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 ! = r ln b ( ra ) • 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 between the plates. where C eq = C1 + C2. 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 +Q C 1 C = 4.0 µF C 1 1 − Q C 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) = . (a) the equivalent capacitance of the combination, C1 C2 Therefore, the total potential difference is: (b) the charge on the positive plate of each ⎛ ⎞ Q Q 1 1 Q capacitor, ∴VBA = V1 + V2 = + = Q⎜ + ⎟ = , C1 C2 ⎝ C1 C2 ⎠ Ceq 1 1 1 (c) the potential difference (voltage) across each providing = + . C eq C1 C2 capacitor, and With more than two capacitors: 1 1 (d) the energy stored in each capacitor. = ∑ . C i C eq i C1 C C 12 C 3 eq C 1 C C C 12 3 eq 200 V C 3 200 V C 3 C 2 C 2 Q (c) By definition V = i . i C (a) Note that C 1 and C 2 are in series: i 1 1 1 1 1 6 Q 0.632 ×10−3 ∴ = + = + = 0.3167 ×10 . ∴V = 1 = =158 V , C C C 4 µF 15 µF 1 −6 12 1 2 C1 4 ×10 200 V ∴ C12 = 3.16 µF. −3 Q2 0.632 ×10 But C and C are in parallel: V2 = = −6 = 42 V, 12 3 C2 15×10 ∴ Ceq = C12 + C3 = 3.16 µF +12 µF = 15.16 µF. and V 3 = 200 V . (b) We have: Q 1 = Q2 (= Q) Q Q ⎛ 1 1 ⎞ and V = V + V = + = Q⎜ + ⎟ 1 1 2 (d) By definition Ui = QiVi, C1 C2 ⎝ C1 C2⎠ 2 6 ∴ 200 = Q × 0.3167 ×10 , ∴ U1 = 0.05 J: U 2 = 0.013 J: U 3 = 0.24 J. 200 −3 Check total stored energy using the equivalent ... i.e., Q = 6 = 0.632 ×10 C ( = Q1 = Q2) 0.3167 ×10 1 2 1 −6 2 U = CeqV = ×15.16 ×10 × 200 = 0.303 J, −6 −3 2 2 Also Q 3 = C3V = 12 ×10 × 200 = 2.4 ×10 C. i.e., the same. d 3 d d d 3 3 d d (a) (b) 3 (a) (b) Case (a) is equivalent to two capacitors in series each d Question 24.5: Two capacitors each have a plate with capacitance C and spacing 3: separation d. A slab of metal is placed between the 1 1 1 2 = + = plates as shown. In case (a) the slab is not connected to C C eq C C C C either plate; in case (b) it is connected to the upper plate. C ∴Ceq = . Which arrangement produces the higher capacitance, or 2 do they have the same capacitance? Case (b) is a single capacitor: C eq = C. Therefore, (b) has the higher capacitance. + +

+ Q 2 µF + Q1 2 µF C 2 + Q2 12 V 4 V − Q − Q1 − Q2 Question 24.6: A 2 .00 µF capacitor is energized to a potential difference of 12.0 V by connecting it across a (a) The initial charge on the 2 µF capacitor is battery. The wires are then disconnected from the battery −6 −6 and connected across of second, initially uncharged Q = C1V = 2 ×10 ×12 = 24 ×10 C. When connected across the second capacitor, this charge capacitor. The potential difference across the 2 .00 µF capacitor then drops to 4.00 V. is redistributed (none is lost!!). The “new” charges are −6 −6 Q 1 = C1V ′ = 2 ×10 × 4 = 8 ×10 C, (a) What is the capacitance of the second capacitor? and

Q 2 = C2V ′ = C2 × 4. (b) What is the energy of the system before and after? But total charge is conserved. so Q 1 + Q2 = Q. ∴ Q2 = Q − Q1, −6 −6 −6 i.e., C 2 × 4 = (24 ×10 C) − (8 ×10 C) = 16 ×10 C. −6 ∴ C2 = 4 ×10 F = 4 µF. + + General case ...

