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 CapacitancePictures from Serway & Beichner A capacitor consists of two conductors (known as plates) carrying charges of equalmagnitude but opposite sign.A potential difference AV exists between the conductors due to the presence of the charges. What is the capacity of the device for storing charge at particularvalue of AV?Definition of CapacitancePictures from Serway & Beichner Experiments show the quantity of electric charge Q on a capacitor is linearly proportional to the potential difference between the conductors, that is Q ~ AV.Or we write Q = C AV 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: C =Q AV SI Unit: farad (F),1F = 1 C/VTypical device have capacitances ranging from microfarad to picofarad.Parallel - Plate Capacitors Pictures from Serway & Beichner 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. Parallel-Plate Capacitors Pictures from Serway & Beichner 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 surface charge density of each plate is o = Q/A. A d 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 Variation with A Parallel-Plate Capacitors Pictures from Serway & Beichner Variation with d A d Potential differenceAV constant across, E field increases as d decreases.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. AV = Ed means that AV decreases.Parallel-Plate Capacitors Pictures from Serway & Beichner Variation with d(contd) A 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 E field result in the wires that drives more charge onto the plates, increasing the potential diff. AV 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 Parallel-Plate Capacitors Pictures from Serway & Beichner (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.Parallel-Plate Capacitors Pictures from Serway & Beichner Assume electric field uniform between the plates, we have (see lecture on Gausss Law) E = Q co o = coA AV = E d = Qd coA C = Q AV Q Qd /coA = C = coA d (As we have argued before) Example Pictures from Serway & Beichner 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. F 10 0 . 60150 = =dACkc( )( )( )1512 12010 0 . 6010 0 . 21 10 85 . 8 1 = =CAdkc

d = 3.10 109m=3.10 nm 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 Pictures from Serway & Beichner 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/Cthroughout 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= coA/d = 8.85x10-12 x (1000)2/800 = 11.1 nF Potential between ground and cloud isAV = Ed = 3.0 x106 x 800 = 2.4 x 109 V Q = C(AV) = 26.6 C Cylindrical Capacitors Pictures from Serway & Beichner 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.L Cylindrical Capacitors Pictures from Serway & Beichner Assume that L is >> a and b, neglect the end effects.E is perpendicular to the long axis of the cylinders and is confined to the region between them.Potential difference between the two cylinders is given byL Vb-Va = -E . ds b a Where E is the E field in the region a < r < b.Our discussion on Gausss Law Er = 2k/r where is the linear charge density of the cylinder. Note that the charge on outer cylinders does not contribute to E field inside it.Cylindrical Capacitors Pictures from Serway & Beichner L Vb-Va = -Er dr b a = -2k b a dr r = -2k ln() b a Using = Q/L , we haveC = =Q AV ln() b a 2kQ L Q C = ln() b a 2k L Cylindrical Capacitors Pictures from Serway & Beichner LC = ln() b a 2k L What is the capacitance per unit length ? ExampleCo-axial Cable. Read the cable, typically 50 pF/m. Is this sensible ?Typically a ~ 0.5 mm, b~ 1.5 mm 50pF/m) 3 ln( 10 99 . 8 21/9= = L CCylindrical Capacitors vs Parallel Plate Capacitors Pictures from Serway & Beichner What are the advantages of cylindrical capacitor over that of a parallel plate capacitor?L Capacitance of an isolated sphere Pictures from Serway & Beichner 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.Q Electric potential of the sphere of radius R is kQ/R and V= 0 at infinity, we haveC =Q AV = Q kQ/R = R k = 4tco R C is proportional to its radius and independent of both the charge on the sphere and the potential difference.Example Pictures from Serway & Beichner 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 = 4tco R R = (8.99 x 109 N m2/C2)(1.00 x 1012 F) = 8.99 mm C = 4t (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 The Spherical Capacitors Pictures from Serway & Beichner 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.The Spherical Capacitors Pictures from Serway & Beichner Gausss law E field outside a spherically symmetric charge distribution is radial and given by Er = kQ/r2. The potential difference between the spheres is Vb-Va = -Er dr b a = -kQ b a dr r2 =kQ 1 b 1 a C = Vb-Va Q = ab k (b-a) What happens to the capacitance of this system when the radius of the outer sphere approaches infinity? Example Pictures from Serway & Beichner SB 26 Q12A 20.0 F 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|.|

