Lecture 20. March 22/24 th, Capacitance (Part I) Chapter , Pages


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1 Lecture 0 March /4 th, 005 Capacitance (Part I) Reading: Boylestad s Circuit Analysis, 3 rd Canadian Edition Chapter , Pages 894 Assignment: Assignment #10 Due: March 31 st, 005
2 Preamble: Capacitance (Part I) The following information has been provided to you in previous lectures in E.P W = Fd Newton meters (Joules) F E = Q Newton / Coulombs W V = Q Joules / Coulomb = Volts kq1q q1q F = = r 4πε 0r Newtons kq1 q1 E = = r 4πε 0r Newton / Coulombs Q I = t Coulombs / second = Amperes dq I ( t) = dt Amperes W P = t Joules / second = Watts Constants: 9 k = 8.99x10 Nm / C 1 Permittivity of free space: ε 0 = 8.854x10 C / Nm March nd, 005 Capacitance (Part I)
3 Note: and W = Fd = QV F E = Q W Fd EQd V = = = = Q Q Q V E = d Ed W VQ V P = = = VI = I R = t t R. March nd, 005 Capacitance (Part I) 3
4 Capacitance: In what follows a new circuit element for use in E.P is going to be introduced. It is called a capacitor. Before it is used, however, it will be developed from information that already is part of E.P The development that follows here is different than that given in the E.P textbook. The result is the same. Consider the following stationary charge scenarios (all performed in a vacuum): A positive point charge See the following diagram. Recall from Professor Salt s early lectures that a positive point charge, q, creates an electric field at a point p, a distance r from q, equal to q q E = k = r 4πε 0r The electric field varies as 1/r. March nd, 005 Capacitance (Part I) 4
5 A long thin wire conductor The length of the conductor is L with charge per unit length of λ. What is the electric field at point p, a distance r from the midpoint of the wire? See the following diagram. Divide wire into infinitesimal segments, dx. The charge dq of a segment dx is dq = λdx The distance s from the segment at position x to point p is 1 ( x r ) s = x + r = + The contribution de to the electric field at point p due to segment dx is 1 1 λdx de = dq = dq = 4πε 0s 4πε0( x + r ) 4πε0( x + r ) In determining the electric field at point p it is necessary to integrate the contributions for all of the segments in the line. While doing this, it would also be advantageous to determine the x and y components. March nd, 005 Capacitance (Part I) 5
6 r x cosθ = sinθ = x + r x + r λrdx de y = de cosθ = 3 4πε ( ) 0 x + r λxdx de x = de sinθ = 3 4πε ( ) 0 x + r Therefore L λrdx E y = 3 L4πε 0 ( x + r ) L λxdx Ex = 3 L4πε 0 ( x + r ) Integrating or using a table of integrals and then substituting the limits λl E y = 1 4πε ( ) 0r L + r E x = 0 Note that there is no resultant electric field in the x direction due to the symmetry of the charges on the wire. If L >> r then λ E y = πε r Newton / Coulombs 0 The electric field varies as 1/r (not 1/r as previous). March nd, 005 Capacitance (Part I) 6
7 An infinite plane sheet of charge Positive charge is distributed uniformly over the entire xyplane. Defining σ = charge per unit area and following a similar development as for the long thin wire problem previous and using those results (i.e., the segments under consideration in this problem would be long lines of charge) to help solve this problem. See the following diagram. The electric field is: σ E z = ε 0 Note that the field is uniform and normal to the plane and does not depend on the distance from the plane. March nd, 005 Capacitance (Part I) 7
8 An infinite plane charged conducting plate Note: A plate as opposed to a plane has some thickness. Positive charge is distributed uniformly over both sides of the plate (charge distributes itself onto the surface of the plate). Why?? See the following diagram. Defining σ = charge per unit area on each side of the plate: What is the electric field outside the plate at point a? Why? What is the electric field outside the plate at point c? Why? What is the electric field inside the plate at point b? Why? March nd, 005 Capacitance (Part I) 8
9 Two oppositely charged parallel plates Positive charge is distributed uniformly on one plate and negative charge is distributed uniformly on the other plate. Defining σ = charge per unit area on each plate. c plate b plate 1 a What is the electric field outside the plate at point a? Why? What is the electric field outside the plate at point c? Why? What is the electric field inside the plate at point b? Why? What was just developed is known as a parallel plate capacitor. The electrical symbol is shown below. C March nd, 005 Capacitance (Part I) 9
10 Now capacitance is defined as Q C = V Coulombs / Volt = Farad For the parallel plate capacitor σ Q E = = ε 0 ε 0 A Q is charge on each plate. Since Qd V = Ed = A ε 0 ε 0 Q A C = = V d (for parallel plates in a vacuum) Note that the units of ε 0 must be F/m. Capacitors separate charge. The amount of charge separated depends on the final voltage across the capacitance and the value of the capacitance. The value of capacitance is proportional to the area and inversely proportional to the spacing between the plates of the capacitor. The process of separating the charge is called charging the capacitor. Let s take a look at what we have. March nd, 005 Capacitance (Part I) 10
