Topic_15_(Capacitance)

Capacitance EE 141 Lecture Notes Topic 15 Professor K. E. Oughstun School of Engineering College of Engineering & Mathematical Sciences University of Vermont 2012 kC7 se4 yy AYN Beate) °°- eosseoo Capacitance of an Isolated Conductor Consider a (perfect) conductor that is removed from any othe r body (i.e., an isolated conductor). Because of the mutual repulsio n between similar charges, energy must be expended to charge t he conductor. Accordingly, the magnitude of its potential Vincreases as charge is added, the magnitude of the change in potential bei ng proportional to the amount of charge added as well as upon the geometrical shape of the conductor body. From Eq. (10) of Topic 14, the electrostatic energy of the cha rged body is given by Ue=1 2/integraldisplay /integraldisplay S̺s(r)V(r)d2r=1 2QV, (1) whereQ=/integraltext/integraltext S̺s(r)d2ris the net charge on the conductor surface and where Vis its constant potential. Capacitance of an Isolated Conductor The ratio Q/Vis defined as the capacitance of the isolated conductor C≡Q VCoul/Volt≡Farad(F) (2) in terms of which its stored electrostatic energy is given by Ue=1 2CV2=Q2 2C. (3) Physically, the capacitance of an isolated body is the elect ric charge that must be added per unit increase in electrostatic potent ial on the body. Capacitance of a Multi-Conductor System Consider a generalized capacitor consisting of two (perfectly) conducting bodies separated in space by dielectric media wi th permittivities ǫjand brought to a static charge state with net charge +Qon body 1 and −Qon body 2, as illustrated. Such a capacitor possesses the following properties: Capacitance of a Multi-Conductor System Free charges ±Qreside on the conductor surfaces, resulting in a surface charge density ̺sj(r) on each body ( j= 1,2) such that ±Q=/contintegraldisplay Sj̺sj(r)d2r, (4) the plus-sign applying for j= 1 and the minus-sign for j= 2. Gauss’ law shows that the E-field lines originate normally from the surface S1of the positively charged body and terminate normally on the surface S2of the negatively charged body, with /contintegraldisplay SD(r)·ˆnd2r=±Q (5) for any surface Senclosing either S1(the plus sign applies) or S2(the minus sign applies). Capacitance of a Multi-Conductor System As a consequence of the perpendicularity of Eat each conductor surface, they are equipotential surfaces , whereV(r) =V1onS1 andV(r) =V2onS2. A single-valued potential difference ∆V=V1−V2then exists between the two conducting bodies, given by ∆V=−/integraldisplayS1 ∞E(r)·d/vectorℓ+/integraldisplayS2 ∞E(r)·d/vectorℓ=−/integraldisplayS1 S2E(r)·d/vectorℓ.(6) In order to make ∆ Vpositive, the potential reference S2is taken on the negatively charged conductor body. Capacitance of a Multi-Conductor System For alinear capacitor , the potential difference ∆ Vis proportional to the charge Q, so that Q=C∆V, (7) whereCis the capacitance of the two-body system. From Eqs. (5) and (6), one obtains C=Q ∆V=−/contintegraltext S1D(r)·ˆnd2r /integraltextS1 S2E(r)·d/vectorℓ(8) Electrostatic Energy & Capacitance The voltage difference due to a charge Qon a linear capacitor with capacitance Cis given by ∆ V=Q/C. The amount of work required to transfer an additional incremental amount of charge dQfromS2 toS1is then given by dWe= ∆VdQ=1 CQdQ. (9) Beginning from an uncharged capacitor and continuing to tra nsfer charge until a final charge Qis reached on S1, the total work expended is given by We=1 C/integraldisplayQ 0QdQ=Q2 2C=1 2C(∆V)2, (10) in agreement with Eq. (3) for the potential energy of an isola ted conductor. Electrostatic Energy & Capacitance The capacitance of a two-conductor system is then seen to be g iven in terms of its electrostatic energy Ueby either of the two equations C=2Ue (∆V)2=1 (∆V)2/integraldisplay /integraldisplay /integraldisplay VD(r)·E(r)d3r,(11) or C=Q2 2Ue=Q2 /integraltext/integraltext/integraltext VD(r)·E(r)d3r, (12) where the integration domain Vcontains