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Higher Physics

Higher Physics. Unit 3.4 Capacitance. 3.4.1 Introduction Capacitors consist of two parallel conducting plates separated by air or an insulator and are used for storing charge and energy.

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Higher Physics

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  1. Higher Physics Unit 3.4 Capacitance

  2. 3.4.1 Introduction Capacitors consist of two parallel conducting plates separated by air or an insulator and are used for storing charge and energy. To store charge a capacitor is connected to a battery – the plate connected to the negative terminal of the battery receives electrons to become negatively charged while the plate connected to the positive terminal of the battery loses electrons to become positively charged. Increasing the charge on the plates is called charging and as the charge on the plates increases a potential difference is created across the gap. Charging stops when this potential difference equals the e.m.f. of the supply.

  3. Symbol: The charge stored by a capacitor refers to the charge stored on either plate. The amount of charge stored depends on the e.m.f. of the battery as well as the design of the capacitor. The ability of a device to store charge is called its capacitance. A capacitor is effectively a break in the circuit and charge cannot flow across it.

  4. B S A A 3.4.2 Capacitors in d.c. Circuits The circuit below is used for investigating the charge and discharge of a capacitor. When a capacitor is filled with electricity it is said to be charged. To charge the capacitor, move the switch to position B. To discharge the capacitor, move the switch to position A.

  5. I I Current is negative because electrons flow off capacitor in opposite direction 0 0 t t When C is fully charged, I = 0 Amps Imax charging = Imax discharging (A) Current/time graphs charging discharging As time increases, the current decreases. This means charge flows onto (off) the plates of the capacitor quickly at the start and very slowly later – charge is not flowing across the gap between the plates. The capacitor is fully charged when the p.d. across the plates is equal to the e.m.f. of the battery and current no longer flows in the circuit.

  6. Not in data booklet! e.m.f. (V) Resistance (Ω) Initial (max) current (A) The initial (maximum) current during charging or discharging can be calculated using; Increasing the resistance in the charge/discharge circuit causes lower charging/discharging currents so the capacitor takes longer to charge/discharge to a certain voltage.

  7. Small R Large R Varying R and/or C Change the value of the resistor: charging I Area under I/t graph = charge stored, areas are equal 0 t

  8. Initial I is the same Small C Large C Change the value of the capacitor: charging I 0 t

  9. Conclusions If the value of R is increased, 1 The initial charge/discharge current decreases. 2 The time taken to charge/discharge increases. 3 The area under the I/t graph is the same each time. If the value of C is increased, 1 Initial current during charging/discharging is the same. 2 The capacitor takes a longer time to charge to the same p.d.as it requires a greater charge to do so. 3 The area under the I/t graph will be greater as the charge stored will be greater.

  10. VR VC (B) Voltage/time graphs The circuit used is shown below. Switch at B for charging, A for discharging. E S B A

  11. VC VR charging V E The max p.d. across C is equal to the e.m.f. when VC = E, VR = 0 and I = 0 0 t

  12. E = VC + VR VR R I = Conclusions The sum of the p.d.’s across C and R is equal to E at all times; As VC ↑ VR↓ Not on data sheet! The charging current at a particular time can be found using; Warning! Do not use VC. No current flows through C

  13. VC VR discharging V E 0 t Since electrons now flow through R in opposite direction, V across R must have changed direction. E V

  14. VR (or VC) R I = Conclusions As the cell is no longer part of the circuit the capacitor now acts as the source of voltage, like a cell. VC = VR (In Size)

  15. charging discharging VR 0 t E Small R Small R Large R Large R E VR 0 t Varying the size of the resistor

  16. charging discharging VR VR E E Small C Small C Large C Large C 0 0 t t Varying the size of the capacitor N.B. • When a capacitor is fully charged, VC = E • I/t and RR/t graphs are identical. VR = IR i.e. VR ∞ I

  17. Example 1 The following information is marked on a capacitor: 47 F, 300 V. This means that the maximum voltage that should be applied across the capacitor is 300 V. Could this capacitor be safely connected to a 230 V a.c. supply? You must justify your answer. VP = √2 Vrms VP = √2 x 230 = 325V 325V > 300V So it is NOT safe!

  18. Example 2 The apparatus below is used to measure how long a ball takes to travel between two strips of metal foil on a track. The ball breaks each foil in turn.