11 Example #1: a) What is the physical size of a parallel plate capacitor of value 1 Farad, if d=1mm? b) What is the physical size of a parallel plate capacitor of value 1 µf, if d=1µm? March nd, 005 Capacitance (Part I) 11
12 Parallel capacitors: Consider the following diagram. Since the voltage across both capacitors is the same the charge on the capacitors must be such that it creates the necessary voltage Q1 Q Qtotal E = = = C1 C C. total The total charge is the sum of the individual charges (why??) Q total = Q 1 + Q. Substituting C E + C E = C Thus C total = C 1 + C. E 1 total. Parallel capacitances are similar to series resistances. Note: Charge divider rule for parallel capacitors Cx Q x = Qtotal C total March nd, 005 Capacitance (Part I) 1
13 Series capacitors: Consider the following diagram. E The charge across both capacitors is the same (why??) Q total = Q1 = Q = Q and the voltage on the capacitors must be such that it creates the necessary charge E = VC + VC = V 1 total. Substituting Q total Q Q 1 1 E = = + = Q + C C1 C C1 C. total Thus = + C total C C or C C C = total C + C. 1 Series capacitances are similar to parallel resistances. Note: Voltage divider rule for series capacitors Ctotal V x = Vtotal C C 1 C x 1 March nd, 005 Capacitance (Part I) 13
14 Example #: What is the capacitance of the following circuit? 10 F C 0 F C3 C1 0 F 50 F C7 C4 5 F C5 C6 5 F 50 F March nd, 005 Capacitance (Part I) 14
15 Example #3: Consider the circuit shown below. The 500nF capacitor C1 initially has 3x108 C of charge on it. The µf capacitor C is initially uncharged and the switch is open. What happens after the switch is closed? March nd, 005 Capacitance (Part I) 15
16 Dielectrics: Up to now it s looking good, but there is a slight problem. How are the parallel plates kept apart (remember they are conducting, so they can t touch or they will cause a short circuit)? The solution is to keep the plates apart by putting something between them that doesn t conduct (i.e., an insulator). When a solid nonconducting insulator is placed between the parallel plates of a capacitor, it is called a dielectric (di for opposing and electric for electric field). The reason for the name is as follows. An ideal dielectric contains no free charges. The individual molecules of the dielectric are either nonpolar or polar. The diagram below shows the difference. NonPolar Polar When either type of molecule is placed in an electric field (i.e., like in a parallel plate capacitor) they line up in the direction of the field (even though they may have originally been randomly oriented). The positive and negative charges within the dielectric cancel each other out, however they do not cancel each other out at the surfaces of the dielectric. One surface becomes positively March nd, 005 Capacitance (Part I) 16
17 charged and the other (i.e., opposite) surface becomes negatively charged. See the diagram below. E d E c This results in a reduced electric field within the capacitor due to the electric field setup by the opposing charge on the dielectric surfaces. The positively charged surface of the dielectric would be next to the negatively charged plate and the negatively charged surface of the dielectric would be next to the positively charged plate, therefore the dielectric reduces the resultant electric field within the capacitor. Recall that V Q E = d and C = V. Since the capacitor s electric field has been reduced, while d has stayed the same, the voltage across the capacitor must have decreased. The charge on the capacitor stays the same, therefore the capacitance must increase over the capacitance without a dielectric between the plates of the capacitor. March nd, 005 Capacitance (Part I) 17
18 The ratio of the value of the capacitance with a dielectric and the value of the capacitance with a vacuum is known as the dielectric constant (or relative permittivity) C K =. C 0 Some relative permittivities are shown in the following table (See Table 10., Boylestad s Circuit Analysis, 3 rd Canadian Edition, page 86 for more). Dielectric Vacuum 1.0 Air Paper (wax).5 Mica 5.0 Glass 7.5 Aluminium Oxide 8.7 Tantalum Oxide 5.0 Ceramic As well the permittivity of the dielectric is defined as ε = K ε = ε 0 rε 0 Thus our equation for a parallel plate capacitance now becomes C = εa d where ε = ε rε 0. Note that the use of a dielectric increases the value of the capacitance (i.e., all ε > 1). r ε r March nd, 005 Capacitance (Part I) 18
19 Example #4: A 5µF capacitor is charged to 5V. It has a dielectric with relative permittivity ε r =45 between its parallel plates. What happens if the dielectric is removed? March nd, 005 Capacitance (Part I) 19
20 Some interesting items: Energy stored in a capacitor The final conditions on a capacitor after it has fully charged is Q = CV where Q is the final charge and V is the final voltage. At a stage of the charging process if the charge on either plate is q the voltage across the plates is q v = C The amount of work required to transfer the next charge dq is qdq dw = vdq = C. The total work is the integral from 0 to Q of dw 1 Q W = dw = qdq C = C 0 Substituting Q 1 1 W = = CV = QV C Q Joules March nd, 005 Capacitance (Part I) 0
21 Example #6: Consider the circuit shown below (from Example #3). The 500nF capacitor C1 initially has 3x108 C of charge on it. The µf capacitor C is initially uncharged and the switch is open. What is the energy stored on the capacitors before and after the switch is closed? March nd, 005 Capacitance (Part I) 1
