Chapter 2: Capacitor And Dielectrics

hapter 2: apacitor And Dielectrics In this chapter, we are going to discuss the different ways that a capacitor could be arranged in a circuit and how its capacitance could be increased.

Overview apacitor and Dielectrics apacitance, V Series ircuit Parallel-Plate apacitor 2 o A d Energy Stored in apacitor U 2 2 V Dielectric 2 V Uniform Electric Field 2 Parallel ircuit harging and Discharging

2. apacitance And apacitor In Series And Parallel Define and use capacitance, V Derive and determine the effective capacitance of capacitors in series and parallel. Derive and use energy stored in a capacitor. U 2 V 2 2 V 2 2 Learning Objectives

apacitor 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 thin insulator (called a dielectric). The symbol for a capacitor is: Uses of capacitor: photoflash unit, giant lasers, on-off switches, smoothen direct current (d.c.) voltages, computer keyboard and etc

apacitance The capacitance of a capacitor is defined as the ratio of the charge on either plate to the potential difference between them. Mathematically, V where : harge on one of the plate V: potential difference across two plates The ability of a capacitor to store charge is measured by its capacitance. It is a scalar quantity and the unit of capacitance is farad (F) OR coulombs per volt ( V ). apacitance is always a positive quantity.

apacitance of Air-filled Parallel Plate apacitor Parallel plate capacitor consists of a pair of parallel plates of area A separated by a small distance d. When the capacitor is charged, its plates have charges of equal magnitudes but opposite signs: + and then the potential difference V across the plates is produced. Since d << A so that the electric field strength E is uniform between the plates.

apacitance of Air-filled Parallel Plate apacitor The capacitance of a parallel-plate capacitor, is proportional to the area of its plates and inversely proportional to the plate separation. Mathematically, where ε A d 0 ε 0 : permittivity of free space A : area of the plate ( d : distance between the two plates Parallel-plate capacitor separated by a vacuum 0 8.850 2 2 N m 2 )

Example The plates of a parallel-plate capacitor are 8.0 mm apart and each has an area of 4.0 cm 2. The plates are in vacuum. If the potential difference across the plates is 2.0 kv, determine a. the capacitance of the capacitor. b. the amount of charge on each plate. c. the electric field strength was produced. (Given 0 = 8.85 x 0-2 2 N - m -2 )

Example Solution

apacitor in Series Equivalent capacitor 2 3 + - + - + - + - V V 2 V 3 V V

apacitor in Series When the circuit is completed, the electron from the battery ( ) flows to one plate of 3 and this plate become negatively charge. This negative charge induces a charge + on the other plate of 3. Because electrons on one plate of 3 are repelled to the plate of 2, this plate is now charged, which then induces a charge + on the other plate of 2. This in turn produces a charge on one plate of and a charge of + on the other plate of capacitor. Hence the charges on all the three capacitors are the same,. 2 3

apacitor in Series The potential difference across capacitor, 2 and 3 are: Total potential difference across capacitor, 2 and 3 are: ; V ; 2 2 2 2 V 3 3 3 3 V 3 2 V V V V 3 2 3 2 V

apacitor in Series If is the equivalent capacitor, and V 2 3 V, then Therefore the equivalent (effective) capacitance eq for n capacitors connected in series is given by... eq 2 3 n capacitors connected in series

apacitor in Parallel 2 + - Equivalent capacitor X 2 + 2-2 Y + - 3 + 3-3 V V

apacitor in Parallel When three capacitors are connected in parallel to a battery, the capacitors are all charged until the potential differences across the capacitors are the same. If not, the charge will flow from the capacitor of higher potential difference to the other capacitors until they all have the same potential difference, V. The potential difference across each capacitor is the same as the supply voltage V. Thus the total potential difference (V) on the equivalent capacitor is V V V2 V3

apacitor in Parallel The charge on each capacitor is V V 2 2V2 2V 3 3V 3 3V Since the total charge on the equivalent capacitor is given by 2 3 V 2V 3V V 2 3

apacitor in Parallel If is the equivalent capacitor, and V 2 3 V, then Therefore the equivalent (effective) capacitance eq for n capacitors connected in parallel is given by... eq 2 3 n capacitors connected in parallel

Example 2 In the circuit shown below, calculate the a) equivalent capacitance b) charge on each capacitor c) the potential difference across each capacitor

Example 2 Solution

Example 3 In the circuit shown in figure above, = 2.00 F, 2 = 4.00 F and 3 = 9.00 F. The applied potential difference between points a and b is V ab = 6.5 V. alculate a) the charge on each capacitor b) the potential difference across each capacitor c) the potential difference between points a and d

Example 3 Solution

Example 4 Determine the equivalent capacitance of the configuration shown in figure below. All the capacitors are identical and each has capacitance of F.

