Capacitors. January 2002 Number 29

Transcription

1 PhysicsFactsheet January 22 Number 29 Capacitors Introuction Capacitors are wiely use in electrical engineering an electronics hey are important in any physics course because of the variety of uses they have very simple capacitor consists of two parallel metal plates o investigate the charge store on a capacitor we can use a coulombmeter When a coulombmeter is connecte to a charge capacitor, it will take all the charge from the capacitor, measure it an isplay the result on a igital reaout left right rocker switch µc coulomb meter Fig 1 he capacitor is first connecte to a c supply an then to a sensitive ammeter (or galvanometer)when the capacitor is connecte to the sensitive ammeter, a momentary eflection is observe his eflection is a brief pulse of charge an illustrates an important iea with capacitors We say a capacitor stores charge We use the symbol to represent the amount of charge involve In this case, charge is taken from the c supply, store on the capacitor plates an then the same charge ischarge, through the ammeter Remember that the unit of charge is the coulomb (abbreviation C) lthough one coulomb is a small amount in current electricity, it is an enormous amount in static electricity In capacitors, the charge store is static an we use much smaller units, typically microcoulombs µc lthough a capacitor can be mae from any two conuctors close to each other, we have consiere the simplest case where the conuctors are two parallel metal plates You shoul also note that the plates are separate by an insulator, in this case air he insulating material is calle the ielectric Because the ielectric is an insulator it is clear that a steay c current cannot pass through a capacitor an this is why we only get the brief pulse of charge referre to above he symbol for a capacitor is simply: so the above iagram can be rerawn with this symbol + - Figure 3 By epressing the left han key of the rocker switch, the capacitor is charge to a potential ifference which can be ajuste by the variable csupply Upon epressing the right han key, the store charge is measure an isplaye by the coulombmeter able 1 shows the charge store for various potential ifferences up to 1 volts able 1 p (olt) charge (µc) he results are isplaye in the graph below You can see that there is a linear relationship charge/µc p/ he ratio is the graient of the straight line, so constant his constant is calle the capacitance of the capacitor Charge store capacitance or C potential ifference he unit of capacitance is the fara (F) o calculate the capacitance, must be in coulombs an in volts Because the coulomb is a large unit so also is the fara In practice you will use submultiples as shown in table 2 Notice that the plates are marke + an an that these signs correspon to those on the c supply You might think that the charge store is 2 his is not so sk yourself how much charge is flowing through the ammeter uring ischarge Only the amount flows from the positive plate, through the ammeter, an neutralises the charge on the negative plate 1 able 2 factor prefix name symbol F 1 3 millifara mf F 1 6 microfara µf F 1 9 nanofara nf F 1 12 picofara pf

2 o calculate the capacitance of the capacitor in the graph, we can use the last point (2 µc, 1) C C F So far we have been thinking that the space between the plates is air or a vacuum What will happen if an insulator, the ielectric, is introuce between the plates? he answer is that the capacitance will be increase by several times he factor by which it is increase is between 2 an 1 for most ielectrics an is calle the relative permittivity ε r Note that ε r oes not have any units, it simply multiplies up the capacitance (see table 3) C F C 2 µf (2 microfaras) ny pair of reaings in table 1 coul have been use giving the same answer, try using one or two yourself In each case you are fining the graient of the line on the charge against potential ifference graph If you o a real experiment, your reaings will probably not increase uniformly as in table 1 In this case you plot all the points an then raw the best straight line passing through the origin he graient of this line gives the average value for the capacitance able 3 Material ir Paper Mica Wax paper Relative permittivity ε r Exam Hint: If you are working out the graient of a line, or oing any calculation, always look carefully at the units to see if metric prefixes are being use Capacitance he size an separation of the plates affects the capacitance he two quantities you nee are area of overlap of the plates an the plate separation area of overlap C ε r ε Expression for capacitance where common area of the plates (m 2 ) separation between the plates (m) ε r relative permittivity (no units) ε constant of proportionality ( ) (Fm -1 ) Worke example capacitor is mae from two parallel metal plates with a common area of 1m 2 an a separation of 1mm (ε Fm 1 ) (a) calculate the capacitance C ε, where 1m 2 an 1 3 m So, C F 1 3 Experiments show that: capacitance varies irectly with the area, C capacitance varies inversely with the separation C 1 Combining these two gives: C o change from a proportionality to equality we introuce a constant of proportionality, in this case ε We can now write the equation C ε We use the subscript when there is nothing between the plates (Strictly there shoul be a vacuum between the plates but the presence of air makes almost no ifference) he term ε, epsilon nought, is calle the permittivity of free space; its value is given by (ε Fm 1 (faras per metre)) he units for ε can be foun by rearranging the equation above an then putting in the known units: ε C units for ε F m Fm 1 m 2 (b) If the plates are now hel apart by a thin sheet of paper 1 mm thick, calculate the new capacitance (Relative permittivity for paper 35) C ε ε r, so C F (31 nf) 1-4 Remember: In the expression for capacitance,c ε ε r he area must be in m 2 (normally small) an separation must be in m Combining capacitors in parallel an series he following are not proofs but aim to help unerstaning he iagram below shows two capacitors an C 2 in parallel You can see that the area of C 2 is ae to that of So, you can see that the total capacitance C + C C 2 nother way to look at capacitors in parallel is to look at their charge In this case the total charge is foun by aing 1 an 2 For capacitors in parallel C + C 2 2

