Parallel and series capacitors—summary Î Capacitors in parallel C eq = C1 + C 2 + ⋅ ⋅ ⋅ Î Capacitors in series 1 1 1 = + +⋅⋅⋅ Ceq C1 C2 It is foolish to connect capacitors in series. Example: 100 μF in series with 10 μF is 9 μF (check yourself). The 100 μF capacitor will be totally wasted. PHY2049: Chapter 25 1 (continued) Î smaller of two Two capacitor in series 1 1 1 = + Ceq C1 C 2 C1C2 C2 = < C2 Ceq = C1 + C 2 1 + C 2 / C1 less than 1 Î Ceq even smaller than smaller of the two n capacitors in series 1 1 1 1 = + + +L Ceq C1 C 2 C3 Ceq 1 1 > Ceq C1 Ceq < C1 smallest even smaller than smallest of all PHY2049: Chapter 25 2 Examples Î Four 1 μF in parallel. Find Ceq. 4 μF Î Four 1 μF in series. Find Ceq. 0.25 μF Î 1.3 μF and 2.0 μF in series. Ceq is: (a) 0.79 μF (b) 1.65 μF (c) 3.3 μF PHY2049: Chapter 25 3 Example: parallel-series combo Î Equivalent capacitance? 1.0 μF 2.0 μF 3.0 μF 1 and 2 in parallel 1.0 + 2.0 = 3.0 μF Together, in series with 3 1 1 1 2 1 = + = = Ceq 3.0 3.0 3.0 1.5 1.5 μF PHY2049: Chapter 25 4 (continued) Î Charge on C1? 2.0 μF 1.0 μF 10 V 3.0 μF Total charge 10 V x 1.5 μF = 15 μC Charge on 1 and 2 Same as the total! Charge on 1 q=q1 + q2 Potential differences across 1 and 2 are the same. q1 and q2 in proportion with C1 and q1=q x 1.0/(1.0+2.0) = 5.0 μC C2. PHY2049: Chapter 25 +q –q +q –q +q –q 5 Energy stored in capacitor Î In terms of charge Derived by considering work dW’ done by a fictitious process which moves infinitesimally small amount of charge +dq’ from conductor 1 to conductor 2 of capacitor, leaving behind –dq’ on conductor 1: q2 U= 2C Î In terms of potential Î Since q=CV (definition of C) Compare with 1 K = mv 2 (kinetic energy) 2 PHY2049: Chapter 25 1 U = CV 2 2 U= 1 2 kx (spring) 2 6 Energy stored in electric field Î Two alternative views Energy is stored in charge configuration in capacitor Energy is stored in E field in capacitor Î Second view (will be important later in dealing with electromagnetic waves) Define energy density Show for parallel-plate capacitor Î u= U volume u= 1 ε0 E 2 2 This equation holds for any E field produced at any point in space by any source Derivation requires vector calculus PHY2049: Chapter 25 7 Equivalence of two views (by example) Spherical conductor ÎView 1 + + + + + + + Energy is stored in capacitor’s charge configuration Q2 U= 2C Generalize definition of capacitance to single conductor and find (see page 661) 2 C = 4πε0 R ÎView 2 Energy Inside is stored in E field E=0 + + Q + R+ + + + + + + Q 8πε0 R 1 u = ε0 E 2 2 1 Q 4πε 0 r 2 Q 2 ∞ dr Q2 2 = 4πr dr = Checks! 2 ∫ R 8πε0 r 8πε0 R Outside E = 1 1 Q U = ∫ udV = ∫ ε0 2 R 2 4πε0 r outside ∞ U= + 2 ( ) PHY2049: Chapter 25 8 Dielectrics ÎDielectric polarized. is insulator. In E field, it becomes partly For microscopic view, read page 671. If dielectric fills the gap of charged capacitor, E0 due to charges +q and –q partly polarizes it, inducing charges –q’ and +q’ near surfaces. These in turn produce field that partly cancels E0. Net field E proportional to, and less than, E0. ÎWhat’s the point? E0 → E = E0/κ less than E0 V0 → V = V0/κ from definition of V C0 → C = κ C0 since C=q/V Larger than C0, which means capacitor stores more charge for given potential difference applied by battery. Beneficial to fill gap with dielectric. PHY2049: Chapter 25 9 (continued) ÎΚis called dielectric constant. Larger than 1. ÎInduced charge q’. So far, general to any capacitor. Now restrict ourselves to parallel-plate capacitor E0 A = q ε0 q + (− q′) EA = ε0 E q − q′ = E0 q κ Gauss’ law 1 q ′ = q 1 − < q κ Induced charge q’ is less than q. κ=1 (vacuum, no dielectric) → q’=0 No induced charge. κ large (strong dielectric) → q’→q PHY2049: Chapter 25 10 REMINDER ÎNext WebAssign due Tomorrow, Thursday ÎTest 1 on Monday in Class (Chapters 21– 25) Study Sample Exams Must bring Gator1 ID card (Will take away points if you forget ;< ) Calculator (No formulae allowed on calculator) One 8.5”x11” formula sheet (May use both sides) Blank scratch paper (Department has no money to provide) Pencil, eraser, and sharpner, as usual PHY2049: Chapter 25 11