all of the charged bodies comprising the multi-conductor capacitor. Example: Parallel Plate Capacitor Consider determining the capacitance of a parallel plate ca pacitor comprised of two parallel conducting plates, each of surfac e areaA separated by a distance dfilled with a dielectric of permittivity ǫ. ε With an applied voltage difference Vbetween the two plates, a charge +Qaccumulates on the surface of the upper plate and a charge−Qaccumulates on the surface of the lower plate, as illustrated. If the plate separation dis much smaller than the plate dimensions, then the effects of fringingmay be neglected and the surface charge may be assumed to be uniformly distributed ov er each capacitor plate with density ̺s=±Q/A. Example: Parallel Plate Capacitor The electrostatic field is then, to a good approximation, con fined between the plates and linearly directed from the positive t o the negative plate. Choosing the z-axis along this direction directed from the negative to the positive plate, one then has that E=−ˆ1zE. From the boundary condition given in Eq. (7) of Topic 13, E=̺s ǫ=Q ǫA. In addition, from Eq. (6), the voltage difference between the two plates is given by V=−/integraldisplayd 0E·ˆ1zdz=E/integraldisplayd 0ˆ1z·ˆ1zdz=Ed. Example: Parallel Plate Capacitor Thecapacitance is then given by C≡Q V=ǫA d(13) From Eq. (3), the stored electrostatic energy in a parallel plate capacitor is given by Ue=1 2CV2=1 2/parenleftbiggǫA d/parenrightbigg (Ed)2=1 2ǫE2Ad, (14) whereAdis the volume of the capacitor. Thebreakdown voltage V=Vbrof the insulating material between the plates occurs at the dielectric strength E=Edsof the material. For quartz, Eds= 30MV/mso that the breakdown voltage when d= 1cmis given by Vbr=Edsd= (30×106V/m)(1×10−2m) = 3×105V. Example: Capacitance of a Coaxial Line Consider a length ℓof an infinitely long coaxial line with inner radius a and outer radius bfilled with a dielectric material with permittivity ǫ. l With an applied voltage difference Vbetween the inner and outer conductors, charges + Q/ℓand−Q/ℓwill be uniformly distributed per unit length along the outer and inner surfaces, respecti vely. The corresponding surface charge densities are then ̺s= +Q 2πbℓon the outer conductor and ̺s=−Q 2πaℓon the inner conductor surface. Example: Capacitance of a Coaxial Line The electrostatic field is then radially directed inwards an d is given by (see Topic 6) E(r) =−ˆ1rQ/ℓ 2πǫr, fora<r<band is zero otherwise. The potential difference Vbetween the outer and inner conductors is then given by V=−/integraldisplayb aE(r)·ˆ1rdr=Q/ℓ 2πǫ/integraldisplayb adr r=Q/ℓ 2πǫln/parenleftbiggb a/parenrightbigg . The capacitance Cis then given by C≡Q V=2πǫℓ ln(b/a)and the capacitance per unit length of the coaxial line is C′≡C ℓ=2πǫ ln(b/a)(15) Problems Problem 16. Determine the stored electrostatic energy per u nit lengthUe/ℓof a coaxial line of inner radius acarrying charge + Q/ℓ and outer radius bcarrying charge −Q/ℓfilled with a uniform dielectric with permittivity ǫ. Problem 17. (a). Apply Gauss’ law and the appropriate bounda ry conditions to determine the electric field in the annular reg ion a<r<cbetween the oppositely charged inner and outer conductors of a coaxial line filled with a dielectric material with permi ttivityǫ1 fora<r<band permittivity ǫ2forb<r<c. (b) Determine the capacitance per unit unit length C′≡C/ℓof this line. Problems Problem 18. (a). Apply Gauss’ law and the appropriate bounda ry conditions to determine the electric field in the spherical s hell a<r<cbetween the oppositely charged concentric inner and outer conducting spheres filled with a dielectric material with pe rmittivity ǫ1fora<r<band permittivity ǫ2forb<r<c. (b) Determine the capacitance Cof this spherical capacitor. Capacitance & Energy of Multi-Conductor Systems The total energy of the electrostatic field produced by a syst em ofn charged ideal conductors embedded in a spatially homogeneo us simple dielectric medium is