1. 1/25/07184 Lecture 111 PHY 184 Spring 2007 Lecture 11 Title: Capacitors

2. 1/25/07184 Lecture 112AnnouncementsAnnouncements  Homework Set 3 is due Tuesday, Jan. 30 at 8:00 am.  Homework Set 4 opened this morning.  Today we will finish up the electric potential and start with capacitors.

3. 1/25/07184 Lecture 113 Review - Electric Potential, V(x) ► If a charge q moves in an electric field… ► For reference V=0 at infinity… ► E is the gradient of V…

4. 1/25/07184 Lecture 114 Review - Electric Potential (2) ► For a point source Q … ► For many sources … (superposition)

5. 1/25/07184 Lecture 115 Problem-Solving Strategies  Given a charge distribution, calculate the electric field and the electric potential: (i) Use Gauss Law to derive the electric field E: (ii) For the potential (V=0 at infinity) use superposition principle  For an arrangement of charges, remember the superposition principle! This holds for the electric force, the electric field, the electric potential and the electric potential energy. Note:

6. 1/25/07184 Lecture 116 Capacitors

7. 1/25/07184 Lecture 117CapacitorsCapacitors  Capacitors are devices that store energy in an electric field.  Capacitors are used in many every-day applications Heart defibrillators Camera flash units  Capacitors are an essential part of electronics. Capacitors can be micro-sized on computer chips or super-sized for high power circuits such as FM radio transmitters.

8. 1/25/07184 Lecture 118CapacitanceCapacitance  Capacitors come in a variety of sizes and shapes.  Concept: A capacitor consists of two separated conductors, usually called plates, even if these conductors are not simple planes.  We will define a simple geometry and generalize from there.  We will start with a capacitor consisting of two parallel conducting plates, each with area A separated by a distance d.  We assume that these plates are in vacuum (air is very close to a vacuum).

9. 1/25/07184 Lecture 119 Parallel Plate Capacitor  We charge the capacitor by placing a charge +q on the top plate a charge -q on the bottom plate  Because the plates are conductors, the charge will distribute itself evenly over the surface of the conducting plates.  The electric potential, V, is proportional to the amount of charge on the plates. e.g., using a battery More precisely, potential difference V(+)-V(-) = V

10. 1/25/07184 Lecture 1110  The proportionality constant between the charge q and the electric potential difference V is the capacitance C.  We will call the electric potential difference V the “potential” or the “voltage” across the plates.  The capacitance of a device depends on the area of the plates and the distance between the plates, but does not depend on the voltage across the plates or the charge on the plates.  The capacitance of a device tells us how much charge is required to produce a given voltage across the plates. Parallel Plate Capacitor (2)

11. 1/25/07184 Lecture 1111 Clicker Question  What is the NET CHARGE on the charged capacitor? A: +q+(-q)=0 B: |+q|+|-q|=2q C: q D: none of the above

12. 1/25/07184 Lecture 1112 Clicker Question What is the NET CHARGE on the charged capacitor? A: +q+(-q)=0 Charges are added with their signs. However, we refer to the charge of a capacitor as being q (the charge of a capacitor is not the net charge!)

13. 1/25/07184 Lecture 1113 Definition of Capacitance  The definition of capacitance is  The units of capacitance are coulombs per volt.  The unit of capacitance has been given the name farad (abbreviated F) named after British physicist Michael Faraday (1791 - 1867)  A farad is a very large capacitance Typically we deal with  F (10 -6 F), nF (10 -9 F), or pF (10 -12 F)

14. 1/25/07184 Lecture 1114 Charging/Discharging a Capacitor  We can charge a capacitor by connecting the capacitor to a battery or to a DC power supply.  A battery or DC power supply is designed to supply charge at a given voltage.  When we connect a capacitor to a battery, charge flows from the battery until the capacitor is fully charged.  If we then disconnect the battery or power supply, the capacitor will retain its charge and voltage.  A real-life capacitor will leak charge slowly, but here we will assume ideal capacitors that hold their charge and voltage indefinitely.

