UNIT 4:Capacitors and Dielectric

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1 UNIT 4:apacitors an Dielectric SF7 4. apacitor A capacitor is a evice that is capable of storing electric charges or electric potential energy. It is consist of two conucting plates separate by a small air gap or a thin insulator (calle a ielectric such as mica, ceramics, paper or even oil). The conucting plates coul be in the shape of ylinrical Spherical Parallel plate The electrical symbol for a capacitor is or 4.. Uses of apacitors apacitors are commonly use in many electronic circuits. Some examples are : Photoflash unit capacitor stores large amount of energy which can be release in a flash when triggere. SF7

2 Giant lasers large amount of energy store in capacitors coul power large lasers. On-off switches low energy capacitors can be use as switches in computer motherboars. Smoothen.c. voltages smoothene.c. voltages are obtaine from the rectification of a.c. voltages. 4.. Types of apacitors There are three commercial capacitor esigns. It is tubular capacitor, high-voltage capacitor an electrolytic capacitor. A tubular capacitor whose layers of metallic foil an ielectric are rolle into a cyliner (figure 4.a). High-voltage capacitor consists of a number of interwoven metallic plates immerse in silicone oil (figure 4.b). An electrolytic capacitor consists of a metallic foil in contact with an electrolyte (figure 4.c). Metal foil Fig. 4.a Paper Fig. 4.b SF7 Fig. 4.c 4. apacitance, The capacitance of a capacitor, Definition is efine as the ratio of the charge on either plate to the potential ifference between them. Mathematically, where :charge on one of the plates : potential ifference across the two plates SF7 4

3 The unit of capacitance is the fara (F). fara is efine as the charge of coulomb store on each of the conucting plates as a result of a potential ifference of volt between the two plates. i.e. By rearranging the equation from the efinition of capacitance, we get where the capacitance of a capacitor, is constant then Note : coulomb fara volt (The charges store, is irectly proportional to the potential ifference, across the conucting plate.) The fara is a very large unit. Therefore in many applications the most convenient units of capacitance are microfara an the picofara where the unit conversion can be shown below : 6 µf F pf F 6 µf µµf SF Parallel-plate apacitors onsier two parallel metallic plate capacitor of equal area A are separate by a istance an the space between plates is vacuum or air as shown in figure 4.a. positive terminal negative terminal Fig. 4.a Area, A One plate carries a charge an the other carries a charge then the potential ifference between this two parallel plates is. Because is small compare to the imensions of each plate so that the electric fiel strength E is uniform between them. The magnitue of the electric fiel strength is given by σ E an σ ε A E () Aε SF7 6 E r

4 Since then equation () can be written as E () Aε Because the fiel between the plates is uniform, the potential ifference between the plates is E Substituting this relation into eq. (), thus the capacitance of a parallelplate capacitor is The capacitance of a parallel-plate plate capacitor is proportional to the area of its plates an inversely proportional to the plate separation where E E Aε ε A or εa Parallel material : permittivity of free space ( ε 8. 85x N m ) A: Area of the plate : istance between the two plates ε Parallel-plate capacitor separate by a vacuum Parallel-plate capacitor separate by a ielectric SF7 7 Example : The plates of a parallel-plate capacitor are 8. mm apart an each has an area of 4. cm. The plates are in vacuum. If the potential ifference across the plates is. k, etermine a. the capacitance of the capacitor. b. the amount of charge on each plate. c. the electric fiel strength was prouce.. the surface charge ensity on each plate. (Given permittivity of free space, ε 8.85 x - N - m - ) Solution: 8.x - m, A4.x -4 m,.x a. By applying the equation of capacitance for parallel-plate capacitor, ε A 4. 4x 44 pf b. From the efinition of the capacitance, the amount of charge store in the capacitor is given by SF7 8 ( 4. 4x 8. 84x )(. x )

