hapter 2: apacitors And Dielectrics 2.1 apacitance and capacitors in series and parallel L.O 2.1.1 Define capacitance and use capacitance apacitor is a device that is capable of storing electric charges or electric potential energy.it is consist of two conducting plates separated by a small air gap or a thininsulator (called a dielectric).the symbol for a capacitor is: The capacitance of a capacitor is defined as the ratio of the charge on either plate to the potential difference between them. V where : harge on one of the plate V: potential difference across two plates It is a scalar quantity The unit of capacitance is farad (F) OR coulombs per volt ( V 1 ) It is always a positive quantity L.O 2.3.1 alculate capacitance of air-filled parallel plate capacitor Parallel plate capacitor consists of a pair of parallel plates of area A separated by a small distance d. If a voltage is applied to a capacitor (connected to a battery), it quickly becomes charged.one plate carries a charge + and the other carries a charge then the potential difference between these two parallel plates is V. Since d<<a so that the electric field strength E is uniform between the plates.the capacitance of a parallel-plate capacitor, is proportional to the area of its plates and inversely proportional to the plate separation. A d where ε : permittivity of free space A : area of the plate d : distance between the two plates ( 8.851 12 2 N 1 m 2 ) 1
L.O 2.1.2 Derive and determine the effective capacitance of capacitors in series and parallel. apacitor in series: V V 1 2 3 1 V2 V3 Derivation: apacitor in parallel: V V 1 2 3 1 V2 V3 Derivation: 2
L.O 2.1.3 Derive and use energy stored in a capacitor When the switch is closed, charges begin accumulate on the plates.a small amount of work (dw) is done in bringing a small amount of charge (d) from the battery to the capacitor. dw Vd dw d and V The total work W required to increase the accumulated charge from zero to is given by: W dw d U 1 W 2 2 2 OR 1 U V 2 OR 1 U V 2 Example uestion The plates of a parallel-plate capacitor are 8. mm apart and each has an area of 4. cm 2. The plates are in vacuum. If the potential difference across the plates is 2. kv, determine a)the capacitance of the capacitor. b)the amount of charge on each plate. c) the electric field strength was produced. (Given = 8.85 x 1-12 2 N -1 m -2 ) Solution In the circuit shown below, calculate the a) equivalent capacitance b) charge on each capacitor c) the potential difference across each capacitor 3
uestion Solution In the circuit shown in figure above, 1 = 2. F, 2 = 4. F and 3 = 9. F. The applied potential difference between points a and b is V ab = 61.5 V. alculate a) the charge on each capacitor b) the potential difference across each capacitor c.) the potential difference between points a and d Determine the equivalent capacitance of the configuration shown in figure below. All the capacitors are identical and each has capacitance of 1 F. A 2 µf capacitor is charged to 2V using a battery. alculate the a) charge delivered by the battery b) energy supplied by the battery. c) energy stored in the capacitor. 4
Exercise uestion An electric field of 2.8 1 5 V m -1 is desired between two parallel plates each of area 21. cm 2 and separated by 25 cm of air. Find the charge on each plate. (Given = 8.85 x 1-12 2 N -1 m -2 ) Four capacitors are connected as shown in figure. alculate a) the equivalent capacitance between points a and b. b) the charge on each capacitor if V ab =15. V. Answer : 5.96 F, 89.5 on 2 F, 63.2 on 6 F, 26.3 on 15 F and on 3 F A 3. µf and a 4. µf capacitor are connected in series and this combination is connected in parallel with a 2. µf capacitor. a) What is the net capacitance? b) If 26. V is applied across the whole network, calculate the voltage across each capacitor. Answer: 3.71 µf, 26. V, 14.9 V, 11.1 V Two capacitors, 1 = 3. F and 2 = 6. F are connected in series and charged with a 4. V battery as shown in figure. alculate a) the total capacitance for the circuit above. b) the charge on each capacitor. c) the potential difference across each capacitor. d) the energy stored in each capacitor. e) the area of the each plate in capacitor 1 if the distance between two plates is.1 mm and the region between plates is vacuum. (Given permittivity of free space, = 8.85 x 1-12 F m -1 ) Answer: 2. µf, 8. µ, V 1 = 2.67 V, V 2 = 1.33 V, U 1 = 1.7 x 1-5 J, U 2 = 5.31 x 1-6 J, 3.39 m 2 5
2.2 harging and discharging of capacitors L.O 2.2.1 Define and use time constant L.O 2.2.2 Sketch and explain the characteristics of -t and I-t graph for charging and discharging of a capacitor L.O 2.2.3 Use for discharging and charging harging Discharging Originally, both plates are neutral When switch S is closed, current I immediately begins to flow through the circuit. Electrons will flow out from the negative terminal of the battery and accumulate on the plate B of the capacitor. Then electrons will flow into the positive terminal of the battery through the resistor R, leaving a positive charges on the plate A As charges accumulate on the capacitor, the potential difference across it increases and the current is reduced until eventually the maximum voltage across the capacitor equals the voltage supplied by the battery, V. At this time, no further current flows (I = ) through the resistor R and the charge on the capacitor thus increases gradually and reaches a maximum value. When switch S is closed, electrons from plate B begin to flow through the resistor R and neutralize positive charges at plate A. Initially, the potential difference (voltage) across the capacitor is maximum, V and then a maximum current I flows through the resistor R. When part of the positive charges on plate A is neutralized by the electrons, the voltage across the capacitor is reduced. The process continues until the current through the resistor is zero. At this moment, all the charges at plate A is fully neutralized and the voltage across the capacitor becomes zero. 6