3 he iagram below shows two capacitors an C 2 in series he values of the capacitors may be ifferent but the charge on each is the same + + C You can see this by looking at the two inner plates, one from each capacitor, an remembering that they are insulate from the rest of the circuit he charge lost by one plate must equal that gaine by the other he total charge store here is Look at iagram below showing the equivalent capacitor with an increase 1 separarion an (because C ), the total capacitance C is reuce It is foun using the formula C C1 C2 + Worke example wo capacitors of 5µF an 2µF are connecte in parallel an a c supply of 5 volts applie to the combination Calculate: (i) the charge on each, (ii) the total charge store, (iii) the total capacitance of the combination, (iv) the charge store on the combination nswer: From the efinition for capacitance, C (i) µC µC (ii) total (25 + 1) µc 35µC 5µF (iii) C + C 2 so, C (5 + 2)µF 7µF (iv) gain use C with C 7µF ( 7 5)µC 35µC Note that the answers to ii) an iv) are the same 2µF 5 Remember: for capacitors in parallel, the potential ifference across each capacitor is the same ypical Exam uestion wo capacitors of 16µF an 48µF are on connecte in series an a c supply of 4 volts applie to the combination Calculate 16 µf 48 µf + + (i) the total capacitance (ii) the total charge store (iii) the charge on each capacitor 4 (iv) the potential ifference across each capacitor nswer: (i) Using, we have, C C C 12 µf C 1 C 2 (ii) Using C, we have store charge 12µ 4 48 µc (iii) he charge on each capacitor (in series) must be the same Hence µc an µc - C 2 Figure 6b 1 2 For capacitors in series 1 1 C1 C2 Energy store in a capacitor s well as storing charge, a capacitor must store energy You can see this is true because work has to be one to charge the capacitor an energy is release uring ischarge For a capacitor, the energy store is the area uner the graph of voltage against charge p f f Charge Here, this is a triangle, so w ½ f f If you use C an substitute for an then, two other expresions are foun: Energy store in a capacitor ½ ½ C 2 2 2C ypical Exam uestion 1µF capacitor, initially uncharge, is connecte to a 2 volt supply Calculate (i) the charge transferre from the supply to the capacitor (ii) the energy taken from the supply (iii) the energy store in the capacitor nswer: (i) From the efinition of capacitance, C So, 1µ 2 2 µc (ii) he supply provies 2µC at a steay p of 2 volts he energy taken from the supply, W, is given by W ½ So, W 2µ 2 4µJ (iii) ny of the three expressions can be use to fin the energy in the capacitor: suppose we use W ½ his gives W ½ 2µ 2 2µJ Note that because we are using W ½ the energy is in joules when the charge is in coulombs an the p in volts With capacitors, it is quite common to have microjoules When a charge coulombs moves through a potential ifference of volts, the work one is Only half of this store in the capacitor! What has happene to the other half? It is har to believe, but the answer is that the missing half is lost as heat in the connecting leas In charging the capacitor there is a current for a short time his current passes through the resistance of the leas an gives the joule heating effect (I 2 R) Note that the resistance of the leas is usually very small an consequently is ignore in most cases (iv) Rearrange C to give 48 µc So, 16 3 an µf (Note that ) 48 µc 48 µf 1 Remember: for capacitors in series, the charge on each capacitor must be the same an the applie potential is ivie 3