5 c. From the relation between uniform E an, thus E 5 E. 5x m. By using the equation of surface charge ensity, then σ σ or E A ε 6 σ. x m Example : A circular parallel-plate capacitor with raius of cm is connecte to a 5 battery. After the capacitor is fully charge, the battery is isconnecte without loss of any of the charge on the plates. If the separation istance between plates is 5 mm an the meium between plates is air. a. Fin the amount of charge on each plate. If their separation is increases to 5 mm after the battery is isconnecte, etermine b. the amount of charge on each plate. c. the potential ifference between plates.. the capacitance of the capacitor. (Given permittivity of free space, ε 8.85 x - N - m - ) SF7 9 σ Eε Solution: rx - m, 5, 5x - m A πr. 4x m a. From the efinition of the capacitance, the amount of charge is ε A an ε A. 9x The separation, 5 mm. b. The amount of charge on each plate is not change because the electric fiel between the plates is uniform then. 9x c. From the relation between uniform E an, thus E E an E ε A ε A. 4 SF7

6 . The capacitance of the capacitor is given by or x Example : (exercise) a. A parallel-plate, air-fille capacitor has circular plates separate by.8 mm. The charge per unit area on each plate has magnitue 5.6 p m -. Fin the potential ifference between the plates of the capacitor. (Young & Freeman,pg.94.no.4.4) Ans. :.4 m b. An electric fiel of.8x 5 m - is esire between two parallel plates each of area. cm an separate by.5 cm of air. Fin the charge on each plate. (Giancoli,pg.68.no. 4) Ans. :5.x -9 (Given permittivity of free space, ε 8.85 x - N - m - ) Example 4 : (exercise) A. µf parallel-plate capacitor with circular plates is connecte to a. battery. alculate a. the charge on each plate. b. the charge on each plate if their separation were twice while the capacitor remaine connecte to the battery. c. the charge on each plate if the capacitor were connecte to the. battery after the raius of each plate was twice without changing their separation (Young & Freeman,pg.94.no.4.5) Ans. : µ, 6 µ, 48 µ SF7 ε A F 4.4 Spherical apacitors onsier a spherical capacitor consists of two thin concentric spherical conucting shells, of raius r a an r b are separate by a vacuum as shown in figure 4.4a. r a r r b E r Gaussian surface The inner shell carries a uniformly istribute charge on its surface an the outer shell an equal but opposite charge. To fin the magnitue of the electric fiel between two spherical shells, a spherical gaussian surface is rawn (figure 4.4a). The magnitue of the electric fiel is constant an normal to the surface at each point everywhere on the gaussian surface. By using the Gauss s law, q Fig. 4.4a A 4πr q enc enc SF7 Φ E EA where an ε

7 E A E( 4πr ) then the magnitue of the electric fiel between two spherical shells is E 4πεr To obtain the potential ifference, ab between the two conucting shell, we apply the relation between E an below, E r Er a ra Er b rb ra a b r an r a b 4πε b r ab 4πε ra rb 4 b a ab SF7 πε ra rb r r ε ab From the efinition of the capacitance, ab then the capacitance of a spherical capacitor is given by where rb ra 4πε ra rb ra rb 4πε rb r or r a : raius of the inner shell SF7 r b : raius of the outer shell 4 a spherical capacitor separate by a vacuum r arb spherical capacitor 4πε rb ra material. ε : permittivity of free space ε : permittivity of ielectric material separate by a ielectric

8 Example 5 : A spherical capacitor is forme from two concentric, spherical, conucting shells separate by vacuum. The inner sphere has raius 5. cm an the capacitance is 6 pf. a. Fin the raius of the outer sphere. b. If the potential ifference between the two spheres is, calculate the magnitue of charge on each sphere. (Young & Freeman,pg.95.no.4.) (Given permittivity of free space, ε 8.85 x - N - m - ) Solution: r a 5.x - m, 6x - F a. By using the equation of capacitance for spherical capacitor, ra rb 4πε rb ra then the raius of the outer shell is given by r b r b ra 4πε. 8 m r a SF7 5 b. Given From the efinition of the capacitance, thus the charge on each sphere is 8. 55x Example 6 : A. µf spherical capacitor is compose of two concentric metal spheres, one having a raius twice as large as the other. The region between the spheres is a vacuum. Determine the volume of this region. (Serway & Jewett,pg.8,no.) (Given permittivity of free space, ε 8.85 x - N - m - ) Solution: r a r, r b r a r,.x -6 F r a r b vacuum By applying the equation of capacitance for spherical capacitor, ra rb 4πε rb ra SF7 6 ( r)( r) 4πε r r 4 r 9. x m