harging Discharging 1 e t R e t R V V 1 e t R V V e t R Note: For calculation of current in discharging process, ignore the negative sign in the formula. The quantity R that appears in the exponent for all equation is called time constantor relaxation time of the circuit or mathematically It is a scalar quantity. Its unit is second (s). It is a measure of how quickly the capacitor charges or discharges. The time constant for a circuit used to charge a capacitor is defined as the time taken for the charging current to decrease to e 1 of its initial value. I I e t R t V R I Ie I R R I R The time constant for a circuit used to discharge a capacitor is defined as the time taken for thecharge (or potential difference across) the capacitor to decrease to e 1 of its initial value. 7
Example uestion The figure shows a simple circuit of the photographic flash used in a camera. The capacitance of the capacitor is 4. µf, and the resistance of the resistor is 45. kω. 45. kω Solution Flash bulb 4. µf a) Explain the function of the capacitor in the application above. b) alculate the time required to charge the capacitor to 65% so that a good flash can be obtained. c) Suggest a way to reduce the charging time of a capacitor. In the R circuit shown in figure below, the battery has fully charged the capacitor. a S R b V Then at t = s the switch S is thrown from position a tob. The battery voltage is 2. V and the capacitance = 1.2 F. The current Iis observed to decrease to.5 of its initial value in 4 s. Determine a)the value of R. b) the time constant, c)the value of on the capacitor at t =. d) the value of on the capacitor at t = 6 s 8
Exercise uestion A 2 µf capacitor and a 1. kω resistor are connected in series. A battery of emf 12 V is connected across the series combination. Determine a) the current at the instant when the battery is connected and current starts to flow in the circuit b) the current at time t = 1 ms c) the time taken by the current to i) decreases to 36.8% of the initial current ii) decrease to 13.5% of the initial current iii) become 2. ma Answer: 12 ma, 7.3 ma, 2 ms, 4 ms, 36 ms A 2 µf capacitor is charged using a battery of emf 1.5 V. The charged capacitor is then discharged through a 6 kω resistor. a) What is the time constant of the discharge circuit? 1 1 b) Find the time taken for the charge on the capacitor to decrease to (i) and (ii) of its e 1 initial value. Answer:.12 s,.12 s,.553 s A charged capacitor is connected in series to a switch and a resistor. A voltmeter of high resistance is connected across the capacitor. The switch is closed, and a set of values for the voltage V and the time t is obtained. The graph of lnv against t is plotted as follows. lnv 2 1 1 2 Determine the time constant of the circuit. Answer: 1 s t (s) 9
2.3 apacitors with dielectrics L.O 2.3.2 Define and use dielectric constant Dielectric is defined as a non-conducting (insulating) material placed between the plates of a capacitor.since dielectric is an insulating material, no free electrons are available in it. The advantages of inserting the dielectric between the plates of the capacitor are a) increase in capacitance b) increase in maximum operating voltage/ prevent capacitor breakdown c) possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing The dielectric strength is defined as the electric field strength at which dielectric breakdown occurs and the material becomes a conductor. Dielectric constant (also known as relative permittivity) is defined as the ratio between the permittivity of dielectric material to the permittivity of free space. r where ε : permittivity of free space ε : permittivity of dielectric material It is dimensionless constant (no unit). Other equations: 1
L.O 2.3.3 Describe the effect of dielectric on a parallel plate capacitor Induced electric field is in the opposite direction of the original electric field; hence reduce the net electric field strength between the plates: Since E E net E d ase 1: Battery disconnected and then dielectrics is inserted ase 2: Battery still connected and then dielectrics is inserted constant V not constant apacitance will increase. Since, V when increase, V decrease, remains unchanged because the circuit is not connected to the power supply (battery). apacitance will increase. The moment you insert the dielectric, increase, V decrease because. Since the circuit is still V connected to the power supply, will be increasing until V = V again. not constant V constant 11
L.O 2.3.4 Use capacitance with dielectric r o o and o o A d Parallel-plate capacitor separated by a vacuum o A d o A d Parallel-plate capacitor separated by a dielectric material Example uestion A vacuum parallel-plate capacitor has plates of area A = 15 cm 2 and separation d = 2 mm. The capacitor is charged to a potential difference V = 2 V. Then the battery is disconnected and a dielectric sheet of the same area A is placed between the plates as shown in Figure. Solution In the presence of the dielectric, the potential difference across the plates is reduced to 5 V. Determine a. the initial capacitance of the capacitor, b. the charge on each plate before the dielectric is inserted, c. the capacitance after the dielectric is in place, d. the relative permittivity, e. the permittivity of dielectric material, f. the initial electric field, g. the electric field after the dielectric is inserted. (Given = 8.85 1 12 2 N 1 m 2 ) 12