+ Q 2 µF + Q1 2 µF C 2 + Q2 12 V 4 V + Q C 1 + Q1 C 1 C 2 + Q2 − Q − Q1 − Q2 V V ′ − Q − Q1 − Q2 (b) 1 2 If C 2 is uncharged initially ... • Energy before ⇒ C1V 2 1 • Initial energy Ui = QV . 1 2 = × 2 ×10−6 ×122 = 1.44 ×10−4J. 2 1 1 • Final energy Uf = Q1V ′ + Q2V ′ 1 2 1 2 2 2 • Energy after ⇒ C1V ′ + C2V ′ 2 2 1 1 = (Q1 + Q2)V ′ = QV ′ . 1 1 2 2 = × 2 ×10−6 × 42 + × 4 ×10−6 × 42 2 2 U V ′ ∴ f = . = 0.48 ×10−4J. Ui V i.e, a loss of 0 .96 ×10−4J! But Q 1 + Q2 = C1V ′ + C2V ′ = (C1 + C2)V ′ . V ′ C1 What? Where Since Q 1 + Q2 = Q = C1V then, = . V C + C has it gone? 1 2 U C ∴ f = 1 i.e., always < 1. Ui C1 + C2 Does it remind you of something in Physics 1? Like an inelastic collision! Dielectrics: Dielectrics (continued): + Q + Q Dielectric d Q − Q Q − Q Three advantages: C = C = ! V! V (a) (b) • maintains plate separation when small, (a) The electric field in an isolated charged parallel plate σ capacitor (in vacuum) is: E! = . • increased capacitance for a given size. ε! σ ∴V! = E!d = d. • dielectric increases the max. electric field possible, ε! and hence potential difference (voltage), between (b) When a dielectric is inserted, ε ⇒ κε , where κ is ! ! plates before breakdown (dielectric strength). σ the dielectric constant, then E = . κε! Material κ Dielectric σ V ∴V = Ed = d = ! , Strength (V/m) κε! κ Air 1.00059 3 ×106 i.e., the potential difference is reduced, but the charge Paper 3.7 16 106 remains the same. × 6 Q κQ ⎛ A⎞ Neoprene 6.9 12 ×10 ∴C = = = κC ⎜ = κε ⎟ . ! ⎝ ! ⎠ 6 V V! d Polystyrene 2.55 24 ×10 Thus, the capacitance increases by a factor of κ. How does a dielectric work? ... +σ −σ − σin + σin +σ −σ − σin + σin + - - + + - + - - + + - + - - + + - + - - + + - + - - + + - - + - - + + d σ σin E = Ein = E − E d no dielectric ! ε ε ! in σ σ E ! − Ein ! ! E = E = in ! in Hence, the induced electric field is: ε! ε! - + E κ −1 - + - + - + - + E = E − ! = ⎛ ⎞ E . - + in ! ⎝ ⎠ ! - + - + - + - + κ κ - + Substituting for E and E , we obtain dielectric becomes polarized in ! ⎛ κ −1⎞ σin = σ. Because the dielectric is polarized, the electric field in ⎝ κ ⎠ the presence of a dielectric is reduced to: Hence σ in ≤ σ. Note: σ in = 0 if κ = 1 (free space, i.e., σ σin σ − σin E! no dielectric). E = E! − Ein = − = = . ε! ε! ε! κ Therefore, the potential difference V (= Ed) is reduced For a conducting slab κ = ∞ and σ in = σ. by a factor of κ also. Since C ∝ 1 V ∴ E = E! − Ein = 0, C > C!. i.e., there is no electric field inside a conductor. + + X Y − − + + X Y (a) When the dielectric is inserted, the capacitance of Y − − increases from C to κ C, where κ is the dielectric constant. But the capacitance of X is unchanged. The Question 24.7: Two, identical capacitors, X and Y, are potential difference across both capacitors remains the connected across a battery as shown. A slab of dielectric same ( = V) and since the charge on a capacitor is given is then inserted between the plates of Y. by Q = CV , if C increases to κ C, then Q increases to κ Q.

(a) Which capacitor has the greater charge or do the Where does the extra charge come from? ... charges remain the same? ... from the battery!