9 The volume of the vacuum region, is given by olume of the outer sphere, b [ r b r ] 4 π a 8 πr 6. x m 4.5. apacitors connecte in series Figure 4.5a shows capacitors connecte in series to a battery of voltage, olume of the inner sphere, a 4.5 ombinations of apacitors,,, equivalent to eq, SF7 7 Fig. 4.5a When the circuit is complete, electrons are transferre onto the plates such that the magnitue of the charge on each plate is the same. Thus the total charge () on the equivalent capacitor is The potential ifference across each capacitor, an are, an respectively. Hence ; ; Since the total potential ifference is given by Therefore the equivalent (effective) capacitance eq for n capacitors connecte in series is given by... SF7 eq n 8 an eq capacitors connecte in series

10 4.5. apacitors connecte in parallel Figure 4.5b shows capacitors connecte in parallel to a battery of voltage,. -, - -, -, equivalent to Fig. 4.5b The potential ifference across each capacitor is the same as the supply voltage (). Thus the total potential ifference () on the equivalent capacitor is The charges store by each capacitor, an are, an respectively. Hence ; SF7 9 ; eq, Since the total charge on the equivalent capacitor is given by an eq Therefore the equivalent (effective) capacitance eq for n capacitors connecte in parallel is given by... eq Example 7 : Determine the equivalent capacitance of the configuration shown in figure below. All the capacitors are ientical an each has capacitance of µf. µf µf n capacitors connecte in parallel µf µf µf µf SF7

11 Solution: Label all the capacitors in the circuit To fin the equivalent capacitance for circuit above, it is easier to solve it from the en of the circuit (left) to the terminal (right) shown by an arrow in figure above. apacitors, an connecte in series, then x SF7 x x µf apacitors x, 4 an 5 connecte in series, then y 6 apacitors y an 6 connecte in parallel, then the eq equivalent capacitance eq is given by y y x 5 µf 4 eq y 6 eq µf µf 5 6 eq µf 5 5 SF7

12 Example 8 : a a In the circuit shown in figure above,. µf, 4. µf an 9. µf. The applie potential ifference between points a an b is ab 6.5. alculate a. the charge on each capacitor. b. the potential ifference across each capacitor. c. the potential ifference between points a an. similar to (Young & Freeman,pg.96.no.4.4) Solution: b,, a x, x b, SF7 b, a. From the figure, apacitors an connecte in parallel then the equivalent capacitance x is given by x µf 4 µf x 6 µf The equivalent capacitance, eq in the circuit is given by eq x eq. 6 µf The total charge store in the equivalent capacitance eq is 6 eq ab (. 6 x )( 6. 5) 4. x Since the capacitors x an connecte in series then the charge store in each capacitor is the same as the total µ µ x The potential ifference across the capacitor is 4. 6 ( µ ) ( 9 µ ) SF7 4

13 then the potential ifference across the equivalent capacitance x is given by x ab x 6. 9 Since the capacitors an connecte in parallel then the potential ifference across each capacitor is the same as x x Therefore ( µ )( 6. 9) 7. 8 µ x or 47 µ b. The potential ifference across the capacitors an is 6. 9 an the potential ifference, is given by 4. 6 c. The potential ifference between points a an is given by a x 6. 9 SF7 5 Example 9 : (exercise) Four capacitors are connecte as shown in figure below. alculate a. the equivalent capacitance between points a an b. b. the charge on each capacitor if ab 5.. (Serway & Jewett,pg.8,no.) Ans. : 5.96 µf, 89.5 µ on µf, 6. µ on 6 µf, 6. µ on 5 µf an on µf Example : (exercise) Fin the equivalent capacitance between points a an b for the group of capacitors connecte as shown in figure below. Take 5. µf,. µf an. µf. (Serway & Jewett,pg.84,no.7) Ans. : 6.4 µf SF7 6

