Physics Electricity and Magnetism Lecture 06 - Capacitance. Y&F Chapter 24 Sec. 1-6

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1 Physics - lectricity and Magnetism Lecture 6 - apacitance Y&F hapter 4 Sec. - 6 Overview Definition of apacitance alculating the apacitance Parallel Plate apacitor Spherical and ylindrical apacitors apacitors in Parallel and Series nergy Stored in an lectric Field Atomic Physics View of Dielectrics lectric Dipole in an lectric Field apacitors with a Dielectric Dielectrics and Gauss Law Summary opyright R. Janow Fall 3

2 What apacitance Measures How much charge does an arrangement of conductors hold when a given voltage is applied? The charge needed depends on a geometrical factor called capacitance. xample: Two conducting spheres: Radii R and R R. Different charges and. Spheres touch and come to the same potential, Apply point charge potential formula, V(infinity) R and also 4πε 4πε R R R apacitance of a single isolated sphere: 4πε R xample: A primitive capacitor The right ball s potential is the same as the side of the battery. Similarly for the ball. How much charge flows onto each ball to produce a potential difference of.5 V? The answer depends on the capacitance..5 V _ charges charges _ opyright R. Janow Fall 3.5 V battery

3 Definition of [ ] oulombs or APAITAN : Volt Measures the charge needed per volt of potential difference Does not depend on applied or charge. Always positive. Depends on geometry (and on dielectric materials) only Units: FARAD oulomb / Volt. - Farads are very large µf -6 F. pf pico-farad - F -6 µf µµf xample - apacitance depends on geometry Move the balls at the ends of the wires closer together while still connected to the battery The potential difference cannot change. But: ds s The distance s between the balls decreased so the field had to increase as did the stored energy. harge flowed from the battery to the balls to increase. The two balls now hold more charge for the same potential difference: i.e. the capacitance increased. av _ charges opyright R. Janow Fall 3 increases charges _.5 V battery constant increases

4 apacitors are charge storage devices Two conductors not in electrical contact lectrically neutral before & after being charged enc net urrent can flow from plate to plate if there is a conducting path (complete circuit) apacitors store charge and potential energy - - d - memory bits - radio circuits - power supplies ommon type: parallel plate, sometimes tubular Method for calculating capacitance from geometry: Assume two conducting plates (equipotentials) with equal and opposite charges and Possibly use Gauss Law to find between the plates alculate between plates using a convenient path apacitance / ertain materials ( dielectrics ) can reduce the field between plates by polarizing - capacitance increases Φ q ε V fi enc f ds opyright R. Janow Fall i 3 S da

5 XAMPL: ALULAT for a PARALLL PLAT APAITOR d S path - Find between plates A plate area. Treat plates as infinite sheets σ - σ σ / A uniform surface charge density is uniform between the plates (d << plate size) Use Gaussian surface S (one plate). Flux through ends and attached conductors is zero. Total flux is A enc σa ε φ ε A σ/ ε (infinite i.e., conducting / ε sheet) Find potential difference : hoose V on negative plate (grounded) hoose path from plate to plate, opposite to field fi path ds ( )( )d ε A d Other formulas for other geometries σ ε d A d ε A DPNDS ONLY ON GOMTRY infinity as plate separation d directly proportional to plate area A opyright R. Janow Fall 3

6 harge on a capacitor - no battery attached 6-: Suppose that we charge an ideal parallel plate capacitor to V, then disconnect the battery. What happens to the charge and voltage on it? A. The charge stays on the plates indefinitely, and the voltage stays constant at V. B. The charge leaks out the bottom quickly, and the voltage goes to V.. The charge jumps quickly across the air gap, and the voltage goes to V. D. The charge stays on the plates, but the voltage drops to V.. The charge instantly disappears, but the voltage stays constant at V. opyright R. Janow Fall 3

7 X 4.3: FIND for a SPHRIAL APAITOR concentric spherical, conducting shells, radii a & b harges are q (inner sphere), -q (outer sphere) All charge on the outer sphere is on its inner surface (by Gauss s Law) hoose Gaussian surface S as shown and find field using Gauss s Law: ε da q q εa ε( 4πr ) S As before: q/( 4πε r To find potential difference use outward radial integration path from r a to r b. V b q 4πε V a b r b r a ds a q q 4πε ) q 4πε a b ba 4πε ab b a r b r a dr r q 4πε Negative For b > a ( ) r b a V b < V a Let b infinity. Then a/b and result becomes the earlier formula for the isolated sphere: 4πε ab opyright 4πεR. ajanow Fall 3 b

