CAPACITANCE: CHAPTER 24 ELECTROSTATIC ENERGY an CAPACITANCE Capacitance an capacitors Storage of electrical energy Energy ensity of an electric fiel Combinations of capacitors In parallel In series Dielectrics Effects of ielectrics Examples of capacitors In the previous chapter, we saw that an object with charge Q, will have a potential V. Conversely, if an object has a potential V it will have a charge Q. The capacitance (C) of the object is the ratio Q V. Example: A charge spherical R conuctor with a charge Q. The potential of the sphere is V = k Q R. Therefore, the capacitance of the charge sphere is C = Q V = Q k Q = R k = 4πε!R R UNITS: Capacitance Coulombs/Volts Fara (F). Example: A sphere with R = 5.0 cm (= 0.05m) C = 5.55 10 12 F ( 5.55pF).
Capacitance is a measure of the capacity that an object has for holing charge, i.e., given two objects at the same potential, the one with the greater capacitance will have more charge. As we have seen, a charge object has potential energy (U); a evice that is specifically esigne to hol or store charge is calle a capacitor. Question 24.1: The Earth is a conuctor of raius 6400 km. If it were an isolate sphere what woul be its capacitance? From before, the capacitance of a sphere is C = 4πε! R = 4π 8.85 10 12 6400 10 3 = 7.1 10 4 F. Earlier, question 22.4, we foun the charge on the Earth was Q = 9.11 10 5 C. So, what is the corresponing potential? By efinition V = Q C = 9.11 105 7.1 10 4 =1.28 109 V which is what we foun in chapter 23 (question 23.7).
Capacitance of two parallel plates: Q We can use a battery or a generator to move an amount of charge Q from one plate to the other. The electric fiel between the plates is: E = σ = Q. (From ch. 22) ε! ε! A Also, the potential ifference (voltage) between the two plates is: Q Area = A V = E. = σ ε!. (From ch. 23) So the capacitance of this pair of plates is C = Q V = σa V = ε A!. Two parallel plates (Practical consierations): Example: A = 5.0 cm 5.0 cm with = 0.5 cm. C = ε! A = 4.4 10 12 F = 4.4pF. The maximum possible value of E in air (from earlier) 3 10 6 V/m. Therefore, the maximum potential ifference (voltage) we can get between this pair of plates (in air) is: V max = E max. 3 10 6 0.005 15,000 V. Note: V max epens only on the spacing. Also, the maximum charge we can achieve is Q max = CV max = 4.4 10 12 15 10 3 = 66 nc. A pair of parallel plates is a useful capacitor. Later, we will fin an expression for the amount of energy store. To have a 1F capacitance the area woul have to be ~ 5.6 10 8 m 2, i.e., the length of each sie of the plates woul be ~ 23.8 km (i.e., about 14 miles!) with a spacing of 0.50 cm.
Two parallel plates: C = ε! A How can we increase the capacitance, i.e, get more charge per unit of potential ifference? Metal foil increase A ecrease Insulating spacer (ielectric) A cylinrical (coaxial) capacitor: Q L Q r a r b The capacitor is rolle up into a cylinrical shape. increase ε! by changing the meium between the plates, i.e., ε! ε = κε! (later). The capacitance of an airfille coaxial capacitor of length L is: C = 2πε! L ln r b ra ( )
A cylinrical (coaxial) capacitor (continue): Example: coaxial (antenna) wire. copper brai copper wire insulation (ielectric) Assume an outer conuctor (brai) raius r b 2.5mm, an an inner conuctor (wire) raius r a 0.5mm, with neoprene insulation (ielectric) ( ε! κε! = 6.9ε! ). The capacitance per meter is: C L = 2πκε! ln r b ra = 2 π 6.9 8.85 10 12 1.609 ( ) = 238.4 pf/m = 2.384 10 10 F/m Question 24.2: What is the relationship between the charge ensity on the inner an outer plates of a cylinrical capacitor? Is it A: larger on the outer plate? B: larger on the inner plate? C: the same on both plates.
Storing energy in a capacitor Q Q When an uncharge capacitor is connecte to a source of potential ifference (like a battery), charges move from one plate to the other. Therefore, the magnitue of the charge ( ±Q) on each plate is always the same. But the charge ensity epens on area (σ = Q A ); because the inner plate has a smaller area than the outer plate, the charge ensity on the inner plate is greater than the charge ensity on the outer plate. Therefore, the answer is B (larger on the inner plate). EFM10VD2.MOV Because work is one to move charges onto the plates of a capacitor, the capacitor stores energy, electrostatic potential energy. The energy is release when the capacitor is ischarge. Where is the energy store?... in the electric fiel (between the plates), which has been prouce uring the charging process!