14 4.6 Energy store in a harge capacitor, U When the switch is close in figure 4.6a, charges begin accumulate on the plates. A small amount of work (W ) is one in bringing a small amount of charge () from the battery to the capacitor. This is given by an W W Fig. 4.6a The total work W require to increase the accumulate charge from zero to is given by W also U W Note : No charges will accumulate on each plate if the capacitor is not charge. SF7 7 U an U Example : Two capacitors,. µf an 6. µf are connecte in series an charge with a 4. battery as shown in figure below. 4. alculate a. the total capacitance for the circuit above. b. the charge on each capacitor. c. the potential ifference across each capacitor.. the energy store in each capacitor. e. the area of the each plate in capacitor if the istance between two plates is. mm an the region between plates is vacuum. (Given permittivity of free space, ε 8.85 x - F m - ) Solution: 4. a. Since capacitors an connecte in series, thus the total capacitance, is given by. µ F 6. µ F SF7 8. µf

15 b. Since capacitors an connecte in series, thus the charge on capacitors an is equal to the total charge in the circuit. Therefore c. The potential ifference across: apacitor,. 67 apacitor,. The energy store in: apacitor, apacitor, (. µ )( 4 ). 8. µ or. U 6. x U. 7x J U U 5. x J SF7 6 9 ( )( ) e. Given. x -6 F,. x - m an ε 8.85 x - F m - By applying the formula of capacitance for parallel plate capacitor, thus the area of the each plate in capacitor is given by ε A A. 9 m Example : onsier the circuit shown in figure below, where 6. µf,. µf an.. apacitor is first charge by the closing of switch S. Switch S is then opene, an the charge capacitor is connecte to the uncharge capacitor by the closing of S. alculate the initial charge acquire by an the final charge on each capacitor. (Serway & Jewett,pg.84,no.) Solution: After the switch S is close. The capacitor is fully charge an the charge has been place on it is given by ( )( ) 6 µ µ SF7..

16 After the switch S is close an S is opene. The capacitors an (uncharge) are connecte in parallel an the equivalent capacitance is eq 9. µf By using the principle of conservation of charge, the total charge on the circuit is given by µ The potential ifference across each capacitor is the same (parallel) an given by ' eq '. Therefore the final charge accumulates on ' capacitor : ' 6. µ. ' µ ' ' ' capacitor : or ' 4. µ ( )( ) ' SF7 4.7 harging an Discharging of a apacitor 4.7. harging a capacitor through a resistor Figure 4.7a shows a simple circuit for charging a capacitor. R switch, S e A B e When the switch S is close, current I immeiately begins to flow through the circuit. Electrons will flow out from the negative terminal of the battery, through the resistor R an accumulate on the plate B of the capacitor. Then electrons will flow into the positive terminal of the battery, leaving a positive charge on the plate A. Fig. 4.7a As charge accumulates on the capacitor, the potential ifference across it increases an the current is reuce until eventually the maximum voltage across the capacitor equals the voltage supplie by the battery,. At this time, no further current flows (I ) through the resistor R an the charge on the capacitor thus increases graually an reaches a maximum value. SF7

17 The charging process can be shown through the graphs in figure 4.7b, 4.7c an τ R Fig. 4.7b : the potential ifference across capacitor increases with time. I time,t τ R time,t Fig. 4.7c : the charge on the capacitor increases with time. I.7 I Fig. 4.7 : the current through the resistor ecreases exponentially with time. τ R time,t SF7 The equations involve in the charging process : harge on charging capacitor : t R e (4.7-) Potential ifference (voltage) t across charging capacitor : e R an urrent in resistor : where I t R e SF7 4 I I e an I R : maximum charge : maximum voltage supply voltage : maximum current R :resistance of the resistor :capacitance of the capacitor t R (4.7-) (4.7-)