8 X 4.4: Find for a YLINDRIAL APAITOR concentric, long cylindrical conductors Radii a & b and length L >> b > neglect end effects harges are q (inner) and -q (outer), λ is uniform All charge on the outer conductor is on its inner surface (by Gauss s Law) hoose Gaussian surface S between plates and find field at radius r. is perpendicular to endcaps > zero flux contribution So: ε Φcyl ε da S q q/(πεrl) λ/(πεr) q ε A ε( π rl) To find potential difference use outward radial integration path from r b to r a. V b V a r b r a ds q πε L r b r a dr r q ln(r) πε L b a q ln(b / a) πε L q/ πε L ln(b / as b/a inf inf as b/a a) V b < V a For b > a depends only on geometrical parameters opyright R. Janow Fall 3

9 xamples of apacitance Formulas apacitance for isolated Sphere Parallel Plate apacitor oncentric ylinders apacitor oncentric Spheres apacitor πε 4πε R 4πε ε A d L ln(b / ab b a a) Units: F (Farad) /Nm / Volt ε length - named after Michael Faraday. [note: ε 8.85 pf/m] All of these formulas depend only on geometrical factors opyright R. Janow Fall 3

10 apacitors in circuits IRUIT SYMBOLS: IRUIT DFINITIONS: urrent i rate of dq/dt - charge flow past point circuit Open ircuit: NO closed path. No current. onductors are equi-potentials losed ircuit: There is/are completed paths through which current can flow. Loop Rule: Potential is a conservative field Potential HANG around ANY closed path xample: HARGING A APAITOR - i S - a in the urrent flows when switch is LOSD, completing circuit Battery (MF) maintains ( MF ), and supplies energy by moving free charges from to terminal, internal to battery onvention: i flows from to outside of battery When switch closes, current (charge) flows until across capacitor equals battery voltage. Then current stops as field in wire DFINITION: UIVALNT APAITAN apacitors can be connected in series, parallel, or more complex combinations The equivalent capacitance is the capacitance of a SINGL capacitor that would have the same capacitance as the combination. The equivalent capacitance can replace the original combination in analysis. opyright R. Janow Fall 3

11 Parallel capacitors - quivalent capacitance The actual parallel circuit... i i is the same for each branch and the equivalent circuit: tot tot eq eq The parallel capacitors are just like a single capacitor with larger plates so... tot harges on parallel capacitors add tot 3... ( Parallel capacitances add directly eq i i 3...) (parallel) (parallel) uestion: Why is the same for all elements in parallel? Answer: Potential is conservative field, for ANY closed loop around circuit: i (Kirchoff Loop Rule) opyright R. Janow Fall 3

12 Series capacitors - equivalent capacitance The actual series circuit... The equivalent circuit... 3 tot 3 tot eq i are NOT necessarily the same for each capacitor in series tot i 3... But... charges on series capacitors are all equal - here s why... so i tot / / i / same Reciprocals of series capacitances add / 3... eq / i tot i / eq eq (series) tot Gaussian surface enc neutral..so - For two capacitors in series: eq opyright eq R. Janow Fall 3

13 xample : xample : xample 3: A 33µF and a 47 µf capacitor are connected in parallel Find the equivalent capacitance Solution: 8 para µ Same two capacitors as above, but now in series connection Solution: 33 x 47 ser 9. 4 µ F A pair of capacitors is connected as shown µf, charged initially to V V i µf, uncharged initially F lose switches. Find final potentials across &. Solution: s are in parallel Same potential V f for each Total initial charge: -3 V. tot i i harge is conserved it redistributes on both & -3 eq tot / Vf Vf -6 Final charge on each: 3 x 33 V. -4 opyright R. Janow Fall 3-4 f Vf 3.3 x. f Vf 6.7 x.

