2049_ch25B - Capacitance PHY2049: Chapter 25 1 What You...

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Unformatted text preview: Capacitance PHY2049: Chapter 25 1 What You Know: Electric Fields Coulomb’s law Electric fields Equilibrium Gauss’ law Electric fields for several charge configurations Point Dipole (along axes) Line Plane (nonconducting) Plane (conducting) Ring (along axis) Disk (along axis) Sphere Cylinder PHY2049: Chapter 25 2 What You Know: Electric Potential Electric potential energy Electric potential Equipotential surfaces Potential of point charge Potential of charge distribution Special cases: dipole, line, ring, disk, sphere Relationship of potential and electric field Calculating the potential from the field Calculating the field from the potential Potential energy from a system of charges PHY2049: Chapter 25 3 Capacitance: Basic Idea Capacitance: Capacity to store charge Like a tank Capacitor is electrically neutral (equal + and − charge regions) q = CV (C is a property of the device, independent of q, V) Units: [C] = “Farad” = Coul/Volt PHY2049: Chapter 25 4 Calculating Potential Difference Electric field lines start on + charges, terminate on − + V+ − V− = − ∫ E ⋅ ds − Follow E field line during integration (note cosθ = −1) + V ≡ V+ − V− = ∫ E ds − E, ds positive ++++++++++++++++ E d --------------------PHY2049: Chapter 25 5 Parallel Plate Capacitor σ q From Gauss’ law (conducting sheet) E = = ε 0 Aε 0 ⎛q⎞ So V = Ed = ⎜ ⎟d ⎝ Aε 0 ⎠ ε0 A Therefore q = V ≡ CV d A C = ε 0 = ε 0 × "length" d Depends only on geometry of device ++++++++++++++++ E d --------------------- PHY2049: Chapter 25 6 Example A = 1m2, d = 1 μm A C = ε 0 = 8.85 × 10-12 × 106 = 8.85 × 10-6 F d C = 8.85 μ F Largish, but somewhat typical value PHY2049: Chapter 25 7 Cylindrical Capacitor Inner = a, outer = b, length = L Gauss’ law: Using λ=q/L + V = ∫ E ds − V = −∫ a b q1 E= 2π Lε 0 r (ds = −dr) b - - + + + - - a + - + - q ln ( b / a ) Edr = 2π Lε 0 C 2π = ε0 L ln ( b / a ) - - + + - + - - Capacitance per unit length, e.g. “coaxial” cable (RF frequencies) PHY2049: Chapter 25 8 Special Case for Cylinder Outer shell “very close” to inner shell: b − a = d (d small) Use ln(1+x) ≅ x (for x small) ⎛b⎞ ⎛a+d⎞ d ln ⎜ ⎟ = ln ⎜ ⎟ a⎠ a⎠a ⎝ ⎝ Asurface 2π L 2π aL ≅ ε0 = ε0 C = ε0 ln ( b / a ) d d A Just like parallel plate capacitor: C = ε 0 d Always true if surfaces are close together PHY2049: Chapter 25 9 Spherical Capacitor Inner radius = a, outer radius = b Coulomb’s law: + V = ∫ E ds − V = −∫ a b E= q b - 1 4πε 0 r 2 + - - a + q b−a Edr = 4πε 0 ab - + + - + - (ds = −dr) - - + + - + - - 4π ab C = ε0 b−a PHY2049: Chapter 25 10 Two Special Cases Isolated sphere: corresponds to b = ∞ 4π ab 4π a = ε0 → ε 0 4π a C = ε0 b−a 1− a / b Outer shell “very close” to inner shell: b − a = d (d small) Asurface 4π ab 4π a ≅ ε0 = ε0 C = ε0 b−a d d A C = ε0 d 2 Again, just like parallel plate: PHY2049: Chapter 25 11 Capacitors in Parallel V1 = V2 = V3 (same potential top and bottom) Total charge: Qtot = Q1 + Q2 + Q3 CeqV = C1V + C2V + C3V Ceq = C1 + C2 + C3 Basic law for combining capacitors in parallel Works for N capacitors PHY2049: Chapter 25 12 Capacitors in Series q1 = q2 = q3 (same current charges all capacitors) Total potential: V = V1 + V2 + V3 q/Ceq = q/C1 + q/C2 + q/C3 1 1 1 1 = + + Ceq C1 C2 C3 Basic law for combining capacitors in series Works for N capacitors PHY2049: Chapter 25 13 ConcepTest Two identical parallel plate capacitors are shown in an end-view in Figure A. Each has a capacitance of C. If the two are joined together at the edges as in Figure B, forming a single capacitor, what is the final capacitance? (a) C/2 (b) C (c) 2C Area is doubled (d) 0 (e) Need more information A PHY2049: Chapter 25 B 14 ConcepTest Each capacitor is the same in the three configurations. Which configuration has the lowest equivalent capacitance? (1) (2) (3) (4) A C/2 (series) B C They all have identical capacitance C C C C C A B