given by [see eq. (15) of Topic 14 ] Ue=−ǫ 2/integraldisplay/integraldisplay/integraldisplay E·∇Vd3r =−ǫ 2/bracketleftbigg/integraldisplay/integraldisplay/integraldisplay ∇·(VE)d3r−/integraldisplay/integraldisplay/integraldisplay V∇·Ed3r/bracketrightbigg =−ǫ 2/integraldisplay/integraldisplay/integraldisplay ∇·(VE)d3r, the second integral vanishing because ∇·E= 0 throughout the regionVexternal to the conductor bodies. Capacitance & Energy of Multi-Conductor Systems The divergence theorem transforms the remaining integral i nto a sum of surface integrals over each surface Sjof the entire system of conductors which spatially bound the electrostatic field pl us an additional integral over an infinitely remote surface. This latter integral vanishes if the charged conductor system is situat ed within a finite region of space so that E∼R−2andV∼R−1asR→ ∞. With the subscript jdenoting the jthconductor with constant potential Vj, the above expression becomes Ue=ǫ 2n/summationdisplay j=1Vj/contintegraldisplay SjE·ˆnda, whereˆnhere denotes the unit outward normal to the conductor surface, directed from the conductor body into the field regi on.1 1Notice that this is opposite to the convention used in the div ergence theorem, as reflected in the change of sign in the above equati on. Capacitance & Energy of Multi-Conductor Systems BecauseQj=−ǫ/contintegraltext SjE·ˆnda, one finally obtains Ue=1 2n/summationdisplay j=1QjVj (16) which is analogous to the expression given in Eq. (8) of Topic 14 for the electrostatic potential energy of a system of point char ges. The charges Qjand potentials Vjof the conductor bodies cannot both be arbitrarily prescribed and consequently must be rel ated in some fashion. Because the field equations are linear & homoge neous, these relations should be linear. The potential of the jthconductor is then directly proportional to the charge Qkof thekthconductor, with V(k) j=pjkQk,j= 1,2,...,n, where the coefficients pjkdepend only upon the geometry of the conductor system. Capacitance & Energy of Multi-Conductor Systems By superposition, the potential on the jthconductor due to the n charged conductors in the system is given by Vj=/summationtextn k=1V(k) j, so that Vj=n/summationdisplay k=1pjkQk (17) forj= 1,2,...,n. The coefficients pjk, which depend only upon the geometry of the multi-conductor system, are called coefficients of potential, wherepjkis the potential of the jthconductor per unit charge on the kthconductor. With this result, Eq. (16) becomes Ue=1 2n/summationdisplay j=1n/summationdisplay k=1pjkQjQk (18) The electrostatic potential energy of a system of charged co nductors is a quadratic function of the charges on the conductors comp rising the system. Capacitance & Energy of Multi-Conductor Systems Theorem The coefficients of potential p jkfor a multi-conductor system satisfy the three fundamental properties: pjk=pkj, pjk≥0, (19) pjj≥pjk,∀k. Capacitance & Energy of Multi-Conductor Systems The relation (17) is a set of nlinear equations giving the potentials Vjon each of the nconductors in terms of the charges Qjresiding on them. This set may be inverted to yield a set of equations givi ngQj in terms of Vjas Qj=n/summationdisplay j=1cjkVk (20) forj= 1,2,...,n. The coefficients cjjare called the coefficients of capacitance or thecapacity coefficients whereas the coefficients cjk withj/negationslash=kare called the electrostatic induction coefficients . Thecapacitance of a conductor is then seen to be given by the total charge on the conductor when it is maintained at unit potenti al, all other conductors in the system being held at zero potential. In that caseQj=cjjVjand cjj=Qj Vj. (21) Capacitance & Energy of Multi-Conductor Systems The coefficients of capacitance and induction form a matrix C= (cjk), with the coefficients of capacitance forming the diagonal and the coefficients of inductance the off-diagonal elements, that is the inverse of the matrix P= (pjk) formed from the