22 Current through a capacitor: Recall that Q = CV and dq dv i ( t) = = C dt dt. If the voltage across the capacitor is constant dv = 0 dt. There is no current through the capacitor. The significance of this is that for a circuit in which the voltage sources are not changing (i.e., a dc circuit), no current flows through a capacitor under steady state. As a result of this, under steady state conditions after the capacitor has charged up to its final voltage, a capacitor is equivalent to an open circuit. This will become important in the lecture material to follow in Capacitance (Part II) March nd, 005 Capacitance (Part I)
23 Electric Flux: Recall that a positive point charge, q, creates an electric field at a point p, a distance r from q, equal to kq q E = = r 4πε0r The electric field lines radiate out in all directions. An example for a positive point charge is shown below. + Note that while the electric field exists at all points in space, we normally draw a small number of lines to indicate the direction of the electric field. By suitable selecting the number of lines we draw we can represent the direction of the electric field as well as the magnitude of the electric field. This is accomplished by spacing the lines in such a way that the number of electric field lines per unit area crossing a surface at right angles to the direction of the electric field is at every point proportional to the electric field. These lines are called flux lines and are represented by the symbol Ψ. Note that while the total number of flux lines is arbitrary, we know that it is proportional to the charge. Therefore it is possible to designate Ψ = Q. We can also define a flux density, D, to be March nd, 005 Capacitance (Part I) 3
24 Q D = A. How are E and D related? Consider again a positive point charge again. If a sphere were placed centered on the positive point charge at a radius r from q, the surface area of the sphere would be A = 4πr The electric field at the surface of the sphere would be q E = 4πε 0r and the flux density would be q D = 4πr. Therefore D = ε E 0 or D ε 0 = E. March nd, 005 Capacitance (Part I) 4
25 Real World Effects: Note: While we discuss the following effects, in E.P we always assume (unless told otherwise) that all capacitors are ideal parallel plate capacitors. Fringing: In reality the infinite plates of the parallel plate capacitor are not infinite. As a result the electric fields at the edges of the capacitor are not uniform and the resultant electric field is smaller than anticipated. They look like this. Since most capacitors have a large area relative to the distance between the plates, this effect is usually ignored. March nd, 005 Capacitance (Part I) 5
26 Leakage current: Consider the following circuit. When the switch is closed the capacitor will charge up to the voltage across the battery s terminals. If the switch is then opened the voltage across an ideal capacitor will stay at V=E. What actually happens is that the voltage across the capacitor slowly decays due to an internal leakage current due to free electrons in the dielectric. A model for this effect is shown below. C R leakage The value of the resistance for good capacitances is on the order of 100 s of Mega ohms (MΩ). March nd, 005 Capacitance (Part I) 6
27 Breakdown voltage (Dielectric strength) Note that the voltage across the capacitor must not be as large as to cause breakdown in the capacitor (as discussed in Prof. Salt s portion of the lectures). This would result in potential catastrophic failure in the dielectric. Some dielectric strengths are shown in the following table (See Table 10. in the E.P textbook: Boylestad s Circuit Analysis, 3 rd Canadian Edition, page 89 for more). Dielectric Breakdown Voltage (V/µm) Air 3 Paper (wax) 50 Mica 00 Glass 10 Aluminium Oxide 1 Tantalum Oxide 400 Ceramic 3 What is needed is a dielectric with a large breakdown voltage and a large relative permittivity since this will give us a small in physical size capacitor with a large capacitance and a large breakdown voltage. What would be a good candidate? Use this table with the previous one. March nd, 005 Capacitance (Part I) 7
28 Example #6: Calculate the physical sizes and breakdown voltages (i.e., maximum voltages) for each of the following capacitors? a) Paper (wax): 1 Farad, if d=100µm? b) Air: 1 µf, if d=1µm? c) Tantalum: 100 µf, if d=1µm? d) Mica 1 µf, if d=1µm? March nd, 005 Capacitance (Part I) 8
29 Electrolytic capacitors A typical electrolytic capacitor is made with a thin layer of extremely thin aluminiumoxide deposited on a layer of aluminium. An electrolyte gel (conducting) contacts the other side of the aluminium oxide. The aluminium is then rolled to form a cylindrical capacitor of small size and relatively high working voltage. The potential problem with this is that the capacitor is polarized. This means that is not bilateral and must be inserted in the circuit such that the polarities of the voltage across it must always correspond to the marked polarities on the capacitor s case. Failure to do so will result in capacitors that may explode due to heating effects caused by high current flow through the capacitor. The symbol for a polarized capacitor is shown below. March nd, 005 Capacitance (Part I) 9
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