14 Three apacitors in Series 6-: The equivalent capacitance for two capacitors in series is: eq Which of the following is the equivalent capacitance formula for three capacitors in series? A. 3 eq D. eq B.. 3 eq 3 eq eq opyright R. Janow Fall Apply formula for eq twice 3 3

15 xample: Reduce circuit to find eq 3 for mixed series-parallel capacitors After Step After Step parallel V V series V Values:. µf, 5.3 µf, µf 3 ( 5.3) 4.5 / ( ) µf 3.57 µf opyright R. Janow Fall 3

16 Series or Parallel? 6-3: In the circuits below, which ones show capacitors and connected in series? 3 A. I, II, III B. I, III. II, IV D. III, IV. None I V 3 II V III 3 IV V V 3 opyright R. Janow Fall 3

17 nergy Stored in a apacitor When charge flows in the sketch, energy stored in the battery is depleted. Where does it go? harge distributions have potential energy. harges that are separated in a neutral body store energy. The electric potential is defined to be V U/q, U qv A small element of charge dq on each plate of a capacitor stores potential energy: du V dq The energy stored by charging a capacitor from charge to is the integral: U du q dq V V q _ V.5 V charges charges _.5 V battery opyright R. Janow Fall 3

18 apacitors Store nergy in the lectrostatic Field The total energy in a parallel plate capacitor is U V d The volume of space filled by the electric field in the capacitor is Ad, so the energy density u is u U vol But for a parallel plate capacitor, V ds d ε A εa V dad V ε V d so u ε nergy is stored in the electric field opyright R. Janow Fall 3

19 What hanges? 6-4: A parallel plate capacitor is connected to a battery of voltage V. If the plate separation is decreased, which of the following increase? A.II, III and IV. B.I, IV, V and VI..I, II and III. D.All except II..All increase. I. apacitance of capacitor II. Voltage across capacitor III. harge on capacitor IV. nergy stored on capacitor q V ε A d V. lectric field magnitude between plates VI. nergy density of field U V u ε V / d opyright R. Janow Fall 3

20 Model for a Molecule that can Polarize A dipole in a uniform external field....feels torque, stores electrostatic potential energy See Lecture 3 p τ qd px torque at θ or θ π torque p at θ /- π/ RSTORING TORU: τ( θ) τ(θ) U p Polarization: An external field aligns dipoles in a material, causing polarization that reduces the field diel - - p pol pol opyright R. Janow Fall 3

21 Polar Material permanent dipole Non-polar Material induced dipole Dielectric materials in capacitors Insulators POLARIZ when an external electric field is applied The NT field inside the material is reduced. MOLULAR VIW NO XTRNAL FILD Dielectrics increase capacitance For a given, more movable charge σ free is needed WITH XTRNAL FILD o Dielectric constant Inside conductors, polarization reduces net to zero vac is field due to free charge Response to vac is the polarization field pol Actual weakened net field inside is diel pol net Κ diel Polarization surface charge density reduces free surface charge density σ net σ free σ pol (σ free σ ext σ vac ) dielectric vacuum pol opyright R. Janow Fall 3

22 Representing Dielectrics ε is the free space permittivity. All materials (water, paper, plastic, air) polarize to some extent and have different permittivities ε κε κ is the dielectric constant - a dimensionless number. Wherever you see ε for a vacuum, you can substitute κε when considering dielectric materials. For example, the capacitance of a parallel plate capacitor increases when the space is filled with a dielectric: diel κεa d κ vac A dielectric weakens the field, compared to what it would be for a vacuum vac/ κ D εκ diel / opyright R. Janow Fall 3

23 opyright R. Janow Fall 3

24 What happens as you insert a dielectric? Initially, charge capacitor to voltage V, charge, field net. With battery detached insert dielectric remains constant, net is reduced Voltage (fixed ) drops to V. Dielectric reduced net and V. V OR With battery attached, insert dielectric. net and V are momentarily reduced but battery maintains voltage harge flows to the capacitor as dielectric is inserted until V and net are back to original values. V V κ ' κv a constant V ds d V a constant ' > opyright R. Janow Fall 3

25 Gauss Law with a dielectric ε Alternatively: ε da S S diel da qfree qpol qnet Κ could vary over Gaussian surface S. Usually it is constant and factors Flux is still measured using field without dielectric: vac Κ diel D/ε ) dφ da Κ da vac ε Κ vac qfree S diel da D da S { free charge on plates field not counting polarization ε vac The lectric Displacement D measures field that would be present due to the free charge only, i.e. without polarization field from dielectric vac Κ diel D ε Only the free charges q free (excluding polarization) are counted as q enc in the above. Using Κ on the left compensates for the polarization. When applying the above include only q free. opyright R. Janow Fall 3 Ignore polarization charges inside the Gaussian surface vac diel ε Κ diel ε OPTIONAL TOPI diel

26 Summary: hapter 5: apacitance Lecture 6 opyright R. Janow Fall 3