\| aba b QbQ kaQ kV Vb a04= =c ta babV VQCb a 04= =tcaaaC0208 =2 4= c ttc08=tcCaThe intervening volume is( ) ( )303 3334 334 334 33487 7c tt t t tCa a b = = =30233847c tCV = = 2.13 x 1016 m3 Example Pictures from Serway & Beichner SB 26 Q16What 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. ( )( ) m 10 37 . 6 m N C 10 85 . 8 446 2 120 ==tt R C eC= 7.08 x 10-4 F A large capacitor ! Combinations of Capacitors Pictures from Serway & Beichner 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 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.Combinations of Capacitors Pictures from Serway & Beichner 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=C1AV, Q2=C2AV Define an equivalent capacitor having Ceq s.t. Q = CeqAV We have CeqAV = C1AV + C2AV And hence Ceq = C1 + C2 (for parallel combination) In general Ceq = C1 + C2+ C3+ .. (for parallel combination) Combinations of Capacitors Combinations of Capacitors Pictures from Serway & Beichner 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.(contd) Combinations of Capacitors Pictures from Serway & Beichner 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.Combinations of Capacitors Pictures from Serway & Beichner Series Combination Voltage AV across battery terminals is split between two capacitors.AV = AV1 + AV2 Where AV1 and AV2 are potential diff across capacitors C1 and C2.Suppose we have equivalent capacitor Ceq = Q/AV For each capacitor, we have AV1 = Q/C1 and AV2=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) Example: Equivalent Capacitance Pictures from Serway & Beichner In parallel use C=C1+C2 In series use 1/C=1/C1+1/C2 6.00 F 20.00 F 2.50 F 8.50 F 20.00 F In series use 1/C=1/C1+1/C2 5.965 F Example: Equivalent Capacitance Pictures from Serway & Beichner In parallel use C=C1+C2 In parallel use C=C1+C2 In series use 1/C=1/C1+1/C2 Example: Equivalent Capacitance 26.22 Pictures from Serway & Beichner In series use 1/CB=1/C+1/C+1/C In series use 1/CA=1/C+1/CC C/2 C/3 In parallel use Ceq=C+C/2+C/3 Example Pictures from Serway & Beichner 26.10 A variable air capacitor used in tuningcircuits 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 u, where u=0 corresponds to the maximum capacitance.Example Pictures from Serway & Beichner S&B 26.10 With u=t, the plates are out of mesh and the overlap area is zero.With u=0, the overlap area is that of a semi-circle, tR2/2.By proportion, the effective area of a single sheet of charge is (tu)R2/2.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-1and the total capacitance is( )( ) ( )22 1 21 220 effective 0distancedR N AN Cu t c c = =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 AV=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) isdW = AV dq =dq q C The total work required to charge the capacitor from q = 0 to some final charge q = Q isW = Q dq 0 q C = 1 C Q 0 q dq = Q2 2C Energy Stored in a Charged Capacitor Work done in charging the capacitor = electric potential energy U stored in the capacitor.U = Q2 2C = 1 2 QAV= 1 2 C (AV)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 Pictures from Serway & Beichner 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 AV 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 = QAV/2.1V = 1 J/C hence the unit for the area is joule J. Energy Density Stored in a Charged Capacitor For parallel plate capacitor, AV = E d,C = coA/d we haveU = 1 2 coAd (E d)2 = 1 2 (coA d) E 2 Ad = Volume occupied by the E field. This lead to a new quantity known as Energy Densityu = U/Volume = U/Ad u = 1 2 co E 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 Pictures from Serway & Beichner 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 = Q2/2Cand C = coA/dand d2 = 2 d1 thenC2= C1/2 and the energy stored doubles. Example Pictures from Serway & Beichner 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 = Q2/2coA. Is this force attractive or repulsive. Does this equation follow from common sense? F =

