Presentation on theme: "Chapter 23: Electrostatic Energy and Capacitance"— Presentation transcript:
1Chapter 23: Electrostatic Energy and Capacitance Capacitors and CapacitanceCapacitorAny two conductors separated by an insulator (or a vacuum) form acapacitorIn practice each conductor initially has zero net charge and electronsare transferred from one conductor to the other (charging the conductor)Then two conductors have charge with equalmagnitude and opposite sign, although the netcharge is still zeroWhen a capacitor has or stores charge Q , the conductorwith the higher potential has charge +Q and the other-Q if Q>0
2Capacitors and Capacitance One way to charge a capacitor is to connect these conductors to oppositeterminals of a battery, which gives a fixed potential difference Vab betweenconductors ( a-side for positive charge and b-side for negative charge). Thenonce the charge Q and –Q are established, the battery is disconnected.If the magnitude of the charge Q is doubled, the electric field becomestwice stronger and Vab is twice larger.Then the ratio Q/Vab is still constant and it is called the capacitance C.-QQWhen a capacitor has or stores charge Q , the conductorwith the higher potential has charge +Q and the other-Q if Q>0
3Calculating Capacitance Parallel-plate capacitor in vacuumCharge density:Electric field:Potential diff.:+Q-QCapacitance:The capacitance depends only on thegeometry of the capacitor.It is proportional to the area A.It is inversely proportional to the separation dWhen matter is present between the plates, itsproperties affect the capacitance.+Q-Q
4Calculating Capacitance Units1 F = 1 C2/N m (Note [e0]=C2/N m2)1 mF = 10-6 F, 1 pF = Fe0 = 8.85 x F/mExample 24.1: Size of a 1-F capacitor
5Calculating Capacitance Example 24.2: Properties of a parallel capacitor
6Calculating Capacitance Example 24.3: A spherical capacitorFrom Gauss’s law:-Q---++ra+--++rb++r+-Q--This form is the same as that for a point charge
7Calculating Capacitance Example 24.4: A cylindrical capacitor (length L)Q-QOuter metal braidrrSignal wireline charge density l
9Capacitors in Series and Parallel Capacitors in series (cont’d)acbThe equivalent capacitance Ceq of the series combination is defined asthe capacitance of a single capacitor for which the charge Q is the sameas for the combination, when the potential difference V is the same.
10Capacitors in Series and Parallel Capacitors in parallelabThe parallel combination is equivalent to a single capacitor with thesame total charge Q=Q1+Q2 and potential difference.
11Capacitors in Series and Parallel Capacitor networks
12Capacitors in Series and Parallel Capacitor networks (cont’d)
13Capacitors in Series and Parallel Capacitor networks 2CCCCCAACCCCCBBCCCCCCCAACBBCC
14Energy Storage and Electric-field Energy Work done to charge a capacitorConsider a process to charge a capacitor up to Q with the final potentialdifference V.Let q and v be the charge and potential difference at an intermediate stageduring the charging process.At this stage the work dW required to transfer an additional element ofcharge dq is:The total work needed to increase the capacitor charge q from zero to Q is:
15Energy Storage and Electric-field Energy Potential energy of a charged capacitorDefine the potential energy of an uncharged capacitor to be zero.Then W in the previous slide is equal to the potential energy U ofthe charged capacitorThe total work W required to charge the capacitor is equal to the totalcharge Q multiplied by the average potential difference (1/2)V duringthe charging process
16Energy Storage and Electric-field Energy We can think of the above energy stored in the field in the region betweenthe plates.Define the energy density u to be the energy per unit volumefield volumeThis relation works for any electric field
17Energy Storage and Electric-field Energy Example 24.9: Two ways to calculate energy storedConsider the spherical capacitor in Example 24.3.The energy stored in this capacitor is:The electric field between two conducting sphere:The electric field inside the inner sphere is zero.The electric field outside the inner surface of the outer sphere is zero.
18Energy Storage and Electric-field Energy Example : Stored energy
19DielectricsDielectric materialsExperimentally it is found that when a non-conducting material(dielectrics) between the conducting plates of a capacitor, thecapacitance increases for the same stored charge Q.Define the dielectric constant k (= K in the textbook) as:When the charge is constant,MaterialkMaterialk3-6vacuum1Micaair(1 atm)Mylar3.1Teflon2.1Plexiglas3.40Polyethelene2.25Water80.4
20DielectricsInduced charge and polarizationConsider a two oppositely charged parallel plates with vacuumbetween the plates.Now insert a dielectric material of dielectric constant k.Source of change in the electric field is redistribution of positiveand negative charge within the dielectric material (net charge 0).This redistribution is called a polarization and it produces inducedcharge and field that partially cancels the original electric field.
22DielectricsMolecular model of induced charge (cont’d)
23DielectricsWhy salt dissolvesNormally NaCl is in a rigid crystal structure, maintained by the electrostatic attraction between the Na+ and Cl- ions.Water has a very high dielectric constant (78). This reduces the field between the atoms, hence their attraction to each other. The crystal lattice comes apart and dissolves.
24Dielectrics Gauss’s law: + + - + conductor dielectric + - + Gauss’s law in dielectricsGauss’s law:++-+conductordielectric+-+enclosed freecharge
25Exercises Problem 1 An air capacitor is made by using two flat plates each with area A separated by a distance d.daIf the distance d is halved, how much doesthe capacitance changes?(b) If the area is doubled, how much does the capacitance changes?For a given stored charge Q, to double the amount of energy storedhow much should the distance d be changed?Now a metal slab of thickness a (< d) and of the same area A is insertedbetween the two plates in parallel to the plates as shown in the figure(the slab does not touch the plates).(d) What is the capacitance of this arrangement?(hint:serial connection)(e) Express the capacitance as a multiple of the capacitance C0 whenthe metal slab is not present.
27Problem 2In this problem you try to measure dielectric constant of a material. Firsta parallel-plate capacitor with only air between the plates is charged byconnecting it to a battery. The capacitor is then disconnected from thebattery without any of the charge leaving the plates.Express the capacitance C0 in terms of the potential difference V0between the plates and the charge Q if air is between the plates.(b) Express the dielectric constant k in terms of the capacitance C0 (air gap)and the capacitance C with material of the dielectric constant k).(c) Using the results of (a) and (b), express the ratio of the potentialdifference V/V0 if Q is the same, where V is the potential differencebetween the plates and a dielectric material dielectric constant is kfills the space between them.(d) A voltmeter reads 45.0 V when placed across the capacitor. Whendielectric material is inserted completely filling the space, the voltmeterreads 11.5 V. Find the dielectric constant of this material.(e) What is the voltmeter read if the dielectric is now pulled partway out sothat it fills only one-third of the space between the plates?(Use the formula for the parallel connection of two capacitors.)
28Problem 2 (a) (b) (c) (d) From (c) In the new configuration the equivalent capacitor iswhere C1/3 is the contribution from the part that has the dielectric materialand C0,2/3 is the part that has air gap.because the capacitance is proportional to the area.Using the results from (c)