26-capacitance

26-capacitance - Physics 4B Lecture Notes Chapter 26 -...

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Physics 4B Lecture Notes 26-1 Chapter 26 - Capacitance Problem Set #5 - due: Ch 26 - 2, 3, 5, 7, 9, 15, 22, 26, 29, 61, 63, 64 The ideas of energy storage in E-fields can be carried a step further by understanding the concept of "Capacitance." Lecture Outline 1. The Definition of Capacitance 2. Capacitors in Circuits 3. Energy Storage in Capacitors and Electric Fields 4. Dielectrics in Capacitors 1. The Definition of Capacitance Consider a sphere with a total charge, Q, and a radius, R. From previous problems we know that the potential at the surface is, V = k Q R . Putting more charge on the sphere stores more energy, but the ratio of energy or potential to charge depends only on R, not on Q or V. That is, Q V = R k . It's true for all charged objects that the ratio of potential to voltage depends only on the shape, so this ratio is defined as the capacitance. C Q V The Defintion of Capacitance The units of capacitance are 1 Coulomb 1 Volt 1 Farad 1 F. Example 1: Calculate the capacitance of two equal but oppositely charged plates of area, A, and separation, d. Neglect any edge effects. The potential difference between the plates is, V = - r E d r s . The field between the plates is just the sum of the fields due to the individual plates (see Ch 23 - example 9), r E = r E + + r E - = σ 2 ε o ˆ k + σ 2 ε o ˆ k = σ ε o ˆ k = q ε o A ˆ k . Using d r s = dz ˆ k , the voltage on the capacitor can be written as, V = - q ε o A ˆ k dz ˆ k = - q ε o A dz = - q ε o A z V = qd ε o A . Applying the definition of capacitance, C Q V = q qd ε o A C = ε o A d . Note that the capacitance only depends on the shape. R Q +q d -q z + E - E
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Physics 4B Lecture Notes 26-2 Example 2: Find the capacitance of two concentric cylindrical conductors of radii a and b with a length, l . Show that the result is consistent with example 1. Assume the cylinders have equal and opposite charges, Q. Then the potential difference between them is, V = - r E d r s where r E = 2k λ r ˆ r = 2kQ r l ˆ r from example 7 of chapter 24. Using d r s = dr ˆ r , the voltage on the capacitor can be written as, V = - 2kQ r l ˆ r dr ˆ r = - 2kQ l 1 r dr a b = - 2kQ l ln b a V = 2kQ l ln b a . Using the definition of capacitance,
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26-capacitance - Physics 4B Lecture Notes Chapter 26 -...

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