18 4.7. Discharging a capacitor through a resistor Figure 4.7e shows a simple circuit for ischarging a capacitor. A B R switch, S When a capacitor is alreay charge to a voltage an it is allowe to ischarge through the resistor R as shown in figure 4.7e. When the switch S is close, electrons from plate B begin to flow through the resistor R an neutralizes positive charges at plate A. Fig. 4.7e Initially, the potential ifference (voltage) across the capacitor is maximum, an then a maximum current I flows through the resistor R. When part of the positive charges on plate A is neutralize by the electrons, the voltage across the capacitor is reuce. The process continues until the current through the resistor is zero. At this moment, all the charges at plate A is fully neutralize an the voltage across the capacitor becomes zero. The ischarging process can be shown through the graphs in figure 4.7f, 4.7g an 4.7h. SF7 5 e e.7.7 Fig. 4.7f : the potential ifference across capacitor ecreases exponentially with time. τ R time,t I τ R Fig. 4.7g : the charge on the capacitor ecreases exponentially with time. τ R time,t time,t.7 I Fig. 4.7h : the current through the resistor ecreases exponentially with time. I SF7 6

19 The equations involve in the ischarging process : harge on ischarging capacitor : e t R (4.7-4) 4) Potential ifference (voltage) across ischarging capacitor : e t R (4.7-5) urrent in resistor : The negative sign inicates that as the capacitor ischarges, the current irection opposite its irection when the capacitor was being charge. Note : For calculation of current in ischarging process, ignore the negative sign in the formula. t R I I e I R an (4.7-6) SF Time constant, τ The quantity R that appears in the exponent for all equation is calle time constant τ or relaxation time of the circuit or mathematically τ R Its imension is the imension of time, then the unit is secon (s). It is a measure of how quickly the capacitor charges or ischarges. harging process From eqs an 4.7-, the charge on the capacitor an the voltage across it, increase from zero at t to maximum values an after a very long time. The time constantτ is efine as the time require for the capacitor to reach (-e - ).6 or 6% of its maximum charge/voltage. either one From eq. 4.7-, the current rops exponentially in time constant equal to τ. The time constantτ is efine as the time require for the current to rop to /e.7 or 7% of its initial value(i ). Discharging process From eqs , an 4.7-6, the charge, the voltage an the current I is seen to ecrease exponentially in time with the same time constant τ.the time constantτ is efine as the time require for the charge on the capacitor/voltage across it/current in the t SF7 resistor ecrease to /e.7 or 7% of its initial value. 8

20 Example : In the R circuit shown in figure below, the battery has fully charge the capacitor. b a S R Then at t s the switch S is thrown from position a to b. The battery voltage is. an the capacitance. µf. The current I is observe to ecrease to.5 of its initial value in 4 µs. Determine a. the value of R. b. the time constant, τ b. the value of, the charge on the capacitor at t. c. the value of at t 6 µs Solution:.,.x -6 F, I.5I, t 4x -6 s a. By applying the equation of current for ischarging process (ignore t the negative sign) : R I I e 6 4 x 6 R(. x. 5I ) Ie then taking natural logs on both sies, thus the value of R is R 57 Ω SF7 9 b. The time constant is given by τ R τ 5. 8x 5 s c. By using the equation of charge for ischarging process an the time, t hence t R e an 5. 4x. By using the equation of charge for ischarging process an the time, t 6x -6 s hence t R e 6 6 x 5 8 x ( 5 4x ) e x 6 SF7 4

21 4.8 Dielectric Definition is efine as the non-conucting (insulating) material place between the plates of a capacitor. When a ielectric (such as rubber, glass or waxe paper) is inserte between the plates of a capacitor, the capacitance increases. This capacitance increases by a factor κ or ε r which is calle the ielectric constant (relative permittivity) of the material. The avantages of inserting the ielectric between the plates of the capacitor are Increase in capacitance Increase in maximum operating voltage. Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby ecreasing an increasing Dielectric constant, κ (ε r ) Definition is efine as the ratio between the capacitance of given capacitor with space between plates fille with ielectric, with the capacitance of same capacitor with plates in a vacuum,. Mathematically, (4.8-) κ (4.8 SF7 4 It is imensionless constant (no unit). For parallel-plates capacitor: εa an then the equation 4.8- can be written as κ ε εa A ε κ or ε κε ε where ε A From the efinition of the capacitance, ε : permittivity of an hence the equation 4.8- can be written as where κ :potential : potential (4.8-) SF7 4 (4.8-) ielectric material is constant ifference across capacitor with ielectric ifference across capacitor in vacuum