Example 4 Solution Label all the capacitors in the circuit. Series 2 4 5 7 3 6 X To calculate the effective capacitance, it is easier to solve it from the end of the circuit (left) to the terminal (right). apacitors, 2 and 3 are connected in series

Example 4 Solution Parallel x Y 4 5 7 6 X x F 3 2 3 apacitors x and 4 are connected in parallel

Example 4 Solution Series 5 Y X 4 y 6 7 Y.33F Z apacitors y, 5 and 6 are connected in series

Example 4 Solution Parallel z 7 z Y 5 6 eq Z 0.36F apacitors z and 7 are connected in parallel

Example 4 Solution eff eq Z 7 eq.36f

Application Keyboard (Extra) When the key is pressed, the space between plates decreases and capacitance increases because ε A d 0 d hange in capacitance is detected, thereby recognizing which key has been pressed

Energy Stored in a harged apacitor, U A charged capacitor stores electrical energy. The energy stored in a capacitor will be equal to the work done to charge it. A capacitor does not become charged instantly. It takes time. Initially, when the capacitor is uncharged, it requires no work to move the first bit of charge over. When some charge is on each plate, it requires work to add more charge of the same sign because of the electric repulsion.

Energy Stored in a harged apacitor, U When the switch is closed in, 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 dw Vd d and V

Energy Stored in a harged apacitor, U Thus the energy stored in a charged capacitor is d W d dw 2 0 2 0 0 2 2 2 W U 2 2 V 2 U V 2 U

Energy Stored in a harged apacitor, U Energy = Area Under The Graph Energy stored by a capacitor V Gradient = / apacitance Energy produced by a cell V U 2 V U V Energy lost = Energy produced by a cell Energy stored in capacitor

Example 5 A 2 µf capacitor is charged to 200V using a battery. alculate the a) charge delivered by the battery b) energy supplied by the battery. c) energy stored in the capacitor. d) energy dissipated as heat.

Example 5 Solution

2.2 harging And Discharging of apacitors Define and use time constant, τ = R Sketch and explain the characteristic of -t and I-t graph for charging and discharging of a capacitor. Use t 0e R for discharging ( ) 0 e R for charging t Learning Objectives

harging Process

harging Process Originally, both plates are neutral When switch S is closed, current I 0 immediately begins to flow through the circuit. Note: The capacitor charges instantaneously without a resistor. Electrons will flow out from the negative terminal of the battery and accumulate on the plate B of the capacitor.

harging Process 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 capacitor equals the voltage supplied by the battery, V 0. At this time, no further current flows (I = 0) through the resistor R and the charge on the capacitor thus increases gradually and reaches a maximum value 0.

The number of charges in the capacitor increases exponentially until fully charged where 0 is the maximum number of charges. 0 e t R As the number of charges increases, the voltage across the capacitor increases at the same rate. V V 0 t e R As the capacitor store more and more charges, the rate of charge flow (current) decreases and finally when the capacitor is full, there will be no more charges flow and the current in the circuit falls to zero. 0 0.63 0 0 V 0.63V 0 0 I I 0 V 0 τ R τ R time,t time,t I I 0 e t R and V R I 0 0.37 I 0 0 0 τ R time,t

Discharging Process

Discharging Process When switch S is closed, electrons from plate B begin to flow through the resistor R and neutralize positive charges at plate A. Note: The capacitor discharges instantaneously without a resistor. (short circuit) Initially, the potential difference (voltage) across the capacitor is maximum, V 0 and then a maximum current I 0 flows through the resistor R.

Discharging Process 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.

The number of charges in the capacitor decreases exponentially until fully discharged when the charges are reduced to zero. 0 As the number of charges decreases, the voltage across the capacitor decreases at the same rate. V 0 e 0 V e t R t R 0.37 0 0 V V 0 0.37V 0 τ R time,t As it discharges more charges, the rate of charge flow (current) increases until all the charges are used up by the resistor. I I e 0 t R and R 0 I 0 Note : For calculation of current in discharging process, ignore the negative sign in the formula. I 0 0.37I 0 I 0 0 τ R τ R time,t time,t

Summary harging & Discharging

Time onstant The quantity R that appears in the exponent for all equation is called time constant or relaxation time of the circuit or mathematically It is a scalar quantity. Its unit is second (s). R It is a measure of how quickly the capacitor charges or discharges.

Time onstant For charging process: The time constant is defined as the time required for the capacitor to reach (-e - ) = 0.63 or 63% of its maximum charge/voltage. OR The time constant is defined as the time required for the current to drop to /e = 0.37 or 37% of its initial value (I 0 ).

harging 0 0 e t R 0.63 0 0 e R R 0.63 0 0 τ R time,t I I I 0 e t R I 0 I 0 e R R 0.37I 0 0.37 I 0 0 τ R time,t

Time onstant For discharging process: The time constant is defined as the time required for the charge on the capacitor/voltage across it/current in the resistor decrease to /e = 0.37 or 37% of its initial value.