(b) What difference (if any) would it make if the (b) If the battery was disconnected before the dielectric battery was disconnected before the dielectric was was inserted, the charge on each capacitor is unchanged. inserted? But, since the capacitance of Y increases to κ C, the potential difference across Y changes from Q Q C to κC, i.e., it gets smaller. Two ways to answer part (a) ...

[1] Algebraically: Let C → C! and V → V! 1 With the dielectric: U = CV2. 2 Question 24.8: A dielectric is placed between the plates 1 2 Without the dielectric: U! = C!V! . of a capacitor. The capacitor is then charged by a 2 C V! battery. After the battery is disconnected, the dielectric But C = κC , i.e., C = , and V = , i.e., V = κV. ! ! κ κ ! is removed. 1⎛ C⎞ 2 ⎛ 1 2⎞ ∴U! = (κV) = κ CV = κU. 2⎝ κ⎠ ⎝ 2 ⎠ (a) Does the energy stored by the capacitor increase, Therefore, the energy increases. decrease or remain the same after the dielectric is removed? [2] Conceptually: You must do work to remove the dielectric. Therefore, the (b) If the battery remains connected when the + + + + + + stored (potential) energy dielectric is removed, does the energy increase, decrease - - - - - + + + + + increases! or remain the same? ------Question 24.9: You are required to fabricate a 0 .01 µF (b) If the battery remains connected, the potential parallel plate capacitor that can withstand a potential difference remains constant ( = V). But the capacitance difference of 2.0 kV. You have at your disposal a dielectric with a dielectric constant of κ = 24 and a changes from C → C!, with C = κC!, i.e., C > C!. The energy changes from dielectric strength of 4.0 ×107 V/m. 1 2 1 2 U = CV to U! = C!V . 2 2 (a) What is the minimum plate separation required? Since C > C!, then U > U!, i.e., the energy decreases. (b) What is the area of each plate at this separation? V V (a) Since E = ⇒ d = . d E V 2000 −6 ∴dmin = = 7 = 50 ×10 m. Emax 4 ×10 A variety of (fixed value) capacitors ε A Cd (b) C = κ ! ⇒ A = . d κε! Cd 0.10 ×10−6 × 50 ×10−6 ∴A = = −12 κε! 24 × 8.854 ×10 = 2.35 ×10−2 m2 ⇒ 235 cm2.

A simple variable (air) capacitor + - “charged” 1 + -

“uncharged” 0

+ - “charged” (a) (b) 1 + - Question 24.10: A parallel plate capacitor is charged by a generator. The generator is then disconnected (a). If + - “charged” the spacing between the plates is decreased (b), what 1 + - happens to: Binary [1 0 1 1] ⇒ Decimal 11 (i) the charge on the plates, (ii) the potential across the plates, and Dynamic random access memory (DRAM) is composed (iii) the energy stored by the capacitor. of banks of capacitors. A “charged” capacitor represents the binary digit “1” and “uncharged” capacitor represents the binary digit “0”. (a) (b)

A parallel plate capacitor is charged by a generator. The generator is then disconnected (a). If the spacing between the plate is decreased (b), what happens to:

(i) the charge on the plates remains the same, where (a) (b) could it go or where could charge come from? Question 24.11: A parallel plate capacitor is charged by a battery (a). When fully charged, and while the battery (ii) the potential across the plates decreases, because the is still connected, the spacing between the plate is electric field remains constant - it’s independent of d decreased (b), what happens to: (chapter 22) - but as spacing decreases, the potential (V= E.d) decreases. (i) the potential across the plates, (ii) the charge on the plates, and (iii) the energy stored by the capacitor decreases, (iii) the energy stored by the capacitor. because the system does work to reduce spacing. Conversely, you would have do work in order increase the spacing. (a) (b) A parallel plate capacitor is charged by a generator (a). When fully charged, and while the generator is still connected, the spacing between the plate is decreased (b).

(i) the potential across the plates remains the same since the source of potential difference is still connected!

(ii) the charge on the plates increases, because the capacitance increases and so, if V is unchanged, Q increases.

(iii) the energy stored by the capacitor increases, 1 1 because Q and C increase ( U = QV = CV2). 2 2