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Physics 2112 Unit 7: Conductors and Capacitance Today’s Concept: Conductors Capacitance Electricity & Magnetism Lecture 7, Slide 1 Comments THERE ARE ONLY THREE THINGS YOU NEED TO KNOW TO DO ALL OF HOMEWORK 1. E = 0 within the material of a conductor: Charges move inside a conductor in order to cancel out the fields that would be there in the absence of the conductor. This principle determines the induced charge densities on the surfaces of conductors. 2. Gauss’ Law: If charge distributions have sufficient symmetry (spherical, cylindrical, planar), then Gauss’ law can be used to determine the electric field everywhere. Qenclosed E d A = 0 U a b = E dl q a b 3. Definition of Potential: Va b CONCEPTS DETERMINE THE CALCULATION ! Electricity & Magnetism Lecture 7, Slide 2 Conductors The Main Points Charges free to move E = 0 in a conductor Surface = Equipotential E at surface perpendicular to surface Electricity & Magnetism Lecture 7, Slide 3 CheckPoint: Two Spherical Conductors 1 Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B. Compare the potential at the surface of conductor A with the potential at the surface of conductor B. A. VA > VB B. VA = VB C. VA < VB Electricity & Magnetism Lecture 7, Slide 4 CheckPoint: Two Spherical Conductors 2 Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B. The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now? A. VA > VB B. VA = VB C. VA < VB Electricity & Magnetism Lecture 7, Slide 5 CheckPoint: Two Spherical Conductors 3 Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B. What happens to the charge on conductor A after it is connected to conductor B by the wire? A. B. C. QA increases QA decreases QA does not change Electricity & Magnetism Lecture 7, Slide 6 Capacitor Electric Circuit Element Same uses as spring in mechanical system • smooth out rough spots • store energy • cause controlled oscillations Unit 7, Slide 7 Capacitor (II) Simplest Example: Parallel plate capacitor d Define capacitance, C, such that: Q CV Q +Q kx | F | E Q V Units of Farad, F = Coulomb/Volt Unit 7, Slide 8 Key Points • +Q and –Q always have same magnitude • Charges don’t more directly from one plate to the other • Charged from the outside Unit 7, Slide 9 Review of Capacitance Example y First determine E field produced by charged conductors: +Q What is ? d E E= Q x o Q = A A = area of plate Second, integrate E to find the potential difference V V = E dy d 0 d d Q V = ( Edy) = E dy = d o A 0 0 As promised, V is proportional to Q ! Q Q C = V Qd / o A C= 0 A d • Method good for all cases • Formula good for parallel plate only Unit 7, Slide 10 Example 7.1 (Capacitor) A flat plate capacitor has a capacitance of C = 10pF and an area of A=1cm2. What is the distance between the plates? Unit 7, Slide 11 CheckPoint Results: Charged Parallel Plates 1 Two parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measured to be V0. An uncharged conducting plate (the green thing in the picture below) is slipped into the space between the plates without touching either one. The charge on the plates is adjusted to a new value Q1 such that the potential difference between the plates remains the same. Compare Q1 and Q0. A. Q1 < Q0 B. Q1 = Q0 C. Q1 > Q0 Electricity & Magnetism Lecture 7, Slide 12 CheckPoint Results: Charged Parallel Plates 1 An uncharged conducting plate (the green thing in the picture below) is slipped into the space between the plates without touching either one. The charge on the plates is adjusted to a new value Q1 such that the potential difference between the plates remains the same. Compare the capacitance of the two configurations in the above problem. A. C1 > C0 B. C1 = C0 C. C1 < C0 Electricity & Magnetism Lecture 7, Slide 13 Example 7.2 (Linear Capacitor) cross-section a a34 a2 a1 metal A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ? metal Conceptual Idea: Plan: • • • • Q C V Find V in terms of some general Q and divide Q out. Put +Q on outer shell and Q on inner shell Cylindrical symmetry: Use Gauss’ Law to calculate E everywhere Integrate E to get V Take ratio Q/V (should get expression only using geometric parameters (ai, L)) Limiting Case: • • L gets bigger, C gets bigger a2 –> a3, C gets bigger Electricity & Magnetism Lecture 7, Slide 14 Example 7.2 (Linear Capacitor) cross-section a a34 a2 a1 metal A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai). What is the capacitance C of this capacitor ? metal Do Limiting Cases Work? • • L gets bigger, C gets bigger a2 –> a3, C gets bigger Unit 7, Slide 15 Energy in Capacitors U is equal to the amount of work took to put all the charge on the two plates: q U = Vdq = C dq Electricity & Magnetism Lecture 7, Slide 16 Example 7.3 (Energy in Capacitor) A 8uF parallel plate capacitor is has a potential different of 120V between its two sides. The distance between the plates is d=1mm. What is the potential stored in the capacitor? What is the energy density of the capacitor? Unit 7, Slide 17