4 ime constant In the iagram the capacitor is first charge from the 6 supply an then ischarge through the 4kΩ resistor R he question is how will the charge leak away from the capacitor through the resistor R? Not only oes the charge fall exponentially but the current an p ecrease exponentially as shown in the two graphs below an they have the same time constant of 8 ms current / m 15 6 R 4KΩ I I e t C 2µF 1 he answer is isplaye in graph below It shows the charge remaining on the capacitor he initial charge is foun by using C: 2µ 6 12µC as shown on the graph 5 t I I 37 charge / µc 12 1 time/ms 1 e p / e t 4 t time / ms ll other values are foun from the equation e -t/ In this equation, e means exponential, an may be foun on a calculator Look for the button marke e x an check for yourself that e an e 1 37 (to 2sf) (see Factsheet 1 Exponentials an Logarithms) he equation an graph escribe an exponential ecay ll you nee to o is to let t in the equation You will then have: e -t/ e 1 37 he term is calle the time constant, it is the time taken for the charge store to fall to 37 or to 37% of the original charge he time constant for the circuit is: t t secons t s (8 ms) 2 t 37 fter the time constant, t,,, an I all fall to 37 of their original value Sometimes we may nee to know when these quantities fall to a half of their original value his time is obviously slightly less than an is approximately 7 secons (he exact value is foun by using the natural logarithm of 2, ln You will meet ln2 in calculations on half life in raioactivity) he time for, an I to fall to half their original values is given by: ½ ln2 ½ 69 approximately 1 time / s he graph above it shows the initial charge 12µC as calculate above he time constant is 8ms, so the charge remaining at that time shoul be 12µC µC 4

5 Exam Workshop In this question, take the permittivity of free space ε Fm -1 stuent has to esign a parallel plate capacitor of value 13 µf (a) Estimate the common area of this capacitor if the ielectric use is air of uniform thickness 1 mm [3]!! C 4 ε m 12 ε his caniate has quote the correct formula an substitute the given values but faile to write C in faras ( ) he high value prouce shoul have given a hint that something was wrong (b) In an improve esign, the stuent fills the space between the plates with an insulator which is only 1 mm thick with a relative permittivity ε R 15 Estimate the new common area [2]! ecf! ε C 2/2 1 εr C m εεr 15 Has correctly recognise an use relative permittivity For a given capacitor reucing will reuce in same ratio Note the caniate has been aware full creit for using the wrong answer from 1 a) (c) capacitor is conecte in parallel with a high resistance voltmeter in a circuit with a 1 volt supply an switch S S 1 c C 13µF Upon closing the switch, calculate; the p across the capacitor an the charge store in it [2] P across P across capacitor capacitor supply supply p p 1 1! C C µ coulombs micro-coulombs! 2/2 Has now correctly worke in micro-coulombs 2/3 () When the switch has been open for 4 secons, the voltmeter reas 6 volts Calculate: (i) the charge remaining in the capacitor [1] e, at t 4s, C, so 4 13µ 6 78µC! 1/1 (ii) the time constant for the capacitor voltmeter combination [4]! t t 78 t e 78 13e e " 13! 2/4 Caniate has calculate ln( 78 13) correctly, but then equate it to Common sense shoul have tol him/her that a negative answer was wrong S/he has also ignore the t Examiner's nswer (a) C ε C! ! ε m 2! (b) C! ε ε R m 2! (c) P 1! C µc () (i) as in answer (ii) e t/! 78 13e t/ ln(78/13) t/! t 4: 4 ln(78/13)! 78 s! ypical Exam uestion a) wo capacitors are available, one of 2µF an one of 5µF Each capacitor is given a charge of 3µC Calculate the potential ifference across each b) he 2µF capacitor is now in the circuit shown below 1 volts c 1 volts c 2µF C 2µF milliammeter Calculate: (i) the charge which flows through the milliammeter when the switch is move from the left to the right; (ii) the current in the milliammeter when the switch oscillates with a frequency of 2 Hz; (iii) the frequency at which the switch shoul vibrate in orer to prouce a current of 1m 5 a) By efinition, Charge store capacitance potential ifference or C! For 2µF, olt! for 5µF, olt! b) (i) From efinition, C, so charge store 1 2µ 2µ C or C! (ii) We (shoul) know that current is the charge circulating in 1 secon Since the switch oscillates 2 times in one secon, the charge circulating in this time is (2 1 6 ) 2 C! his is coulombs in 1 secon! coulombs in 1 secon 4 millicoulombs in 1 secon Hence, current 4 m! (iii) require current 1 m which is coulombs in 1 secon! Working in microcoulombs, this becomes (1 1 3 ) 1 6 microcoulombs in 1 secon, 1 4 µc in 1 secon Charge is still being elivere in pulses of 2 µc! so the number of pulses per secon or frequency will be 1 4 µ 5 Hz! 2 µ