Discharging 0 e 0 t R 0.37 0 e 0 R R 0.37 0 0 τ R time,t I I I 0 e t R 0 τ R time,t I 0 e R R 0.37I 0 0.37I 0 I 0

Example 6 The figure shows a simple circuit of the photographic flash used in a camera. The capacitance of the capacitor is 40.0 µf, and the resistance of the resistor is 45.0 kω. 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.

Example 6 Solution

Example 7 In the R circuit shown in figure below, the battery has fully charged the capacitor. V 0 b a S R Then at t = 0 s the switch S is thrown from position a to b. The battery voltage is 20.0 V and the capacitance =.02 F. The current I is observed to decrease to 0.50 of its initial value in 40 s. Determine a) the value of R. b) the time constant, c) the value of on the capacitor at t = 0. d) the value of on the capacitor at t = 60 s

Example 7 Solution

2.3 apacitors With Dielectrics alculate the capacitance of air-filled parallel plate capacitor Define and use dielectric constant Describe the effect of dielectric on a parallel plate capacitor Use capacitance with dielectric Learning Objectives

Dielectric 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.

Dielectric 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

Dielectric onstant, κ (ε r ) When a dielectric (such as rubber, glass or waxed paper) is inserted between the plates of a capacitor, the capacitance increases. This capacitance increases by a factor or r which is called the dielectric constant (relative permittivity) of the material. Dielectric constant is defined as the ratio between the permittivity of dielectric material, ε to the permittivity of free space, ε 0. OR Dielectric constant is defined the ratio between the capacitance of given capacitor with space between plates filled with dielectric, to the capacitance of same capacitor with plates in a vacuum, 0.

Dielectric onstant, κ (ε r ) Mathematically, 0 A 0 d and εa d r OR r 0 0 where ε 0 : permittivity of free space ε : permittivity of dielectric material where 0 : capacitance of capacitor with vacuum between plates : capacitance of capacitor with dielectric between plates It is dimensionless constant (no unit).

Dielectric onstant, κ (ε r ) From the definition of the capacitance, 0 V 0 and is constant V r V V 0 where V 0 : potential difference across the capacitor in vacuum V : potential difference across the capacitor with dielectric

Dielectric onstant, κ (ε r ) From the relationship between E and V for uniform electric field, V Ed and V0 E0d r E E 0 where E 0 : electric field strength of the capacitor in vacuum E : electric field strength of the capacitor with dielectric SUMMARY r 0 0 V V 0 E E 0

Dielectric onstant, κ (ε r ) The dielectric constant depends on the insulating material used. Material Dielectric constant, r Dielectric Strength (0 6 V m ) Air.00059 3 Mylar 3.2 7 Paper 3.7 6 Silicone oil 2.5 5 Water 80 - Teflon 2. 60 The dielectric strength is defined as the electric field strength at which dielectric breakdown occurs and the material becomes a conductor.

Dielectric Effect On The Parallel Plate apacitors (without battery) d Initially the plates are separated by a vacuum and connected to a battery, giving the charge on the plates + and. The battery is now removed and the charge on the plates remains constant. The electric field between the plates is uniform and has a magnitude of E 0. Meanwhile the separation between plates is d. E 0

Dielectric Effect On The Parallel Plate apacitors (without battery) E When a dielectric is placed in the electric field between the plates, the molecules of the dielectric tend to become oriented with their positive ends pointing toward the negatively charged plate and their negative ends pointing toward the positively charged plated E 0

Dielectric Effect On The Parallel Plate apacitors (without battery) The result is a buildup of positive charge on one surface of the dielectric and of negative charge on the other. E The number of field lines within the dielectric is reduced thus the applied electric field E 0 is partially canceled.

Dielectric Effect On The Parallel Plate apacitors (without battery) Because the new electric field strength (E < E 0 ) is less then the potential difference, V across the plates is less as well. V Ed, when E, V Since V is smaller while remains the same, the capacitance,, when V, V is constant is increased by the dielectric.

Dielectric Effect On The Parallel Plate apacitors (with battery) If the battery is connected when dielectric is inserted, then the charge is not constant. Initially, when dielectric is inserted, capacitance is increased. Since the capacitor is still connected to the battery, when increase, increase., when, V is constant V will be increasing until V = V battery again.

apacitance With Dielectric r o o and o o d A Parallel-plate capacitor separated by a vacuum o A d o A d Parallel-plate capacitor separated by a dielectric material

Example 8 A vacuum parallel-plate capacitor has plates of area A = 50 cm 2 and separation d = 2 mm. The capacitor is charged to a potential difference V 0 = 2000 V. Then the battery is disconnected and a dielectric sheet of the same area A is placed between the plates as shown in Figure. d dielectric In the presence of the dielectric, the potential difference across the plates is reduced to 500 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 0 = 8.85 0 2 2 N m 2 )

Example 8 Solution

Summary apacitor and Dielectrics apacitance, V Series ircuit Parallel-Plate apacitor 2 o A d Energy Stored in apacitor U 2 2 V Dielectric 2 V Uniform Electric Field 2 Parallel ircuit harging and Discharging