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Topic Outline Definition of Capacitance Calculating Capacitance Combinations of Capacitors Energy Stored in a Charged Capacitor Capacitors with Dielectrics An Atomic Description of a Dielectric Capacitance and Dielectric Capacitors: Device that store electric charge A capacitor consists of two conductors separated by an insulator. Capacitance: Depends on its geometry and on the material, called a dielectric, that separates the conductors. Definition of Capacitance A capacitor consists of two conductors (known as plates) carrying charges of equal magnitude but opposite sign. A potential difference DV exists between the conductors due to the presence of the charges. What is the capacity of the device for storing charge at particular value of DV? Pictures from Serway & Beichner Definition of Capacitance Experiments show the quantity of electric charge Q on a capacitor is linearly proportional to the potential difference between the conductors, that is Q ~ DV. Or we write Q = C DV The capacitance C of a capacitor is the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between them: Q C= DV SI Unit: farad (F), 1F = 1 C/V Typical device have capacitances ranging from microfarad to picofarad. Pictures from Serway & Beichner Parallel - Plate Capacitors A parallel-plate capacitor consists of two parallel conducting plates, each of area A, separated by a distance d. When the capacitor is charged, the plates carry equal amounts of charge. One plate carries positive charge, and the other carries negative charge. The plates are charged by connection to a battery. Describe the process by which the plates get charged up. Pictures from Serway & Beichner Parallel-Plate Capacitors d Two parallel metallic plates of equal area A separated by a distance d as shown. One plate carries a charge Q and the other carries a charge –Q. And A surface charge density of each plate is s = Q/A. Variation with A If plates are large, then charges can distribute themselves over a substantial area, and the amount of charge that can be stored on a plate for a given potential diff increases as A is increased. Thus we expect C to be proportional to A C~A Pictures from Serway & Beichner Parallel-Plate Capacitors d Variation with d Potential difference DV constant across, E field increases as d decreases. A Imagine d decreases and consider situation before any charges have had a chance to move in response to this change. Because no charge move E the same but over a shorter distance. DV = Ed means that DV decreases. Pictures from Serway & Beichner Parallel-Plate Capacitors d Variation with d (cont’d) The difference between this new capacitor voltage and the terminal voltage of the battery now exists as a potential difference across the wires connecting the battery to the capacitor. A A E field result in the wires that drives more charge onto the plates, increasing the potential diff. DV until it matches that of the battery. potential diff. Across wire = 0 flow of charges stop. More charges has accumulated at the capacitor as a result. We have d decrease, Q increases. Similarly d increases Q decreases. Capacitance inversely proportional to d. C ~ 1/d Pictures from Serway & Beichner Parallel-Plate Capacitors (a) The electric field between the plates of a parallel-plate capacitor is uniform near the center but nonuniform near the edges. (b) Electric field pattern of two oppositely charged conducting parallel plates. Pictures from Serway & Beichner Parallel-Plate Capacitors Assume electric field uniform between the plates, we have (see lecture on Gauss’s Law) s Q E= = eo eoA Qd DV = E d = eoA Q Q C= = DV Qd /eoA eoA C= d (As we have argued before) Pictures from Serway & Beichner Example SB 26 Q8 A 1-megabit computer memory chip contains many 60.0-fF capacitors. Each capacitor has a plate area of 21.0 x 10-12 m2. Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The characteristic atomic diameter is 10-10 m =0.100nm. Express the plate separation in nanometers. e 0 A C 60.0 1015 F d e 0 A 18.85 1012 21.0 1012 d C 60.0 1015 9 d 3.10 10 m 3.10 nm Pictures from Serway & Beichner What is a lightning discharge ? Friction forces between the air molecules within a cloud result in positively charged molecules moving to the lower surface, and the negative charges moving to the upper surface . The lower surface induces a high concentration of negative charges in the earth beneath. The resultant electric field is very strong, containing large amounts of charge and energy. If the electric field is greater than than breakdown strength of air, a lightning discharge occurs, in which air molecules are ripped apart, forming a conducting path between the cloud and the earth. Example: Lightning SB 26 Q6 Regarding the Earth and a cloud layer 800 m above the Earth as the “plates” of a capacitor, calculate the capacitance if the cloud layer has an area of 1.00 x 1.00 km2. Assume that the air between the cloud and the ground is pure and dry. Assume that charge builds up on the cloud and on the ground until a uniform electric field of 3.00 x 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold? C= eoA/d = 8.85x10-12 x (1000)2/800 = 11.1 nF Potential between ground and cloud is DV = Ed = 3.0 x106 x 800 = 2.4 x 109 V Q = C(DV) = 26.6 C Pictures from Serway & Beichner Cylindrical Capacitors L A solid cylindrical conductor of radius a and charge Q is coaxial with a cylindrical shell of negligible thickness, radius b > a, and charge –Q. Find the capacitance of this cylindrical capacitor if its length is L. Pictures from Serway & Beichner Cylindrical Capacitors Assume that L is >> a and b, neglect the end effects. E is perpendicular to the long axis L of the cylinders and is confined to the region between them. Potential difference between the two cylinders is given by b Vb-Va = - E . ds Where E is the E field in the region a < r < b. a Our discussion on Gauss’s Law Er = 2kl/r where l is the linear charge density of the cylinder. Note that the charge on outer cylinders does not contribute to E field inside it. Pictures from Serway & Beichner Cylindrical Capacitors b b dr b Vb-Va = - Er dr = - 2kl r = - 2kl ln( a ) a a Using l = Q/L , we have Q Q L C= = DV 2kQ b ln( ) L a L C= b 2k ln( ) a Pictures from Serway & Beichner Cylindrical Capacitors L C= L b 2k ln( ) a What is the capacitance per unit length ? Example Co-axial Cable. Read the cable, typically 50 pF/m. Is this sensible ? Typically a 0.5 mm, b 1.5 mm 1 C/L 50pF/m 2 8.99 10 ln(3) 9 Pictures from Serway & Beichner Cylindrical Capacitors vs Parallel Plate Capacitors L What are the advantages of cylindrical capacitor over that of a parallel plate capacitor? Pictures from Serway & Beichner Capacitance of an isolated sphere Calculate the capacitance of an isolated spherical conductor of radius R and charge Q by assuming that the second conductor making up the capacitor is a concentric hollow sphere of infinite radius. Electric potential of the sphere of radius R is Q kQ/R and V= 0 at infinity, we have Q Q R C= = = = 4peo R DV kQ/R k C is proportional to its radius and independent of both the charge on the sphere and the potential difference. Pictures from Serway & Beichner Example SB 26 Q4 (a) If a drop of liquid has capacitance 1.00 pF, what is its radius ? (b) If another drop has radius 2.00 mm, what is its capacitance ? (c) What is the charge on the smaller drop if its potential is 100V ? C = 4peo R R = (8.99 x 109 N · m2/C2)(1.00 x 10–12 F) = 8.99 mm C = 4p (8.85 x 10-12) x 2.0x10-3 = 0.222 pF Q =CV = 0.222 pF x 100 V = 2.22 x 10-11 C Pictures from Serway & Beichner The Spherical Capacitors A spherical capacitor consists of a spherical conducting shell of radius b and charge –Q concentric with a smaller conducting sphere of radius a and charge Q. Find the capacitance of this device. Pictures from Serway & Beichner The Spherical Capacitors Gauss’s law E field outside a spherically symmetric charge distribution is radial and given by Er = kQ/r2. The potential difference between the spheres is b b dr 1 1 Vb-Va = - Er dr = - kQ r2 = kQ b a a a Q ab C= = Vb-Va k (b-a) What happens to the capacitance of this system when the radius of the outer sphere approaches infinity? Pictures from Serway & Beichner Example SB 26 Q12 A 20.0 mF spherical capacitor is composed of two metallic spheres, one having a radius twice as large as the other. If the region between the spheres is a vacuum, determine the volume of this region. Let the radii be b and a with b = 2a. Put charge Q on the inner conductor and –Q on the outer. Electric field exists only in the volume between them. The potential of the inner sphere is Va=kQ/a; that of the outer is Vb=kQ/b. Then kQ kQ Q ba Q 4pe 0 ab Va Vb = = C= = a b 4p e 0 ab Va Vb ba 4pe 0 2a 2 C C= =8p e 0 a a= a 8pe 0 C3 4 p b 3 4 p a 3 74 pa 3 74 p 3 3 3 The intervening