15. 1/25/07184 Lecture 1115 Charging/Discharging a Capacitor (2) Illustrate the charging processing using a circuit diagram. Lines represent conductors The battery or power supply is represented by The capacitor is represented by the symbol This circuit has a switch (ab) (open) When the switch is between positions a and b, the circuit is open (not connected). (pos a) When the switch is in position a, the battery is connected across the capacitor. Fully charged, q = CV. (pos b) When the switch is in position b, the two plates of the capacitor are connected. Electrons will move around the circuit – a current will flow -- and the capacitor will discharge.

16. 1/25/07184 Lecture 1116 Demo: Big Spark  Energy stored in this particular capacitor: 90 J  This is equivalent to the kinetic energy of a mass of 1 kg moving at a velocity of 13.4 m/s!

17. 1/25/07184 Lecture 1117 Parallel plate capacitor – -- a simple ideal model

18. 1/25/07184 Lecture 1118 Parallel Plate Capacitor  Consider two parallel conducting plates separated by a distance d  This arrangement is called a parallel plate capacitor.  The upper plate has +q and the lower plate has –q.  The electric field between the plates points from the positively charged plate to the negatively charged plate.  We will assume ideal parallel plate capacitors in which the electric field is constant between the plates and zero elsewhere.  Real-life capacitors have fringe field near the edges.

19. 1/25/07184 Lecture 1119 Parallel Plate Capacitor (2) We can calculate the electric field between the plates using Gauss’ Law We take a Gaussian surface shown by the red dashed line Flux (through bottom surface of the top plate only!) = EA Enclosed charge = q

20. 1/25/07184 Lecture 1120 Parallel Plate Capacitor (3)  Now we calculate the electric potential across the plates of the capacitor in terms of the electric field.  We define the electric potential across the capacitor to be V and we carry out the integral in the direction of the blue arrow. V = Ed Note: V = V(+)  V(  )

21. 1/25/07184 Lecture 1121 Remember the definition of capacitance… … so the capacitance of a parallel plate capacitor is Parallel Plate Capacitor (4) Variables: A is the area of each plate d is the distance between the plates Note that the capacitance depends only on the geometrical factors and not on the amount of charge or the voltage across the capacitor.

22. 1/25/07184 Lecture 1122 Demo: Parallel Plate Capacitor  The voltage of a charged capacitor as function of the distance between the plates  For constant charge, the voltage is proportional to the distance between the plates

23. 1/25/07184 Lecture 1123 Example - Capacitance of a Parallel Plate Capacitor  We have a parallel plate capacitor constructed of two parallel plates, each with area 625 cm 2 separated by a distance of 1.00 mm.  What is the capacitance of this parallel plate capacitor? Result: A parallel plate capacitor constructed out of square conducting plates 25 cm x 25 cm separated by 1 mm produces a capacitor with a capacitance of about 0.5 nF. C = 0.553 nF

24. 1/25/07184 Lecture 1124 Example 2 - Capacitance of a Parallel Plate Capacitor  We have a parallel plate capacitor constructed of two parallel plates separated by a distance of 1.00 mm.  What area is required to produce a capacitance of 1.00 F? Result: A parallel plate capacitor constructed out of square conducting plates 10.6 km x 10.6 km (6 miles x 6 miles) separated by 1 mm produces a capacitor with a capacitance of 1 F.

25. 1/25/07184 Lecture 1125 Example - Capacitance, Charge and Electrons …  A storage capacitor on a random access memory (RAM) chip has a capacitance of 55 nF. If the capacitor is charged to 5.3 V, how many excess electrons are on the negative plate? Idea: We can find the number of excess electrons on the negative plate if we know the total charge q on the plate. Then, the number of electrons n=q/e, where e is the electron charge in coulomb. Second idea: The charge q of the plate is related to the voltage V to which the capacitor is charged: q=CV.