Storing energy in a parallel plate capacitor: q q q To store energy in a capacitor we charge it, proucing an electric fiel between the plates. We o work moving charges from plate A to plate B. If the plates alreay have charge ±q an q is then move from A to B, the incremental work one is W = q(v A V B ), where V A an V B are the potentials of plates A an B, respectively. This, then is the incremental increase in potential energy, U, of the capacitor system. If V = V B V A ( V B > V A ), then U = Vq. But, by efinition: V = q C, so the incremental increase in energy when q of charge is taken from A B is: U = B A q q. C So, in charging a capacitor from 0 Q the total increase in potential energy is: Q q U = U = q = 1 C Potential ifference V V = q C 0 [ ] 0 Q = 1 2C q2 1 = 2 QV = 1 2 CV2 q q. 2 Q 2 Note, U is the area uner the V Q plot. This potential energy can be recovere when the capacitor is ischarge, i.e., when the store charge is release. (Note also, this is the same expression we obtaine earlier for a charge conucting sphere.) Slope = V Q = 1 C Q C Area = U = Vq Charge
Q Q Q Q D Question 24.3: A parallel plate capacitor, with a plate separation of, is charge by a battery. After the battery is isconnecte, the capacitor is ischarge through two wires proucing a spark. The capacitor is recharge exactly as before. After the battery is isconnecte, the plates are pulle apart slightly, to a new istance D (where D > ). When ischarge again, is the energy of the spark it was before the plates were pulle apart? A: greater than B: the same as C: less than D A [1] Algebraically: Put C 1 = ε! an C A 2 = ε! D Since D >, then C 2 < C 1. The charge must be the same in each case (where can it go or come from?) U 1 = 1 Q 2 an U 2 C 2 = 1 Q 2, 1 2 C 2 i.e., U 2 > U 1. Therefore, the store energy increases when the plates are pulle apart so the spark has more energy. Answer A. [2] Conceptually: You o (positive) work to separate the plates (because there is an attractive force between them). The work goes into the capacitor system so the store energy increases. Answer A.
Energy ensity of an electric fiel... Combining capacitors (parallel): Assume we have a parallel plate capacitor, then the store energy is: V B C 2 C 1 V BA = V B V A V BA C eq But C = ε! A Area = A an V = E., U = 1 2 ε! U = 1 2 CV2. A (E.)2 = 1 2 ε!(a.)e 2. But A. volume of the electric fiel between the plates. So the energy ensity is: u e = U volume = 1 2 ε!e 2. This result is true for all electric fiels. The potential ifference is the same on each capacitor. The charges on the two capacitors are: an the total charge store is: where C eq = C 1 C 2. Q 1 = C 1 V BA an Q 2 = C 2 V BA Q = Q 1 Q 2 = (C 1 C 2 )V BA = C eq V BA So, this combination is equivalent to a single capacitor with capacitance C eq = C 1 C 2. When more than two capacitors are connecte in parallel: V A C eq = i C i. Equivalent capacitor
Combining capacitors (series): The charge on the two capacitors is the same: ±Q. If V B > V A, the iniviual potential ifferences are: V 1 = (V B V m ) = Q C 1 an V 2 = (V m V A ) = Q C 2. Therefore, the total potential ifference is: V BA = V 1 V 2 = Q Q = Q 1 1 = Q, C 1 C 2 C 1 C 2 C eq proviing V B V A C 1 V BA = V B V A V m C 2 1 C eq = 1 C 1 1 C 2. Q Q Q Q With more than two capacitors: 1 1 = i. C eq C i V BA C eq Equivalent capacitor Question 24.4: In the circuit shown above, the capacitors were completely ischarge before being connecte to the voltage source. Fin (a) the equivalent capacitance of the combination, (b) the charge on the positive plate of each capacitor, (c) the potential ifference (voltage) across each capacitor, an C 1 C 1 = 4.0 µf 200 V C 2 C 3 C 2 = 15.0 µf C 3 = 12.0 µf () the energy store in each capacitor.
C C 1 1 C 12 C eq 200 V C C 3 C C 12 C C 3 eq 200 V 3 3 C C 2 2 (a) Note that C 1 an C 2 are in series: 1 C 12 = 1 C 1 1 C 2 = 1 4 µf 1 15 µf = 0.3167 106. C 12 = 3.16 µf. But C 12 an C 3 are in parallel: C eq = C 12 C 3 = 3.16 µf 12 µf = 15.16 µf. (b) We have: Q 1 = Q 2 (= Q) an V = V 1 V 2 = Q Q = Q 1 1 C 1 C 2 C 1 C 2 (c) By efinition V i = Q i C i. V 1 = Q 1 0.632 10 3 = C 1 4 10 6 =158 V, V 2 = Q 2 0.632 10 3 = C 2 15 10 6 = 42 V, an V 3 = 200 V. () By efinition U i = 1 2 Q iv i, 200 V i.e., Q = 200 = Q 0.3167 10 6, 200 0.3167 10 6 = 0.632 10 3 C ( = Q 1 = Q 2 ) Also Q 3 = C 3 V = 12 10 6 200 = 2.4 10 3 C. U 1 = 0.05 J: U 2 = 0.013 J: U 3 = 0.24 J. Check total store energy using the equivalent... U = 1 2 C eqv 2 = 1 2 15.16 10 6 200 2 = 0.303 J, i.e., the same.