coefficients of potential pjk, so that C=P−1. (22) Theorem The coefficients of capacitance and induction c jkfor a multi-conductor system satisfy the three fundamental prop erties: cjk=ckj, cjj>0, (23) cjk≤0,j/negationslash=k. Capacitance & Energy of Multi-Conductor Systems Substitution of Eq. (20) into Eq. (16) for the electrostatic potential energy yields Ue=1 2n/summationdisplay j=1n/summationdisplay k=1cjkVjVk (24) The electrostatic potential energy of a system of charged co nductors is thus seen to be a quadratic function of either the charges o r the potentials on the various conductors comprising the system . Capacitance & Energy of Multi-Conductor Systems If two ideal conductors S1andS2form a capacitor, application of (17) to that arrangement yields V1=p11(−Q)+p12Q+VE, V2=p21(−Q)+p22Q+VE, with +QonS2and−QonS1, whereVEis the common potential due to any external charges. The potential difference ∆ V=V2−V1 is then given by ∆V= (p11+p22−2p12)Q, (25) where the potential reference is taken on the negatively cha rged conductor in order to make ∆ Vnon-negative. The capacitance of the capacitor is then given by [cf. Eq. (2)] C≡Q ∆V= (p11+p22−2p12)−1(F) (26) Capacitance & Energy of Multi-Conductor Systems The mutual capacitance of a two conductor system can also be expressed in terms of the capacity and induction coefficients cjk. From Eq. (18) the electrostatic potential energy of a two con ductor capacitor is given by Ue=1 2Q2(p11+p22−2p12). Because P=C−1, then /parenleftbiggp11p12 p12p22/parenrightbigg =/parenleftbiggc11c12 c12c22/parenrightbigg−1 =1 c11c22−c2 12/parenleftbiggc22−c12 −c12c11/parenrightbigg , so that Ue=1 2Q21 c11c22−c2 12(c11+c22+2c12). Comparison of this expression with that given in Eq. (3) then yields C=c11c22−c2 12 c11+c22+2c12(27) for the mutual capacitance of a two conductor system. Capacitance & Energy of Multi-Conductor Systems Consider now determining the change in the electrostatic en ergyof a system of conductors that is caused by an infinitesimal chang e in either their charges or their potentials . Beginning with Eq. (16) of Topic 14, the variation of the tota l electrostatic energy of a system of ncharged conductors is given by δUe=1 2ǫ/integraldisplay /integraldisplay /integraldisplay δ(E·E)d3r=ǫ/integraldisplay /integraldisplay /integraldisplay E·δEd3r.(28) Upon setting E=−∇Vand using the fact that ∇·E= 0 in the dielectric so that ∇·δE= 0, the above result becomes δUe=−ǫ/integraldisplay /integraldisplay /integraldisplay ∇·(VδE)d3r=ǫn/summationdisplay j=1Vj/contintegraldisplay SjδE·ˆnd2r after application of the divergence theorem, where ˆndenotes the outward unit normal vector to the conductor surface Sj. Capacitance & Energy of Multi-Conductor Systems The variation in charge on the jthconductor is obtained from Gauss’ law asδQj=ǫ/contintegraltext SjδE·ˆnd2r, so that δUe=n/summationdisplay j=1VjδQj. (29) This expression then gives the change in electrostatic energy due to a change in the charges on the conductors . This is simply the work required to bring a set of ninfinitesimal charges δQjfrom infinity to the various conductor bodies in the system in the presence of the (fixed) potential Vj. Capacitance & Energy of Multi-Conductor Systems The expression (28) for the change in electrostatic energy c an also be written as δUe=−ǫ/integraltext/integraltext/integraltext E·∇(δV)d3r=−ǫ/integraltext/integraltext/integraltext ∇·(EδV)d3r. Because infinitesimal changes in the potentials, just like t he potentials themselves, are constant over the surface of eac h conductor, this expression becomes δUe=ǫn/summationdisplay j=1δVj/contintegraldisplay SjE·ˆnd2r after application of the divergence theorem. Each surface i ntegral here is recognized as the charge Qj=ǫ/contintegraltext SjE·ˆnd2ron the corresponding conductor, so that δUe=n/summationdisplay j=1QjδVj, (30) which expresses the change in electrostatic energy in terms of the change in the potentials of the conductor bodies. Capacitance & Energy of Multi-Conductor Systems The relations given in Eqs. (29) & (30) show that, by different iating the electrostatic energy Ueof a system of charged conductors with respect to the charges on the conductors, the potentials on t he individual conductors are obtained as Vj=∂Ue ∂Qj, (31) whereas the derivatives of Uewith respect to the potentials gives the charges on the conductors as Qj=∂Ue ∂Vj. (32) The symmetry relation cjk=ckjfor the coefficients of capacitance then follows from the fact that ∂2Ue/∂Vj∂Vk=∂2Ue/∂Vk∂Vj. The remaining properties in that theorem follow from the positive-definiteness of the quadratic form given in Eq. (24 ). Capacitance & Energy of Multi-Conductor Systems The electrostatic force that acts between charged bodies ca n be obtained through a consideration of the change in the total electrostatic energy of the system under a small virtual dis placement. As an illustration, consider the force per unit area acting o n the surfaceSof an ideal conductor carrying surface charge density ̺s(r). An element of surface charge ̺s(r)daexperiences an electrostatic force due to the electrostatic field of all the other charges i n the system. Because the E-field is normal to the surface of a perfect conductor, this force is perpendicular to the conductor sur faceS, and because the element of charge ̺s(r)dais bound to the conductor by internal forces, the force acting on it is directly transfer red to the body of the conductor. In the immediate neighborhood of the conductor surface the electrostatic energy density is give n by ue(r) =ǫ 2/vextendsingle/vextendsingleE(r)/vextendsingle/vextendsingle2=1 2ǫ̺2 s(r). (33) Capacitance & Energy of Multi-Conductor Systems An infinitesimally small virtual displacement ∆ ζof an elemental area ∆aof the conductor surface will then result in a decrease in the electrostatic energy by an amount given by the product of the energy densityue(r) at that point and the excluded volume ∆ a∆ζ, so that ∆Ue=−1 2ǫ̺2 s(r)∆a∆ζ. Because the magnitude of the force F(r) is given by ∆ Ue/∆ζ, this result means that there is an outward force per unit area equal in magnitude to dF(r) da=1 2ǫ̺2 s(r) =ue(r) (34) at the surface of the conductor. Capacitance & Energy of Multi-Conductor Systems By Gauss’ law, the total E-field flux emerging from the surface elementdawith surface charge density ̺s(r) is given by1 ǫ̺s(r)da, half of it directed into the body and half directed out of the b ody of the conductor. The electrostatic field intensity due to the l ocal surface charge density alone is then comprised of two parts d irected along±ˆnwith equal magnitude1 2ǫ̺s(r). Because the electric field intensity at an exterior point infi nitesimally close to the surface of the conductor is given by E(r) =ˆn̺s(r)/ǫ, it is then seen that the element of surface charge ̺s(r)daat that point produces exactly half of the total field external to that poin t. The electric field intensity acting on the element of surface charge ̺s(r)dadue to all of the other charges in the system is then given by E0(r) =ˆn̺s(r)/2ǫ. Capacitance & Energy of Multi-Conductor Systems The force dF(r) acting on the element of surface charge ̺s(r)da, and consequently acting on the surface element daof the conductor itself, is then given by dF= (̺sda)E0, so that dF(r) da=1 2ǫ̺2 s(r)ˆn=ue(r)ˆn, (35) in agreement with the expression given in Eq. (34), where ˆnis the outward unit normal vector to the conductor surface at the po int r∈ S. Hence, the electrostatic force on a conductor tends to pull the conductor into the field; that is, an electrostatic field exerts a negative pressure on a conductor with magnitude equal to the energy density in the field. Capacitance & Energy of Multi-Conductor Systems The total force acting on a conductor body is then obtained by integrating the expression for the force per unit area given in in Eq. (35) over the entire surface Sof the conductor body as F=1 2ǫ/contintegraldisplay S̺2 s(r)ˆnda=1 