W=U= Fdx}||.|

\|=|.|

\|=Ax QdxdCQdxddxdU02 22 2 cF = Q2/2coA Rewiring Two Charged Capacitors Pictures from Serway & Beichner Two capacitors C1 and C2 (where C1 > C2) are charged to the same initial potential difference AVi, 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 AVf 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. Rewiring Two Charged Capacitors Pictures from Serway & Beichner Before switches are closed:Q1i = C1 AVi andQ2i = - C2 AVi (negative sign for plate 2) Total Q = Q1i + Q2i = (C1-C2) AVi After switches are closed:Total charge in system remain the sameTotal Q = Q1f + Q2f Charges redistribute until the entire system is at the same potential AVf. And this potential is the same across both the capacitors. Q1f = C1 AVf and Q2f = C2 AVf Rewiring Two Charged Capacitors Pictures from Serway & Beichner After switches are closed (contd):Q = Q1f + Q2f Q1f / Q2f = C1/C2 Q1f = [ C1/C2 ] Q2f Hence Q = Q1f + Q2f =[ 1+ C1/C2 ] Q2f We have Q2f = QC2 + C1 C2 and C2 + C1 C1 Q1f = Q AV1f = Q1f C1 = C2 + C1 C1 Q C1 = Q C2 + C1 And =AV2f =AVf AVf = Q C2 + C1 = (C1-C2) AVi C2 + C1 Rewiring Two Charged Capacitors Energy Before switches are closed: Ui = C1 (AVi)2/2 + C2 (AVi)2/2 = ( C1 + C2 ) (AVi)2/2 After switches are closed: Uf = C1 (AVf)2/2 + C2 (AVf)2/2 = ( C1 + C2 ) (AVf)2/2 Uf =( C1 + C2 ) 1 2 Q C2 + C1 2 = 1 2 Q2 C2 + C1 Uf = 1 2 C2 + C1 (C1-C2)2 (AVi )2 Ui = 1 2 ( C1 + C2 ) (AVi)2 We haveUf Ui = C1 - C2 C1+ C2 2 Example Pictures from Serway & Beichner S&B Chapter 26 Q 23 Consider the circuit as shown, where C1 = 6.00F and C2= 3.00 F and AV =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.Example Pictures from Serway & Beichner S&B Chapter 26 Q 23 S1 close, S2 open C=Q/V Q=120 C After S1 open, S2 close Q1+Q2=120 C Same potential Q1/C1=Q2/C2 (120-Q2)/C1= Q2/C2 (120-Q2)/6= Q2/ 3 Q2 = 40 C Q1 = 80 C Capacitors with Dielectrics Pictures from Serway & Beichner 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 k , which is called the dielectric constant. Dielectric constant is a property of a material and varies from onematerial to another.Capacitors with Dielectrics Pictures from Serway & Beichner 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 AVo to AV = AVo/k. Thus the capacitance increases from Co to k Co.Note no battery is involved in this example.Capacitors with Dielectrics Pictures from Serway & Beichner Capacitance increases by the factor kwhen 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 = k Qo . The additional charge is supplied by the battery and the capacitance again increases by the factor k. For parallel plate capacitor:C = kcoA d Dielectric Strength Pictures from Serway & Beichner 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.Dielectric Constant and Dielectric Strength of Various Materials at Room Temperature Pictures from Serway & Beichner Material Dielectric Constant kDielectric Strength (V/m) Air (dry)1.000593 x 106 Bakelite4.924 x 106 Fused quartz3.788 x 106 Neoprene rubber6.712 x 106 Nylon3.414 x 106 Paper3.716 x 106 Polystyrene2.5624 x 106 Polyvinyl Chloride3.440 x 106 Porcelain612 x 106 Pyrex Glass5.614 x 106 Silicone Oil2.515 x 106 Strontium Titanate2338 x 106 Teflon2.160 x 106 Vacuum1.00000- Water80- Capacitors with Dielectric MaterialPictures from Serway & Beichner 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.Types of Capacitors Pictures from Serway & Beichner (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.Energy Stored Before and After Pictures from Serway & Beichner 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 k is inserted between the plates as shown (b). Find the energy stored in the capacitor before and after the dielectric is inserted.Energy Stored Before and After Pictures from Serway & Beichner Before (a) Uo =Qo2 2 Co Charge on the capacitor the same before and after. (Why?) U == =Qo2 2 C Qo2 2 k Co Uo k 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 Capacitors with Dielectric Pictures from Serway & Beichner 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.The H2O Molecule (Example of a Polar Molecule) Pictures from Serway & Beichner 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 positive charges Center of negative charges (why not at the center of the oxygen atom?) (Hint think about the effect of the Hydrogen atoms) How does a microwave work?Pictures from Serway & Beichner 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.Induced polarization Pictures from Serway & Beichner 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. An Atomic Description of Dielectrics Pictures from Serway & Beichner Potential difference AVo between the plates of a capacitor is reduced to AVo/k 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/k. I.e. The field is reduced.How is this possible?An Atomic Description of Dielectrics Pictures from Serway & Beichner (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? An Atomic Description of Dielectrics Pictures from Serway & Beichner (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 oind on the right face and an equal negative surface charge density oind on the left face.

An Atomic Description of Dielectrics Pictures from Serway & Beichner 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

What is the magnitude of the induced charge density?Pictures from Serway & Beichner 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.For parallel plate capacitor,External field Eo = o/ co Induced Field Eind = oind/ co And E = Eo/k = o/ k co Substitute into E = Eo-Eind gives o k co = o co co oind Andoind = k 1 k o Effect of a Metallic Slab between the plates Pictures from Serway & Beichner 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.Effect of a Metallic Slab between the plates Pictures from Serway & Beichner 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.(a) Effect of a Metallic Slab between the plates Pictures from Serway & Beichner (b)As a 0, d-a d andC = 1/C = 1/C1+ 1/C2 coA (d-a)/2 coA (d-a)/2 + 1 1 = 1 C C = coA (d-a) (a) coA d The original capacitance What happens when a d? Why?Effect of a Metallic Slab between the plates Pictures from Serway & Beichner What are the differences in the result as compared to the previous example if we insert the metallic slab as shown?A Partially Filled Capacitor Pictures from Serway & Beichner 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 k and thickness d/3 is inserted between the plates as shown.A Partially Filled Capacitor Pictures from Serway & Beichner Two capacitors in series 1/C = 1/C1 + 1/C2 where kcoA C1 = d/3 coA C2 = 2d/3 and C 1 = d/3 kcoA + 2d/3 coA = d 3coA 1 k + 2 C =3k 2k+ 1 coA d