3 3 3 3 (a) (b) (a) (b) Question 24.5: Two capacitors each have a plate separation. A slab of metal is place between the plates as shown. In case (a) the slab is not connecte to either plate; in case (b) it is connecte to the upper plate. Which arrangement prouces the higher capacitance, or o they have the same capacitance? Case (a) is equivalent to two capacitors in series each with capacitance C an spacing 3 : C C 1 C eq = 1 C 1 C = 2 C Case (b) is a single capacitor: C eq = C. C eq = C 2. Therefore, (b) has the higher capacitance.
Question 24.6: A 2.00 µf capacitor is energize to a potential ifference of 12.0 V by connecting it across a battery. The wires are then isconnecte from the battery an connecte across of secon, initially uncharge capacitor. The potential ifference across the 2.00 µf capacitor then rops to 4.00 V. (a) What is the capacitance of the secon capacitor? (b) What is the energy of the system before an after? (a) The initial charge on the 2 µf capacitor is Q = C 1 V = 2 10 6 12 = 24 10 6 C. When connecte across the secon capacitor, this charge is reistribute (none is lost!!). The new charges are an Q Q 2 µf 12 V Q 1 = C 1 V = 2 10 6 4 = 8 10 6 C, Q 2 = C 2 V = C 2 4. Q 1 2 µf C 2 Q 2 4 V Q 1 Q 2 But total charge is conserve. so Q 1 Q 2 = Q. Q 2 = Q Q 1, i.e., C 2 4 = (24 10 6 C) (8 10 6 C) = 16 10 6 C. C 2 = 4 10 6 F = 4 µf.
(b) Q Q 2 µf 12 V Energy before 1 2 C 1V 2 Q 1 2 µf C 2 Q 2 4 V Q 1 Q 2 = 1 2 2 10 6 12 2 = 1.44 10 4 J. Energy after 1 2 C 1 V 2 1 2 C 2 V 2 = 1 2 2 10 6 4 2 1 2 4 10 6 4 2 i.e, a loss of 0.96 10 4 J! = 0.48 10 4 J. What? Where has it gone? Does it remin you of something in Physics 1? General case... Q Q C 1 V If C 2 is uncharge initially... Initial energy U i = 1 2 QV. Q 1 C 1 C 2 Q 2 V Q 1 Q 2 Final energy U f = 1 2 Q 1 V 1 2 Q 2 V = 1 2 (Q 1 Q 2 ) V = 1 2 Q V. U f = V U i V. But Q 1 Q 2 = C 1 V C 2 V = (C 1 C 2 ) V. Since Q 1 Q 2 = Q = C 1 V then, Like an inelastic collision! V V = C 1. C 1 C 2 U f C = 1 i.e., always <1. U i C 1 C 2
Dielectrics: Q Q Dielectric C = Q Q C! = Q Q V! V (a) (b) (a) The electric fiel in an isolate charge parallel plate capacitor (in vacuum) is: E! = σ. ε! Dielectrics (continue): Three avantages: maintains plate separation when small, increase capacitance for a given size. V! = E! = σ ε!. (b) When a ielectric is inserte, ε! κε!, where κ is the ielectric constant, then E = σ. κε! V = E = σ κε! = V! κ, i.e., the potential ifference is reuce, but the charge remains the same. C = Q V = κq V! = κc! = κε! A. Thus, the capacitance increases by a factor of κ. ielectric increases the max. electric fiel possible, an hence potential ifference (voltage), between plates before breakown (ielectric strength). Material κ Dielectric Strength (V/m) Air 1.00059 3 10 6 Paper 3.7 16 10 6 Neoprene 6.9 12 10 6 Polystyrene 2.55 24 10 6
How oes a ielectric work?... σ σ σ in σ in no ielectric E! = σ ε! E in = σ in Because the ielectric is polarize, the electric fiel in the presence of a ielectric is reuce to: E = E! E in = σ ε! σ in ε! = σ σ in ε! = E! κ. Therefore, the potential ifference V (= E) is reuce by a factor of κ also. Since C 1 V C > C!. ε! E! E in ielectric becomes polarize σ σ σ in σ in E! = σ ε! E in = σ in ε! Hence, the inuce electric fiel is: E in = E! E! κ = κ 1 κ E!. Substituting for E in an E!, we obtain σ in = κ 1 κ σ. E! E in Hence σ in σ. Note: σ in = 0 if κ = 1 (free space, i.e., no ielectric). For a conucting slab κ = an σ in = σ. E = E! E in = 0, i.e., there is no electric fiel insie a conuctor.