2/contintegraldisplay SD(r)·E(r)ˆnda. (36) Capacitance & Energy of Multi-Conductor Systems Consider now an isolated electrostatic system (comprised o f conductors, dielectrics, and point charges) when one of its parts undergoes a differential displacement drunder the influence of the electrostatic forces acting upon it. The mechanical work do ne by the system is then given by dWm=F·dr=Fxdx+Fydy+Fzdz. (37) Because the system is isolated, this work is done at the expen se of the electrostatic energy Ue, so that ( conservation of energy ) dUe+dWm= 0. (38) Capacitance & Energy of Multi-Conductor Systems Eqs. (37)–(38) then show that Fx=−/parenleftbigg∂Ue ∂x/parenrightbigg Q(39) with analogous expressions for FyandFz. The subscript Qindicates that the total charge of the system remains constant during t he displacement. Capacitance & Energy of Multi-Conductor Systems If the body under consideration is constrained to rotate abo ut an axis, then Eq. (37) is replaced by dWm=/vector τ·d/vectorθ, (40) where/vector τ= (τ1,τ2,τ3) is theelectrical torque &d/vectorθ= (dθ1,dθ2,dθ3) is the differential angular displacement. Eqs. (38) & (40) th en show that τ1=−/parenleftbigg∂Ue ∂θ1/parenrightbigg Q(41) with analogous expressions for τ2andτ3. Capacitance & Energy of Multi-Conductor Systems Consider next an isolated electrostatic system in which all of the free charge resides on the surfaces of the conductors in the syste m which are maintained at fixed potentials by means of an external ene rgy source (e.g., by batteries). If one of its parts undergoes a diff erential displacement drunder the influence of the electrostatic forces acting upon it, Eq. (37) still holds but conservation of energy now r equires that [cf. Eq. (38)] dUe+dWm=dWext, (42) wheredWextis the energy supplied by the external source. Capacitance & Energy of Multi-Conductor Systems From Eq. (16) which gives the electrostatic energy in terms o f the chargesqkand potentials Vkon a system of charged conductors as Ue=1 2/summationtext kVkqk, if some part of the system is displaced while the potentials Vkare held fixed, then dUe=/summationdisplay kVkdqk. (43) The energy dWextsupplied by the external sources is the work required to move each of the charge increments dqkfrom zero potential to the potential Vkof thekthconductor, so that dWext=/summationdisplay kVkdqk, (44) and consequently dWext= 2dUe. (45) Capacitance & Energy of Multi-Conductor Systems Substitution of this result in Eq. (42) to eliminate dWextand combining the result with Eq. (37) then gives dUe=Fxdx+Fydy+Fzdz, (46) so that Fx=/parenleftbigg∂Ue ∂x/parenrightbigg V(47) with analogous expressions for FyandFz. The subscript Vindicates that all potentials of the system are maintained constant du ring the displacement dr. In a similar manner, analogous expressions are obtained for the electric torque, viz. τ1= (∂Ue/∂θ1)V, etc. Capacitance & Energy As an example, consider a parallel plate capacitor where eac h plate has width wand length bwith plate separation dfilled with a dielectric block of permittivity ǫ. Let the capacitor plates be maintained at the constant potential difference ∆ V. V ε ε0 If the dielectric block is withdrawn from the capacitor alon g the direction of the b-dimension until the length xremains between the plates, as illustrated, determine the force tending to pull the dielectric block back into place. Capacitance & Energy From Eq. (15) of Topic 14, the energy of the capacitor system i s given by Ue=1 2/integraldisplay/integraldisplay/integraldisplay VǫE2(r)d3r With fringing effects at the edge of the capacitor plates negl ected so thatE= ∆v/dbetween the capacitor plates and is zero outside, the above integration gives Ue=ǫ 2/parenleftbigg∆V d/parenrightbigg2 dwx+ǫ0 2/parenleftbigg∆V d/parenrightbigg2 dw(b−x). The force pulling the dielectric block back into place is the n obtained using Eq. (47) as Fx=/parenleftbigg∂Ue ∂x/parenrightbigg V=w 2d(ǫ−ǫ0)(∆V)2 in the direction of increasing x.