22 From the relationship between E an for uniform electric fiel, E an E thus the equation 4.8- can be written as where E E E κ : electric fiel strength of the capacitor in vacuum E :electric fiel strength of the capacitor with ielectric The ielectric constant epens on the insulating material use. Table below shows the value of ielectric constant an the ielectric strength fro several materials. Material Air Mylar Paper Silicone oil Water Teflon SF7 4 E E κ (4.8-4) 4) Dielectric constant, κ Dielectric Strength ( 6 m - ) The ielectric strength is efine as the electric fiel strength at which ielectric breakown occurs an the material becomes a conuctor. Since E for a uniform electric fiel, the ielectric strength etermines the maximum potential ifference that can be applie across a capacitor per meter of plate spacing. Summary : ε E κ ε E Example 4 : A parallel-plate capacitor has plates of area A x - m an separation cm. The capacitor is charge to a potential ifference. Then the battery is isconnecte an a ielectric sheet of the same area A is place between the plates as shown in figure below. ielectric In the presence of the ielectric, the potential ifference across the plates is reuce to. Determine a. the initial capacitance of the air-fille capacitor. b. the charge on each plate before the ielectric is inserte. c. the capacitance after the ielectric is in place. SF7 44

23 . the relative permittivity. e. the permittivity of ielectric sheet. f. the initial electric fiel. g. the electric fiel after the ielectric is inserte. (Given permittivity of free space, ε 8.85 x - F m - ) Solution: x - m, Ax - m,, a. Before the ielectric is in place, the capacitance is given by ε A. 77 x b. The charge on each plate is 5. x 9 c. In the presence of the ielectric, the charge on each plate is the same as before the ielectric was inserte. Therefore the capacitance is 5. x 6 9 F F SF7 45. From the efinition of ielectric constant, thus relative permittivity is κ κ e. The permittivity of ielectric sheet is ε κε - ε. 66 x F N m f. By applying the relationship between E an for uniform electric fiel, the magnitue of the initial electric fiel is g. The magnitue of the final electric fiel is E 5 - E x m E κ E 5 ( x ) E 5 E x N - - SF7 46

24 4.8. Dielectric effect on the parallel-plate capacitor (a) Polar ielectrics The molecules of some ielectrics like water have permanent electric ipole moments where the concentration of positive an negative charges are separate. When no electric fiels is present the polar molecules are oriente ranomly as shown in figure 4.8a. The electric ipoles ten to line up when the external electric fiel is applie to them as in figure 4.8b. Fig. 4.8a Fig. 4.8b The alignment of the electric ipoles prouces an electric fiel that is irecte opposite the applie fiel an smaller in magnitue. SF7 47 (b) Non-polar ielectrics Non-polar molecules such as glass or paraffin oil have their positive an negative charge centres at the same point in the absence of an external electric fiel as shown in figure 4.8c. When the non-polar molecules are place in an external electric fiel, these centres become separate slightly an the molecules acquire inuce ipole moments. These inuce ipole moments ten to align with the electric fiel an the ielectric is polarize as shown in figure 4.8. Fig. 4.8c Fig. 4.8 SF7 48

25 (c) Dielectric in a parallel-plate capacitor onsier a capacitor whose plates are separate by a ielectric material (either polar or non-polar). This capacitor has a charge on one plate an on the other, so that the electric fiel E is prouce between the plates. Because of the electric fiel, all the ielectric molecules ten to become oriente as shown in figure 4.8e. Fig. 4.8e The net effect in either case polar or non-polar is as if there were a net negative charge on the outer ege of the ielectric facing the positive plates an a net positive charge on the opposite sie as shown in figure 4.8f. E r SF7 49 E r Fig. 4.8f Fig. 4.8f The electric fiel lines o not pass through the ielectric but instea en on charges inuce on the surface of the ielectric as shown in figure 4.8f. Therefore the electric fiel within the ielectric is less than in air. E r E r SF7 5 in Fig. 4.8g

26 Accoring to figure r 4.8g, r the r electric fiel within the ielectric E is given by since where E E E κ E in E E E in E E E in then E E Ein κ or E κ Ein :electric fiel ue to the charges on the plates :electric fiel ue to the inuce charge on the surfaces of the ielectric SF7 5

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