EE468G NOTES (5A)Capacitance: Defined for one or more conducting structures Free charge on conductor = Q Voltage: V Capacitance VQC = Unit: [F]=Faraday Calculation of capacitance: Step 1: Apply a voltage source to the conductors V Step 2: Calculate free-charge amount Q: Step 3: C=Q/V Example 1: Parallel plate capactor (ignore fringe effect): Plate surface area= Sd 0V ε Electric field in the dielectric: the electric field intensity and electric flux are: ()zadVEˆ/0−=r ()zadVEDˆ/0εε−==rr Free charge density and total charge on the upper plate: dVDazS/ˆ0ερ=⋅−=r dSVSQS/0ερ== Capacitance ][/00FdSVdSVVQCεε=== Property: dCSCC /1,,∝∝∝εInfinite plate: Electric field intensity vector is uniform between the plates and pointing downward (if the upper plate has higher potential). Fringe effect (effect of finite plate size): Electric field intensity is almost uniform between the two plates, especially away from the edges. The approximation to infinite plate case is good when the dimension in x and y directions are much larger than that of the z-directrion. By “ignoring the fringe effect” or “ignoring the edge effect”, we mean that the fields between the two plates are of the same as that of the infinite plate case.Example 2: Parallel plate capacitor with two layers of dielectrics, as shown (surface area = S). Calculate the capacitance. d 0V 04εε= 02εε= Solution: We have found the electric field intensity and the flux: ()()()()dzdadVDadVEdzadVDadVEzzzz<<−=−=<<−=−=2/,ˆ3/8,ˆ3/42/0,ˆ3/8,ˆ3/20020200101εεrr The charge density and total charge on upper plate are: () ()dSVSQdVDaSzS3/8,38ˆ00002ερερ===⋅−=r The capacitance is ()dSVQC 3/8/00ε== Consider the single layer parallel plate capacitor 04ε 02ε 2d 2d (2//401dSC)ε= )2//(202dSCε= ()CdSSdSdSdCC13/818322/42/11000021===+=+εεεεParallel connection of capacitors d 0V 04εε= 02εε= Surface Area= 1SSurface Area= 2S From the solution of simple capacitors we know that the electric field between two conducting plates is given by 0ˆzVEad=−r This means that the electric field is not a function of dielectric permittivity (as long as it is uniform). Hence we can use this to solve the problem of parallel connection. Total charge on upper plate of part 1: ()01101111 1 1垐,zzVSVSQS a a Cddεdεε⎛⎞=⋅− ⋅ − = =⎜⎟⎝⎠ Similarly, the total charge the upper plate of part 2: ()02202222 2 2垐,zzVSVSQS a a Cddεεε⎛⎞=⋅−⋅ − = =⎜⎟⎝⎠d Total charge on both plates: ()11 2 2 0220 22012SSVSV SVQQQdd dεεεε+=+= + = Capacitance: 11 2 2 11 2 212SS S SCCCdddεεε+==+=+εCapacitance of a coaxial cable per unit length a b ε V Potential=0 Solution: Assuming that the inner line has potential 0, and the outer line has potential V. The electric flux is found to be: ()ρρεaabVDˆln−=r Surface charge density at b=ρis: ()abbVDaaSlnˆερρρ=⋅−==r The surface charge per unit length is: ()abVbQSln22περπ== Capacitance: ()abVQCln2πε== [F/m], for coaxial cable If mm,533.1 mm,406.0,26.2=== barε ()12122 2.26 8.854 1093.73 10 [ / ] 93.73 [pF/m]ln 1.533/0.406CFπ−−×× ×==×=mTwo-wire lines: Radius of each wire = a, Center-to-center distance of wires is D D 2a Capacitance per unit length: 21,[F/m]coshln 1222CDDDaaaπεπε−==⎡⎤⎛⎞⎛⎞⎜⎟⎢⎥+−⎜⎟⎝⎠⎢⎥⎝⎠⎣⎦ Example: calculate the capacitance per unit length of a two-wire line with a=0.01 [m], and D=0.1 [m]. Solution: 1222D0.152a 0.028.854 1012.1[pF/m]ln 5 5 1ln 122CDDaaπε π−==××===⎡⎤⎡⎤+−⎛⎞⎣⎦⎢⎥+−⎜⎟⎢⎥⎝⎠⎣⎦Example: Find the center-to-center distance of a two-wire line that has capacitance of 30 [pF/m], assuming that the radius of each wire is 0.8 [cm]. Solution: ()()()12022201220.927222DLet u=2a30 10ln 1ln 122ln 1 0.92730 101 2.5271 2.527 6.386 5.0545.054 6.386 1 7.3867.3861.4615.054 22 2 0.8 1.461 2.34CuuDDaauuuu euu uuuDuaDauπεπεπε−−==⎡⎤+−⎛⎞⎢⎥+−⎜⎟⎢⎥⎝⎠⎣⎦+−= =×+−= =−= − = − +=+=====×=×× = [cm]=×Capacitance of two concentric conducting spheres b a Laplace equation in spherical system 22222 2211 1sin 0sin sinVVVrrr r r rθθθ θ θφ∂∂ ∂ ∂ ∂⎛⎞ ⎛ ⎞∇= + + =⎜⎟ ⎜ ⎟∂∂ ∂ ∂ ∂⎝⎠ ⎝ ⎠2V Spherical symmetry: V has no angle dependences: 2211122210(CCVVVrrC Vrrr r r r r r∂∂ ∂ ∂⎛⎞=⇒ =⇒ = ⇒ =+⎜⎟∂∂ ∂ ∂⎝⎠)C Potential on inner sphere is V, and one outer sphere is 0: 0()1111120 2 1 02,0 ,CCCV C C a b VCVab−−−+= += ⇒ = − =0 Hence ()0001121 11垐() ,rrVVVVr V E V a aabr rra b−−−−∂=+=−∇=−=−∂−rCharge density and total charge on inner sphere is: 20021 1 14ˆ.,4()Sr SVVaE Q aaa b a b1επερε πρ−− −−== = =−−r Capacitance: 1104[]QCFVabπε−−==−Capacitance of one spherical conducting surface with radius = a This can be obtained by assuming that the outer sphere has radius of infinitely large: b →∞ 1144limbCaabπεπε−−→∞⎛⎞==⎜⎟−⎝⎠ You can use this formula to calculate the capacitance of the earth if it is assumed that the earth is of perfect conductor. Example: A conducting sphere of radius 0.2 [m] is immersed in a dielectric material with 2.56rε=. Calculate its capacitance. Solution: 124 4 2.56 8.854 10 0.2 57.0 [pF]Caπε π−==××××=Circuit model of capacitors (I-V equation) C I + - V ()dQ d CV dVICdt dt dt== = Capacitors with lossy dielectrics C I + - V C R ε When 0σ≠, conduction current will be induced in the material which causes energy dissipation. We introduce a resistor to account for this effect. ,,RC C RdV VII I I C Idt RdV VICdt R=+ = ==+ For homogeneous dielectrics, we have a relation between C and R: RCεσ=Example: A capacitor filled with homogeneous dielectric with . If 402, 1.010 [S/m]εεσ−==×10[k ]R=Ω, find the capacitance C. Solution: From: RCεσ=, we have 1234112 8.854 1010 10 101.77 10 [ ]17.7[pF]CRFεσ−−−××==××=×= (HW problem 6-10)Energy: Work must be done to create a charge distribution. By conservation of energy, the energy is stored in the charge distribution. 1Q 1Qis introduced 1Q 2Qis introduced in 1Qsystem 2Q Consider moving a point charge to the location at P1, the work done is . Then we move a second charge to the location at P2. This action needs to