X Y Question 24.7: Two, ientical capacitors, X an Y, are connecte across a battery as shown. A slab of ielectric is then inserte between the plates of Y. X Y (a) When the ielectric is inserte, the capacitance of Y increases from C to κc, where κ is the ielectric constant. But the capacitance of X is unchange. The potential ifference across both capacitors remains the same ( = V) an since the charge on a capacitor is given by Q = CV, if C increases to κc, then Q increases to κq. (a) Which capacitor has the greater charge or o the charges remain the same? Where oes the extra charge come from?...... from the battery! (b) What ifference (if any) woul it make if the battery was isconnecte before the ielectric was inserte? (b) If the battery was isconnecte before the ielectric was inserte, the charge on each capacitor is unchange. But, since the capacitance of Y increases to κc, the potential ifference across Y changes from Q C to Q κc, i.e., it gets smaller.
Two ways to answer part (a)... [1] Algebraically: Let C C! an V V! With the ielectric: U = 1 2 CV2. Question 24.8: A ielectric is place between the plates of a capacitor. The capacitor is then charge by a battery. After the battery is isconnecte, the ielectric is remove. (a) Does the energy store by the capacitor increase, ecrease or remain the same after the ielectric is remove? (b) If the battery remains connecte when the ielectric is remove, oes the energy increase, ecrease or remain the same? Without the ielectric: U! = 1 2 C!V! 2. But C = κc!, i.e., C! = C κ, an V = V! κ, i.e., V! = κv. U! = 1 C ( κv) 2 κ 2 = κ 1 2 CV2 = κu. Therefore, the energy increases. [2] Conceptually: You must o work to remove the ielectric. Therefore, the store (potential) energy increases!
(b) If the battery remains connecte, the potential ifference remains constant ( = V). But the capacitance changes from C C!, with C = κc!, i.e., C > C!. The energy changes from U = 1 2 CV2 to U! = 1 2 C!V 2. Since C > C!, then U > U!, i.e., the energy ecreases. Question 24.9: You are require to fabricate a 0.01 µf parallel plate capacitor that can withstan a potential ifference of 2.0 kv. You have at your isposal a ielectric with a ielectric constant of κ = 24 an a ielectric strength of 4.0 10 7 V/m. (a) What is the minimum plate separation require? (b) What is the area of each plate at this separation?
(a) Since E = V = V E. min = (b) C = κ ε!a V = 2000 E max 4 10 7 = 50 10 6 m. A = C κε!. A variety of (fixe value) capacitors A = C κε! = 0.10 10 6 50 10 6 24 8.854 10 12 = 2.35 10 2 m 2 235 cm 2. A simple variable (air) capacitor
charge 1 uncharge 0 charge 1 charge 1 Binary [1 0 1 1] Decimal 11 Dynamic ranom access memory (DRAM) is compose of banks of capacitors. A charge capacitor represents the binary igit 1 an uncharge capacitor represents the binary igit 0. (a) (b) Question 24.10: A parallel plate capacitor is charge by a generator. The generator is then isconnecte (a). If the spacing between the plates is ecrease (b), what happens to: (i) the charge on the plates, (ii) the potential across the plates, an (iii) the energy store by the capacitor.
(a) (b) A parallel plate capacitor is charge by a generator. The generator is then isconnecte (a). If the spacing between the plate is ecrease (b), what happens to: (i) the charge on the plates remains the same, where coul it go or where coul charge come from? (ii) the potential across the plates ecreases, because the electric fiel remains constant it s inepenent of (chapter 22) but as spacing ecreases, the potential (V= E.) ecreases. (iii) the energy store by the capacitor ecreases, because the system oes work to reuce spacing. Conversely, you woul have o work in orer increase the spacing. (a) (b) Question 24.11: A parallel plate capacitor is charge by a battery (a). When fully charge, an while the battery is still connecte, the spacing between the plate is ecrease (b), what happens to: (i) the potential across the plates, (ii) the charge on the plates, an (iii) the energy store by the capacitor.
(a) (b) A parallel plate capacitor is charge by a generator (a). When fully charge, an while the generator is still connecte, the spacing between the plate is ecrease (b). (i) the potential across the plates remains the same since the source of potential ifference is still connecte! (ii) the charge on the plates increases, because the capacitance increases an so, if V is unchange, Q increases. (iii) the energy store by the capacitor increases, because Q an C increase ( U = 1 2 QV = 1 2 CV2 ).