  19. A computer monitors and displays the voltage across the capacitor. (a) What will be the voltage across the capacitor before foil 1 is broken? The circuit is complete so C will be fully charged VC = 9V (b) What happens in the circuit after foil 1 is broken? C is now cut off from power supply and starts to discharge through R

  20. 9 2 0 t Foil 1 broken Foil 2 broken (c) When foil 2 is broken, the voltage across C is 2 V. Draw a voltage/time graph for the computer display from A to B. VC

  21. 3.4.3 Charge, Voltage and Capacitance

  22. I = Constant! 0 t 3.4.3Charge, Voltage and Capacitance The relationship between charge, Q and current, I is given by Q = I t. This is only true if the current is constant. Current is not constant during the charging of a capacitor. I Solution: Use a variable resistor to keep I = constant. Initially set R on a high value and then gradually reduce to keep I = constant.

  23. S R C A V Charging a Capacitor with a Constant Current Circuit diagram Procedure Initially C is uncharged, R is on a large value. Close the switch and constantly reduce the value of R to keep charging current constant.

  24. Charge Voltage 0 Q V k = Results Conclusions Q ∞ V  Q = kV (k = constant) Constant k is written as C and defines the capacitance of the capacitor. Capacitance is the ratio of charge to potential difference.

  25. charge capacitance p.d. across C Q = VC Q V C = Unit: If charge is measured in coulombs and the p.d. in volts, the capacitance is measured in farads (F). 1 farad = 1 coulomb per volt 1 F = 1 C V–1 1 farad is a very large capacitance – practical capacitors have capacitances in the microfarad (1 F = 1  10–6 F), nanofarad (1 nF = 1  10–9 F) or picofarad (1 pF = 1  10–12 F).

  26. S R = 6.0 kΩ C = 10.0 µF 12.0 V d.c 12 6x103 E R = I = Example 1 In the circuit below the capacitor is initially uncharged. Switch S is now closed. (a) What is the initial current in the circuit? = 2mA

  27. (b) When there is a current of 050 mA in the circuit what is the p.d. across the capacitor? p.d. across R = VR = IR = 0.5x10-3 x 6x103 = 3V E = VC + VR 12 = VC + 3  VC = 9V (c) What is the charge on the capacitor when fully charged? Q = V C Q = 12 x 10µ = 120µC

  28. VR R Not VC!! Be Careful!!! During charging I = 12-8 6x103 I = Or 67mA I = 6.7x10-4 A (d) What is the current in the circuit when the voltage across the capacitor is 80 V? (e) The charged capacitor is removed from the circuit and a 10  resistor connected across its terminals. What happens to the energy stored in the capacitor? The stored energy will be changed to heat in the resistor.

  29. (f) If the charged capacitor had been connected to a 50  resistor, what effect would this have had on the amount of energy released? No Affect. Energy released would be the same but time taken to release it would be longer. Warning In some questions you are given values for I and t and asked to calculate Q. There is a tendency to want to use Q = I t, however Q = I t can only be used in situations where there is a constant current, I, for a time, t, or if the question asks for you to calculate the average current. During charging/discharging if the current changes, Q = I t cannot be used.

  30. Q V Q = C V  C = 5µC 1V  5 µF = Example 2 A capacitor has a value of 5 F. Explain in terms of charge what this means. To raise the p.d. across the capacitor by 1 volt, 5 µC of charge need to be added.

  31. R 6.0 V d.c C A V Constant current! Q = VC = It = 0.2x10-3 x 30 = 6.0x10-3C Q V 6x10-3 5 Q = VC  C = = Example 3 The resistance of R is carefully reduced to keep the current constant at 020 mA. (a) Calculate the charge on C after 30 seconds if V = 50 V and hence calculate a value for the capacitance of C. = 1.2x10-3 C = 1.2 mF

  32. E I R = 6 0.2x10-3 = (b) Calculate a suitable range for R. = 30 000Ω Range = 0 - 30kΩ

  33. Example 4 The circuit shown is used to investigate the charging and discharging of a capacitor. The capacitor is repeatedly charged and discharged by switching S between contacts 1 and 2.

  34. (a) For one setting of R1 and R2 the following trace is obtained on the oscilloscope. The time base of the oscilloscope is set at 5 ms per cm and the Y gain is set at 2 V per cm. Calculate the frequency of the vibrating switch. T = 4 x 5 ms = 20ms f = 1/T = 1/20x10-3 = 50Hz

  35. (b) With the settings of the oscilloscope unaltered, a new trace is produced when the resistance of one of the variable resistors is changed. State which one of the variable resistors was changed and whether its resistance was increased or decreased. Justify your answer. C is still charging to 8V but is now not discharging completely  R in discharge circuit has increased  R2 has increased

  36. (c) Calculate the charge lost by the capacitor each time the switch is in the discharge position. Initial charge = Q = VC = 8 x 2.2µ = 17.7µC Final charge = Q = VC = 2 x 2.2µ = 4.4µC Charge lost = 17.6µ - 4.4µ = 13.2µC

  37. Example 5 The circuit below is used to investigate the charging and discharging of a capacitor.

  38. The graph below shows how the power supply voltage varies with time after switch S is closed. The capacitor is initially uncharged. The capacitor charges fully in 03 s and discharges in 03 s. Sketch a graph of the reading on the voltmeter for the first 25 s after switch S is closed.