PHY2049: Chapter 25 C 15 Energy in a Capacitor Capacitors have energy associated with them Grab a charged capacitor with two hands and find out! Calculate energy Continually move charge from – to + surface dU = V dQ Q Q2 1 2 V=Q/C U = Q / C dQ = = CV ∫0 ( ) 2C 2 Q2 1 U= = 2 CV 2 2C So capacitors store and release energy as they acquire and release charge This energy is available to drive circuits PHY2049: Chapter 25 16 Where is the Energy Stored? Answer: Energy is stored in the electric field itself!! Example: Find energy density of two plate capacitor E field is constant ε 0 ( A / d ) ( Ed ) U CV = = = 1 ε0E 2 u= 2 2 Ad Ad 2 Ad 2 2 Energy density depends only on E field! A general result, independent of geometry Can be shown more generally by Maxwell’s equations u = 1 ε0E 2 2 PHY2049: Chapter 25 17 Example: Spherical Charged Conductor Capacitance: C = ε 0 4π R Q2 Q2 = Total energy of spherical conductor: U = 2C 8πε 0 R Calculate directly: integrate energy density over volume u= 1 ε E2 20 U =∫ U = ∫ u dV R dV = 4π r 2dr 2 ⎛Q⎞ Q2 ε0 ⎜ 4π r 2dr = 2 ⎜ 4πε r 2 ⎟ ⎟ 8πε 0 R 0 ⎝ ⎠ ∞1 R ∞ PHY2049: Chapter 25 Checks! 18 Dielectric Materials and Capacitors Insulating material that can be polarized in E field Induced charges +++++++++++++++++++++++++ ------------------------------------------- κ Dielectric material +++++++++++++++++++++++++ ------------------------------------------- Induced charges at dielectric surface partially cancel E field E → E / κ κ > 1 is “dielectric constant” V → V / κ (since V = Ed) C → κC (since C = Q / V) “Good” dielectric requires more than high κ value Good insulator High breakdown voltage Low cost (no charge leakage) (no arcing at high voltage) (affordable) PHY2049: Chapter 25 19 Dielectric Mechanism is Due to Polarization E = 0, Dipoles randomly aligned E applied, partially aligns dipoles Aligned dipoles induce surface charges Surface charges partially cancel E field Yow! http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html PHY2049: Chapter 25 20 ConcepTest Two identical capacitors are given the same charge Q, then disconnected from a battery. After C2 has been charged and disconnected it is filled with a dielectric. Compare the voltages of the two capacitors. Voltage lowered to V/κ (1) V1 > V2 (2) V1 < V2 (3) V1 = V2 C1 PHY2049: Chapter 25 C2 21 ConcepTest When we fill the capacitor with the dielectric, what is the amount of work required to fill the capacitor? (1) W > 0 (2) W < 0 (3) W = 0 Energy lowered to U/κ C1 C2 If U is total energy in capacitor Positive work: One “pushes in” dielectric ΔU > 0 Negative work: Capacitor “sucks in” dielectric ΔU < 0 PHY2049: Chapter 25 22 Capacitors in Circuits Multi-step process 2 + 3 in series C o Combine 2 & 3 ⇒ C/2 Ceq 1 + (2+3) in parallel Combine 1 & (2 & 3) ⇒ 3C/2 2 1C 3C o PHY2049: Chapter 25 23 Example: Find qi and Vi on All Capacitors C1 is charged in position A, then S is thrown to B position Initial voltage across C1: Initial charge on C1: V0 = 12 q10 = 12 x 4 = 48μC After switch is thrown to B: V1 = V23 (parallel branches) q2 and q3 in series: q2 = q3 = q23 (C23 = 2μF) Charge conservation: q10 = q1 + q23 48 = C1V1 + C23V1 (V1 = V23) Find V1: V1 = 48 / (C1 + C23) = 8 V Find q1: q1 = C1V1 = 32μC 12V q23 = q2 = q3 = 48 – 32 = 16μC V2 = q2 / C2 = 2.67 V V3 = q3 / C3 = 5.33 V PHY2049: Chapter 25 A B 6μF 4μF 3μF 24 Another Example Each capacitor has capacitance 10μF. Find the total capacitance Do it in stages 2&3 Add 4 Add 5 Add 1 ⇒ ⇒ ⇒ ⇒ 5 μF 15 μF 6 μF 16 μF 2 1 4 3 5 PHY2049: Chapter 25 25 Find Charges on All Capacitors Each capacitor has capacitance 10μF. V = 10 volts q1 = 10 x 10 = 100μC C2345 = 6μF q2345 = q234 = q5 (series) q2345 = q234 = q5 = 10 x 6 = 60μC V5 = q5 / C5 = 6 Find q4, V4 V234 = V4 = 10 – 6 = 4 q4 = C4 x 4 = 40μC 2 1 4 3 5 Find q2, q3, V2, V3 (C23 = 5μF) q2 = q3 = q23 = C23 x 4 = 5 x 4 = 20μC V2 = 20 / C2 = 2 V3 = 20 / C3 = 2 PHY2049: Chapter 25 26 ...
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