apacitance PHY049: hapter 5 1
oulomb s law Electric fields Equilibrium Gauss law What You Know: Electric Fields Electric fields for several charge configurations Point Dipole (along axes) Line Plane (nonconducting) Plane (conducting) Ring (along axis) Disk (along axis) Sphere ylinder PHY049: hapter 5
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 alculating the potential from the field alculating the field from the potential Potential energy from a system of charges PHY049: hapter 5 3
apacitance: Basic Idea apacitance: apacity to store charge Like a tank apacitor is electrically neutral (equal and charge regions) q = V ( is a property of the device, independent of q, V) Units: [] = Farad = oul/volt PHY049: hapter 5 4
alculating Potential Difference Electric field lines start on charges, terminate on V V = E d s Follow E field line during integration (note cosθ = 1) = V V V Eds E, ds positive E d PHY049: hapter 5 5
Parallel Plate apacitor From Gauss law (conducting sheet) So q V = Ed = d Aε 0 ε q= 0 A V V d Therefore A = ε0 = ε0 "length" d Depends only on geometry of device E σ = = ε E q Aε 0 0 d PHY049: hapter 5 6
Example A = 1m, d = 1 μm A 1 6 6 = ε0 = 8.85 10 10 = 8.85 10 F d 8.85μF = Largish, but somewhat typical value PHY049: hapter 5 7
Inner = a, outer = b, length = L Gauss law: Using λ=q/l ylindrical apacitor E = q π Lε 0 0 1 r V = Eds (ds = dr) a q V = Edr = ln b/ a b πlε L = ε 0 π ln b/ a ( ) ( ) b a apacitance per unit length, e.g. coaxial cable (RF frequencies) PHY049: hapter 5 8
Special ase for ylinder Outer shell very close to inner shell: b a = d (d small) Use ln(1x) x (for x small) b a d d ln = ln a a a π L π al A = ε0 ε0 = ε0 ln b/ a d d ( ) Just like parallel plate capacitor: Always true if surfaces are close together = ε surface 0 A d PHY049: hapter 5 9
Inner radius = a, outer radius = b oulomb s law: V V = a b Eds E = Edr = Spherical apacitor = q 4πε 0 (ds = dr) 0 r 1 q b a 4πε ab b a = ε 0 4π ab b a PHY049: hapter 5 10
Two Special ases Isolated sphere: corresponds to b = 4π ab 4π a = ε0 = ε0 ε04πa b a 1 a/ b Outer shell very close to inner shell: b a = d (d small) 4πab 4πa A = ε0 ε0 = ε0 b a d d A = ε0 d Again, just like parallel plate: surface PHY049: hapter 5 11
apacitors in Parallel V 1 = V = V 3 (same potential top and bottom) Total charge: Q tot = Q 1 Q Q 3 eq V= 1 V V 3 V eq = 1 3 Basic law for combining capacitors in parallel Works for N capacitors PHY049: hapter 5 1
apacitors in Series q 1 = q = q 3 (same current charges all capacitors) Total potential: V = V 1 V V 3 q/ eq = q/ 1 q/ q/ 3 1 1 1 1 = eq 1 3 Basic law for combining capacitors in series Works for N capacitors PHY049: hapter 5 13
onceptest Two identical parallel plate capacitors are shown in an endview in Figure A. Each has a capacitance of. If the two are joined together at the edges as in Figure B, forming a single capacitor, what is the final capacitance? (a) / (b) (c) Area is doubled (d) 0 (e) Need more information A B PHY049: hapter 5 14
onceptest Each capacitor is the same in the three configurations. Which configuration has the lowest equivalent capacitance? (1) A () B / (series) (3) (4) They all have identical capacitance A B PHY049: hapter 5 15
Energy in a apacitor apacitors have energy associated with them Grab a charged capacitor with two hands and find out! alculate energy ontinually move charge from to surface du = V dq Q Q V = Q / ( / ) 0 U = Q dq= = V U So capacitors store and release energy as they acquire and release charge This energy is available to drive circuits Q = = 1 V 1 PHY049: hapter 5 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 u ( A/ d)( Ed) Energy density depends only on E field! U V ε = = = = Ad Ad Ad 0 1 ε 0 A general result, independent of geometry an be shown more generally by Maxwell s equations E u = 1 ε 0 E PHY049: hapter 5 17
Example: Spherical harged onductor apacitance: = ε 0 4π R Total energy of spherical conductor: U Q Q = = 8πε 0 R alculate directly: integrate energy density over volume 1 ε 0 4π R u = E U = udv dv = r dr U 1 Q = ε 0 4πr dr = R 4πε0r Q 8πε 0 R hecks! PHY049: hapter 5 18
Dielectric Materials and apacitors 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) κ (since = Q / V) Good dielectric requires more than high κ value Good insulator (no charge leakage) High breakdown voltage (no arcing at high voltage) Low cost (affordable) PHY049: hapter 5 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.phyastr.gsu.edu/hbase/electric/dielec.html PHY049: hapter 5 0
onceptest Two identical capacitors are given the same charge Q, then disconnected from a battery. After has been charged and disconnected it is filled with a dielectric. ompare the voltages of the two capacitors. Voltage lowered to V/κ (1) V 1 > V () V 1 < V (3) V 1 = V 1 PHY049: hapter 5 1
onceptest When we fill the capacitor with the dielectric, what is the amount of work required to fill the capacitor? (1) W > 0 () W < 0 (3) W = 0 Energy lowered to U/κ 1 If U is total energy in capacitor Positive work: One pushes in dielectric ΔU > 0 Negative work: apacitor sucks in dielectric ΔU < 0 PHY049: hapter 5
Multistep process apacitors in ircuits 3 in series o ombine & 3 / 1 (3) in parallel eq 1 3 ombine 1 & ( & 3) 3/ o PHY049: hapter 5 3
Example: Find q i and V i on All apacitors 1 is charged in position A, then S is thrown to B position Initial voltage across 1 : V 0 = 1 Initial charge on 1 : q 10 = 1 x 4 = 48μ After switch is thrown to B: V 1 = V 3 (parallel branches) q and q 3 in series: q = q 3 = q 3 ( 3 = μf) harge conservation: q 10 = q 1 q 3 48 = 1 V 1 3 V 1 (V 1 = V 3 ) Find V 1 : V 1 = 48 / ( 1 3 ) = 8 V Find q 1 : q 1 = 1 V 1 = 3μ A B 6μF q 3 = q = q 3 = 48 3 = 16μ 1V V = q / =.67 V V 3 = q 3 / 3 = 5.33 V 4μF 3μF PHY049: hapter 5 4
Another Example Each capacitor has capacitance 10μF. Find the total capacitance Do it in stages & 3 5 μf Add 4 15 μf Add 5 6 μf Add 1 16 μf 1 3 4 5 PHY049: hapter 5 5
Find harges on All apacitors Each capacitor has capacitance 10μF. V = 10 volts q 1 = 10 x 10 = 100μ 345 = 6μF q 345 = q 34 = q 5 (series) q 345 = q 34 = q 5 = 10 x 6 = 60μ V 5 = q 5 / 5 = 6 1 3 4 Find q 4, V 4 V 34 = V 4 = 10 6 = 4 q 4 = 4 x 4 = 40μ 5 Find q, q 3, V, V 3 ( 3 = 5μF) q = q 3 = q 3 = 3 x 4 = 5 x 4 = 0μ V = 0 / = V 3 = 0 / 3 = PHY049: hapter 5 6