capacitor is connected to a source EFM10VD2.MOV of potential difference (like a battery), charges move from one plate to the other. Therefore, the magnitude of Because work is done to move charges onto the plates of the charge ( ± Q) on each plate is always the same. But a capacitor, the capacitor stores energy, electrostatic Q potential energy. The energy is released when the the charge density depends on area ( σ = A); because capacitor is discharged. the inner plate has a smaller area than the outer plate, the charge density on the inner plate is greater than the Where is the energy stored? charge density on the outer plate. ... in the electric field (between the plates), which Therefore, the answer is B (larger on the inner plate). has been produced during the charging process! Storing energy in a parallel plate capacitor: So, in charging a capacitor from 0 → Q the total increase in potential energy is: + q + + + + + + + B To store energy in a Q 2 capacitor we “charge” it, ⎛ q⎞ 1 2 Q 1 Q + dq U = dU = ⎜ ⎟ dq = q = ∫ ∫ [ ]0 producing an electric field 0⎝ C⎠ 2C 2 C − q - - - - - - - A between the plates. We do ⎛ 1 1 ⎞ ⎜ = QV = CV2⎟ . work moving charges from plate A to plate B. If the ⎝ 2 2 ⎠ plates already have charge ± q and dq is then moved from Potential A to B, the incremental work done is difference V 1 Slope = Q = C V dW = −dq(VA − VB), Area = dU = Vdq where V and V are the potentials of plates A and B, q A B V = C respectively. This, then is the incremental increase in dq Charge potential energy, dU, of the capacitor system. q Q If V = VB − VA ( V B > VA), then dU = Vdq . q Note, U is the area under the V Q plot. This potential But, by definition: V = , − C energy can be recovered when the capacitor is so the incremental increase in energy when dq of charge discharged, i.e., when the stored charge is released. is taken from A → B is: (Note also, this is the same expression we obtained earlier ⎛ q⎞ ∴dU = ⎜ ⎟ dq. for a charged conducting sphere.) ⎝ C⎠ + Q − Q + Q − Q + − + − d D A A [1] Algebraically: Put C1 = ε! and C2 = ε! d D d D Since D > d, then C 2 < C1. The charge must be the same in each case (where can it go or come from?) Question 24.3: A parallel plate capacitor, with a plate 1 Q2 1 Q2 separation of d, is charged by a battery. After the battery ∴U = and U = , 1 2 C 2 2 C is disconnected, the capacitor is discharged through two 1 2 wires producing a spark. The capacitor is re-charged i.e., U 2 > U1. exactly as before. After the battery is disconnected, the Therefore, the stored energy increases when the plates plates are pulled apart slightly, to a new distance D are pulled apart so the spark has more energy. (where D > d). When discharged again, is the energy of Answer A. the spark ___________ it was before the plates were pulled apart? [2] Conceptually: You do (positive) work to separate the plates (because there is an attractive force between A: greater than them). The work goes into the capacitor system so the B: the same as stored energy increases. C: less than Answer A. Energy density of an electric field ... Combining capacitors (parallel): V B Assume we have a parallel plate capacitor, then the C C 1 C 2 eq stored energy is: V BA = VB − VA V BA 1 U = CV2. 2 V A Equivalent A But C = ε! and V = E.d, capacitor d The potential difference is the same on each capacitor. 1 A 2 1 2 ∴U = ε! (E.d) = ε!(A.d)E . The charges on the two capacitors are: 2 d 2 Q 1 = C1VBA and Q 2 = C2VBA and the total charge stored is: But A .d ⇒ volume of Q = Q + Q = (C + C )V = C V the electric field 1 2 1 2 BA eq BA where C C C . between the plates. eq = 1 + 2 Area = A So the energy density So, this combination is equivalent to a single capacitor with capacitance d is: U 1 2 C eq = C1 + C2. ue = volume = ε!E . 2 When more than two capacitors are connected in parallel: This result is true for all electric fields. C eq = ∑iCi . Combining capacitors (series): V B C + Q 1 C C1 = 4.0 µF C 1 − Q eq V = V − V V V 200 V C 3 C2 = 15.0 µF BA B A m +Q BA C 2 C 12.0 F C 3 = µ 2 − Q V A Equivalent capacitor Question 24.4: In the circuit shown above, the capacitors were completely discharged before being The charge on the two capacitors is the same: ±Q. connected to the voltage source. Find If V B > VA, the individual potential differences are: Q Q V1 = (VB − Vm) = and V2 = (Vm − VA) = .
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