6 uestions 1 In the circuit shown below, the capacitor is initially uncharge an then the switch s is close Supply 1 ypical Exam uestion he iagram below represents a thuner clou 5 km 5 km an 1km above the groun 5km 5km Clou I C 2 µf 1km R 5 MΩ (a) Fin the initial current I charging the capacitor (b) Fin the current charging the capacitor when the charge store on the capacitor is (i) 4 µc (ii) 19 µc (c) Fin the maximum charge store by the capacitor () Sketch a graph to show how the charge store on the capacitor varies with the time from when the switch is close (e) What is the graient at the origin on the graph you have rawn? 2 In the iagrams all the resistors are of equal value an all the capacitors are of equal value he circuit in Fig 1 has a time constant of What are the time constants for Fig 2 an Fig 3? Fig 1 Fig 2 Fig 3 3 he circuit below has a 12 volt supply an two capacitors an C 2 he switch S is connecte to terminal a S a b 12 (a) Calculate the charge store on 5µF C 2 1µF (b) he switch S is then connecte irectly to terminal b Calculate: (i) the capacitance of the capacitor combination (ii) the potential ifference across the capacitors an, (iii) the charge on (c) he circuit is now moifie by replacing C 2 with resistor R 16 MΩ he switch is again move irectly from a to b S a b Earth (a) Estimate the capacitance of the clou-earth system stating any assumptions you make (b) If the maximum potential ifference between the clou an earth is 1 G calculate the maximum charge store an the corresponing energy (ε Fm 1 ) nswer: (a) ssumptions are to treat the clou - earth system as a parallel plate capacitor with imensions given air as ielectric so that ε r is approximately 1 he capacitance is ε Α foun from C Points to watch are the units So area (5 1 3 ) (5 1 3 ) m 2 an 1 3 m (b) C F or, F to 2sf 1 3 In calculating charge, remember that 1G 1 9 In this case,using C, coulomb o calculate the energy, any of the three expressions may be use Using W ½ we have W ½ J Note that in questions of this type, we are only estimating our final value Hence it seems sensible to give answers to no more than 2 significant figures s a rule, be guie by the figures supplie In the above, ε is given to 3 sigfig but the rest only to 2 sig, fig So the answer must be to 2 sig fig You may woner what the ifference is between lengths of 5km, 5km, an 5km he answer is that the calculation will show no ifference but the information supplie is telling you the number of significant figures to which you shoul work nswers 1 (a) 2 µ (b) 16 µ (c) 1µ () see page 4 (e) graient initial current I 2 µ 2 all three have the same time constant 3 (a) 6 µc (b) (i) 6 µf (ii) 1 (iii) 5 µc -6 (c) e µC 12 R 16 MΩ 5µF Calculate the charge on 6 secons after closing S 6 cknowlegements: his was researche an written by Keith Cooper he Curriculum Press,Unit 35B, he Big Peg,12 yse Street, Birmingham, B18 6NF s may be copie free of charge by teaching staff or stuents, provie that their school is a registere subscriber No part of these Factsheets may be reprouce, store in a retrieval system, or transmitte, in any other form or by any other means, without the prior permission of the publisher ISSN