  39. Voltage/V 6V 0 0.5 0.8 1.0 1.3 1.5 1.8 2.0 2.3 2.5 Time/s Answer

  40. 3.4.4 Energy Stored in a Capacitor

  41. Electron current - + VS 3.4.4 Energy Stored in a Capacitor A charged capacitor can be used to light a bulb for a short time, therefore the capacitor must contain a store of energy. The charging of a parallel plate capacitor is considered below. When the circuit is completed there is an initial surge of electrons from the negative terminal of the cell onto one of the plates (and electrons out of the other plate towards the positive terminal of the cell).

  42. VC - - + + Electron current - + VS Once the capacitor starts to charge any further electrons which the –ve terminal of the cell wants to add to the –ve plate will experience a repulsive force from the electrons already on the –ve plate. Similarly, to make the +ve plate more positively charged the +ve terminal of the cell will have to exert an attractive force on the electrons remaining on the +ve plate. To move electrons against these attractive and repulsive forces the cell must use up some of its energy i.e. work is done in charging the capacitor. The energy is being stored in the electric field between the plates.

  43. VC = VS - - + - + + - + VS When the capacitor is fully charged (the p.d. across the plates of the capacitor is equal and opposite to the supply voltage) no more electrons are transferred so the battery is not adding energy to the capacitor. The quantity of charge held by a fully charged capacitor depends on the voltage across its plates when fully charged. Increasing the voltage will force the capacitor to store more charge and so more work must be done by the battery.

  44. charge Q 0 V voltage Energy Stored/Work Done W = Q V, if V = constant. For a capacitor V  constant during charging. The quantity of work done (or the energy stored) can be found from the area under the Q versus V graph. Area of triangle = ½  base  height = ½ VQ = ½ Q V

  45. W = Q x final V 2 W = Q x Initial V + final V 2 Ew = ½ Q V = area under Q/V graph Alternatively: energy needed to charge a capacitor = Q x V (V = average voltage) Bit initial V = 0 volts, W = ½ Q V EW = the energy stored on a capacitor, or the work done charging a capacitor. V = the final p.d. across the capacitor

  46. On data sheet Or substitute V = Q C Q C Ew = ½ Q x Q2 C Ew = energy = ½ Alternative formulae: Ew = ½ Q V Substitute Q = V C Ew = ½ VC x V Ew = energy = ½ C V2 Ew = ½ Q V The above are all equivalent expressions. The most appropriate one to use depends on the information given in any particular example.

  47. work done in moving charge Q between two points in an electric field at a fixed potential difference (see section 2.1.) e.g. electron gun of a TV, C.R.O. work done in adding charge Q to a capacitor (energy stored in a capacitor). V = potential difference across the capacitor which is not fixed. Warning The following two formulae are very similar and are often used incorrectly. 1. W = Q V 2. W = ½ Q V

  48. R Neon bulb 0.4 µF 100 V d.c. S Example 1 A neon lamp lights when the voltage across it rises to 80 V and goes out when the voltage falls to 60 V. In the following circuit, the 04 F capacitor is connected in series with a 500 k resistor to a 100 V d.c. supply. When switch S is closed the neon lamp, connected across the capacitor, flashes at a frequency of 8 Hz.

  49. Flash V 80 60 Discharge Charge t (a) Determine the energy dissipated in each flash of the lamp. E before = ½ Q V = ½ C V2 = ½ x 0.4µ x 802 = 1280µJ E after = ½ Q V = ½ C V2 = ½ x 0.4µ x 602 = 720µJ E dissipated = Before – After = 560µJ (b) Sketch the waveform obtained on a C.R.O screen connected across the neon lamp.

  50. (c) If the value of R is increased, the frequency of the flashing neon lamp changes. How would the value of C require to be changed to restore the frequency to its original value? If R increases, Charging I decreases, so time to charge C increases, therefore the value of C must be decreased. (d) The frequency could also have been restored to its original value by increasing the supply potential difference. Explain. If supply p.d. increases, Charging I increases, therefore C charges more quickly restoring the original frequency.

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