volume is 3 8 p e0 3 3 3 7C 3 V = 2.13 x 1016 m3 384p 2e 0 3 Pictures from Serway & Beichner Example SB 26 Q16 What is the capacitance of the Earth ? Think of Earth spherical conductor and the outer conductor of the “spherical capacitor” may be considered as a conducting sphere at infinity where V approaches zero. C 4p e 0 R 4p 8.85 10 12 C N m 2 6.37 106 m C= 7.08 x 10-4 F A large capacitor ! Pictures from Serway & Beichner Combinations of Capacitors Parallel Combination The individual potential differences across capacitors connected in parallel are all the same and are equal to the potential difference applied across the combination. Pictures from Serway & Beichner Combinations of Capacitors Parallel Combination When the capacitors are first connected, electrons transfer between wires and plates. Leave left plates positively charged and right plates negatively charged. Energy source for this charge transfer is internal chemical energy stored in the battery. Flow of charges ceases when the voltage across the capacitors is equal to that across the battery terminals. Capacitors reach their maximum charge when the flow of charges ceases. Pictures from Serway & Beichner Combinations of Capacitors Parallel Combination Let the maximum charges on the two capacitors Q1 and Q2. Total charge Q stored by two capacitors is Q = Q1+Q2. Voltage across are the same Q1=C1DV, Q2=C2DV Define an equivalent capacitor having Ceq s.t. Q = CeqDV We have CeqDV = C1DV + C2DV And hence Ceq = C1 + C2 (for parallel combination) In general Ceq = C1 + C2+ C3+ ………….. (for parallel combination) Pictures from Serway & Beichner Combinations of Capacitors Series Combination Start with uncharged situation and follow what happen just after a battery is connected to the circuit. When a battery is connected, electrons transferred out of the left plate of C1 and into the right plate of C2. As this charge accumulates on the right plate of C2, an equivalent amount of negative charge is forced off the left plate of C2 and this left plate therefore has an excess positive charge. (cont’d) Pictures from Serway & Beichner Combinations of Capacitors Series Combination The negative charge leaving the left plate of C2 travels through the connecting wire and accumulates on the right plate of C1. As a result, all right plates end up with a charge –Q and all the left plates end up with a charge +Q. Thus the charges on capacitors connected in series are the same. Pictures from Serway & Beichner Combinations of Capacitors Series Combination Voltage DV across battery terminals is split between two capacitors. DV = DV1 + DV2 Where DV1 and DV2 are potential diff across capacitors C1 and C2. Suppose we have equivalent capacitor Ceq = Q/DV For each capacitor, we have DV1 = Q/C1 and DV2=Q/C2 Q/Ceq = Q/C1 + Q/C2 1/Ceq = 1/C1 + 1/C2 (series combination) In general 1/Ceq = 1/C1 + 1/C2 + 1/C3 + ….. (series combination) Pictures from Serway & Beichner Example: Equivalent Capacitance In series use 1/C=1/C1+1/C2 2.50 mF 20.00 mF 6.00 mF In series use 1/C=1/C1+1/C2 8.50 mF In parallel use C=C1+C2 20.00 mF 5.965 mF Pictures from Serway & Beichner Example: Equivalent Capacitance In parallel use C=C1+C2 In parallel use C=C1+C2 In series use 1/C=1/C1+1/C2 Pictures from Serway & Beichner Example: Equivalent Capacitance 26.22 In parallel use Ceq=C+C/2+C/3 In series use 1/CA=1/C+1/C C C/2 C/3 In series use 1/CB=1/C+1/C+1/C Pictures from Serway & Beichner Example 26.10 A variable air capacitor used in tuning circuits is made of N semicircular plates each of radius R and positioned a distance d from each other. As shown, a second identical set of plates is enmeshed with its plates halfway between those of the first set. The second set can rotate as a unit. Determine the capacitance as a function of the angle of rotation q, where q=0 corresponds to the maximum capacitance. Pictures from Serway & Beichner Example S&B 26.10 With qp , the plates are out of mesh and the overlap area is zero. With q0, the overlap area is that of a semi-circle, pR2/2. By proportion, the effective area of a single sheet of charge is (pqR2/2. When there are two plates in each comb, the number of adjoining sheets of positive and negative charge is 3, as shown in the sketch. When there are N plates on each comb, the number of parallel capacitors is 2N-1 and the total capacitance is e 0 Aeffective 2 N 1e 0 p q R 2 2 C 2 N 1 distance d 2 Pictures from Serway & Beichner Energy Stored in a Charged Capacitor Suppose q is the charge on the capacitor at some instant during the charging process. At the same instance, the potential difference across the capacitor is DV=q/C. The work necessary to transfer an increment of charge dq from the plate carrying charge –q to the plate carrying charge q (which is at the higher electric potential) is q dW = DV dq = dq C The total work required to charge the capacitor from q = 0 to some final charge q = Q is Q Q q 1 Q2 W= dq = q dq = C C 2C 0 0 Energy Stored in a Charged Capacitor Work done in charging the capacitor = electric potential energy U stored in the capacitor. Q2 1 1 U= = QDV = C (DV)2 2C 2 2 This result applies to any capacitors, regardless of its geometry. Is there a limit to the maximum energy (or charge) that can be stored in a capacitor? If so, what is the limiting factor? Energy Stored in a Charged Capacitor A plot of potential difference versus charge for a capacitor is a straight line having a slope 1/C. The work required to move charge dq through the potential difference DV across the capacitor plates is given by the area of the shaded rectangle. The total work required to charge the capacitor to a final charge Q is the triangular area under the straight line, W = QDV/2. 1V = 1 J/C hence the unit for the area is joule J. Pictures from Serway & Beichner Energy Density Stored in a Charged Capacitor For parallel plate capacitor, DV = E d, C = eoA/d we have 1 eoA 1 U= (E d)2 = (eoA d) E 2 2 d 2 Ad = Volume occupied by the E field. This lead to a new quantity known as Energy Density u = U/Volume = U/Ad 1 u= eo E 2 2 Although above equation was derived for parallel-plate capacitor, the expression is generally valid. Energy Density in any electric field is proportional to the square of the magnitude of the electric field at a given point. Example 26.33: A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled ? U = Q2/2C and C = eoA/d and d2 = 2 d1 then C2= C1/2 and the energy stored doubles. Pictures from Serway & Beichner Example 26.35: A parallel-plate capacitor has a charge Q and plates of area A. Show that the force exerted on each plate by the other is F = Q2/2eoA. Is this force attractive or repulsive. Does this equation follow from common sense? W U F dx dU d Q d Q x 2 2 F= 2e A dx dx 2C dx 0 F = Q2/2eoA Pictures from Serway & Beichner Rewiring Two Charged Capacitors Two capacitors C1 and C2 (where C1 > C2) are charged to the same initial potential difference DVi, but with opposite polarity. The charged capacitors are removed from the battery, and their plates are connected as shown. (a) Find the final potential difference DVf between a and b after the switches are closed. (b) Find the total energy stored in the capacitors before and after the switches are closed and the ratio of the final energy to the initial energy. Pictures from Serway & Beichner Rewiring Two Charged Capacitors Before switches are closed: Q1i = C1 DVi and Q2i = - C2 DVi (negative sign for plate 2) Total Q = Q1i + Q2i = (C1-C2) DVi After switches are closed: Total charge in system remain the same Total Q = Q1f + Q2f Charges redistribute until the entire system is at the same potential DVf. And this potential is the same across both the capacitors. Q1f = C1 DVf and Q2f = C2 DVf Pictures from Serway & Beichner Rewiring Two Charged Capacitors After switches are closed (cont’d): Q = Q1f + Q2f Q1f / Q2f = C1/C2 Q1f = [ C1/C2 ] Q2f Hence Q = Q1f + Q2f =[ 1+ C1/C2 ] Q2f C2 C1 We have Q2f = Q and Q1f = Q C2 + C1 C2 + C1 C1 Q Q1f C2 + C1 Q DV1f = = = = DV2f = DVf C1 C1 C2 + C1 And Q (C1-C2) DVi DVf = = C2 + C1 C2 + C1 Pictures from Serway & Beichner Rewiring Two Charged Capacitors Energy Before switches are closed: Ui = C1 (DVi)2/2 + C2 (DVi)2/2 = ( C1 + C2 ) (DVi)2/2 After switches are closed: Uf = C1 (DVf)2/2 + C2 (DVf)2/2 = ( C1 + C2 ) (DVf)2/2 2 Q2 1 Q 1 Uf = ( C1 + C2 ) = 2 C2 + C1 2 C2 + C1 1 (C1-C2)2 (DVi )2 1 Uf = Ui = ( C1 + C2 ) (DVi)2 2 C2 + C1 2 2 We have Uf C1 - C2 = Ui C1+ C2 Example S&B Chapter 26 Q 23 Consider the circuit as shown, where C1 = 6.00mF and C2= 3.00 mF and DV =20.0V. Capacitor C1 is first charged by closing of switch S1. Switch S1 is then opened and the charged capacitor is connected to the uncharged capacitor by the closing of S2. Calculate the initial charge acquired by C1 and the final charge on each. Pictures from Serway & Beichner Example S&B Chapter 26 Q 23 S1 close, S2 open C=Q/V Q=120 mC After S1 open, S2 close Q1+Q2=120 mC Same potential Q1/C1=Q2/C2 (120-Q2)/C1= Q2/C2 (120-Q2)/6 = Q2/ 3 Q2 = 40 mC Q1 = 80 mC Pictures from Serway & Beichner Capacitors with Dielectrics A dielectric is a nonconducting material, such as rubber, glass, or waxed paper. When a dielectric is inserted between the plates of a capacitor, the capacitance increases. If the dielectric completely fills the space between the plates, the capacitance increases by a dimensionless factor , which is called the dielectric constant. Dielectric constant is a property of a material and varies from one material to another. Pictures from Serway & Beichner Capacitors with Dielectrics A charged capacitor (a) before and (b) after insertion of a dielectric between the plates. The charge on the plates remains unchanged, but the potential difference decreases from DVo to DV = DVo/. Thus the capacitance increases from Co to Co. Note no battery is involved in this example. Pictures from Serway & Beichner Capacitors with Dielectrics Capacitance increases by the factor when dielectric completely fills the region between the plates. If the dielectric is introduced while the potential difference is being maintained constant by a battery, the charge increases to a value Q = Qo . The additional charge is supplied by the battery and the capacitance again increases by the factor . For parallel plate capacitor: eoA C= d Pictures from Serway & Beichner Dielectric Strength For any given separation d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the dielectric strength (maximum electric field) of the dielectric. If magnitude of the electric field in the dielectric exceeds the dielectric strength, then the insulating properties break down and the dielectric begins to conduct. Pictures from Serway & Beichner Dielectric Constant and Dielectric Strength of Various Materials at Room Temperature Material Dielectric Constant Dielectric Strength (V/m) Air (dry) 1.00059 3 x 106 Bakelite 4.9 24 x 106 Fused quartz 3.78 8 x 106 Neoprene rubber 6.7 12 x 106 Nylon 3.4 14 x 106 Paper 3.7 16 x 106 Polystyrene 2.56 24 x 106 Polyvinyl Chloride 3.4 40 x 106 Porcelain 6 12 x 106 Pyrex Glass 5.6 14 x 106 Silicone Oil 2.5 15 x 106 Strontium Titanate 233 8 x 106 Teflon 2.1 60 x 106 Vacuum 1.00000 - Water 80 - Pictures from Serway & Beichner Capacitors with Dielectric Material What are the advantages of dielectric material in a capacitor? • Increase the capacitance • Increase the maximum operating voltage • Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C. Pictures from Serway & Beichner Types of Capacitors (a) A tubular capacitor, whose plates are separated by paper and then rolled into a cylinder. (b) A high-voltage capacitor consisting of many parallel plates separated by insulating oil. (c) An electrolytic capacitor. Pictures from Serway & Beichner Energy Stored Before and After A parallel-plate capacitor is charged with a battery to a charge Qo as shown (a). The battery is then removed and a slab of material that has a dielectric constant is inserted between the plates as shown (b). Find the energy stored in the capacitor before and after the dielectric is inserted. Pictures from Serway & Beichner Energy Stored Before and After Before Qo2 Uo = 2 Co Charge on the capacitor the same before and after. (Why?) (a) Qo2 Qo2 Uo U= = = 2C 2 Co Energy reduced, where does the “missing” energy go to? Dielectric, when inserted, gets pulled into the device. External agent do negative work to keep dielectric from accelerating. Work = U-Uo Pictures from Serway & Beichner Capacitors with Dielectric The nonuniform electric field near the edges of a parallel-plate capacitor causes a dielectric to be pulled into the capacitor. Note that the field acts on the induced surface charges on the dielectric, which are nonuniformly distributed. Pictures from Serway & Beichner The H2O Molecule (Example of a Polar Molecule) Molecules are said to be polarized when a separation exists between the average position of the negative charges and the average position of the positive charges. Center of negative charges (why not at the center of the oxygen atom?) (Hint think about the effect of the Hydrogen atoms) Center of positive charges Pictures from Serway & Beichner How does a microwave work? Microwave ovens take advantage of the polar nature of the water molecule. When in operation, microwave ovens generate a rapidly changing electric field that causes the polar molecules to swing back and forth, absorbing energy from the field in the process. Because the jostling molecules collide with each other, the energy they absorb from the field is converted to internal energy, which corresponds to an increase in temperature of the food. Pictures from Serway & Beichner Induced polarization This molecule is non-polar since the center of negative charges coincide with the center of positive charges. The presence of an external E field causes the charges to change their positions and such that the center of negative charges does not coincide with the center of positive charges. We say that a dipole moment has been induced due to the presence of the E field. This is called induced polarization. Pictures from Serway & Beichner An Atomic Description of Dielectrics Potential difference DVo between the plates of a capacitor is reduced to DVo/ when a dielectric is introduced. How is this possible? Think about the E field, if Eo is the E field without the dielectric, then the field in the presence of E field is E = Eo/. I.e. The field is reduced. How is this possible? Pictures from Serway & Beichner An Atomic Description of Dielectrics (a) Polar molecules are randomly oriented in the absence of an external electric field. (b) When an external field is applied (to the right as shown), the molecules partially align with the field. (dielectric is polarized!) What is the effective E field inside the dielectric? Pictures from Serway & Beichner An Atomic Description of Dielectrics (a) When a dielectric is polarized, the dipole moments of the molecules in the dielectric are partially aligned with the external field Eo. (b) This polarization causes an induced charge on the opposite side. This separation of charge results in a reduction in the net electric field within the dielectric. The net effect on the dielectric is the formation of an induced positive surface charge density sind on the right face and an equal negative surface charge density –sind on the left face. Pictures from Serway & Beichner An Atomic Description of Dielectrics The induced surface charge give rise to an induced electric field Eind in the direction opposite the external field Eo. Therefore, the net electric field E in the dielectric has a magnitude E = Eo-Eind Pictures from Serway & Beichner What is the magnitude of the induced charge density? For parallel plate capacitor, External field Eo = s/ eo Induced Field Eind = sind/ eo And E = Eo/ = s/ eo Substitute into E = Eo-Eind gives s s sind = eo eo eo 1 And sind = s Induced charge on a dielectric placed between the plates of a charged capacitor. Note that the induced charge density on the dielectric is less than the charge density on the plates. Pictures from Serway & Beichner Effect of a Metallic Slab between the plates A parallel-plate capacitor has a plate separation d and plate area A. An uncharged metallic slab of thickness a is inserted midway between the plates. (a) Find the capacitance of the device. (b) Show that the capacitance is unaffected if the metallic slab is infinitesimally thin. Pictures from Serway & Beichner Effect of a Metallic Slab between the plates (a) Charge on one plate must induce a charge of equal magnitude but opposite sign on the near side of the metallic slab. Net charge on the slab is zero, electric field inside the slab is zero. Capacitor = two capacitor in series, each having a plate separation of (d-a)/2 as shown. Pictures from Serway & Beichner Effect of a Metallic Slab between the plates (a) 1/C = 1/C1+ 1/C2 1 1 1 = + C eoA eoA (d-a)/2 (d-a)/2 eoA C = (d-a) What happens when a d? Why? eoA (b) As a 0, d-a d and C = The original capacitance d Pictures from Serway & Beichner Effect of a Metallic Slab between the plates What are the differences in the result as compared to the previous example if we insert the metallic slab as shown? Pictures from Serway & Beichner A Partially Filled Capacitor A parallel-plate capacitor with a plate separation d has a capacitance Co in the absence of a dielectric. What is the capacitance when a slab of dielectric material of dielectric constant and thickness d/3 is inserted between the plates as shown. Pictures from Serway & Beichner A Partially Filled Capacitor Two capacitors in series 1/C = 1/C1 + 1/C2 where eoA eoA C1 = and C2 = d/3 2d/3 1 d/3 2d/3 = + C eoA eoA d 1 = +2 3eoA 3 eoA C= 2+ 1 d Pictures from Serway & Beichner