 # Chapter IX Conductors, Capacitors

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1.1 The balance of charges on conductors: In conductors there are charged particles which can be freely move under any small force. Therefore the balance of charges on conductors can be observed under these circumstances:  The electric field equals zero everywhere inside the conductor E = 0 The electric potential is constant inside the conductor V = const  The electric field vector on the surface of conductors direct along the normal of the surface at each point E = En The surface of conductors is equipotential Inside conductors there is no charge. This conclusion can be proved by applying the Gauss’s law for any arbitrary closed surface inside conductor. All the charge...

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## Nội dung Text: Chapter IX Conductors, Capacitors

1. GENERAL PHYSICS II Electromagnetism & Thermal Physics 2/20/2008 1
2. Chapter IX Conductors, Capacitors §1. Charges and electric field on conductors §2. Capacitance of conductors and capacitors §3. Energy storage in capacitors and electric field energy §4. Electric current, resistance and electromotive force 2/20/2008 2
3. §1. Charges and electric field on conductors: 1.1 The balance of charges on conductors: In conductors there are charged particles which can be freely move under any small force. Therefore the balance of charges on conductors can be observed under these circumstances:  The electric field equals zero everywhere inside the conductor E=0 The electric potential is constant inside the conductor V = const  The electric field vector on the surface of conductors direct along the normal of the surface at each point E = En The surface of conductors is equipotential  Inside conductors there is no charge. This conclusion can be proved by applying the Gauss’s law for any arbitrary closed surface inside conductor. All the charge is distributed on the surface of conductors. 2/20/2008 3
4.  Since the distribution of charge on conductors does not depend on the distribution of the matter the distribution of charges is the same for hollow and solid conductors.  The fact that the distribution of charges only on the surface of conductors can be understood as follows: Suppose that we provide the conductor with an amount of charges charges repulse mutually and tend to leave as far as possible each from other. 2/20/2008 4
5. 1.2 The electric field at the surface of a conductor: Consider the red small cylindrical surface σ with the base dS. Applying the Gauss’s law for this closed surface we have The electric field in vacuum near the surface The electric field in dielectric (σis the surface charge density environment near the surface at the considered point on the surface) + Near convexes of the surface: the equipotential surfaces are dense E is large the charge density is large + Near deepenings of the surface: the equipotential surfaces are 2/20/2008 rare E is smaller 5 the charge density is small
6. §2. Capacitance of conductors and capacitors: 2.1 Charging by induction: Negatively charged Metal sphere rod A charge body (rod) can give another body a charge of opposite sign, without losing Isulating any of its own charge. stand a) b) Pictures: wire a) Metal sphere is initially uncharged b) Charged rod brought nearby c) Wire allows piled-up electrons to flow to ground d) Wire is disconnected from sphere c) d) e) Charged rod is removed: Electrons on sphere rearrange themselves. e) 2/20/2008 6
7. 2.2 Capacitors and capacitance:  A capacitor is a device whose purpose is to store electrical energy which can then be released in a controlled manner during a short period of time.  A capacitor consists of 2 spatially separated conductors which can be charged to +Q and -Q respectively.  Definition: The capacitance of the capacitor is the ratio of the charge on one conductor of the capacitor to the potential difference between the conductors: Q C [The unit of capacitance is the V Farad: 1 F = 1C/V] • The capacitance belongs only to the capacitor, independent of the charge and voltage. 2/20/2008 7
8. Example 1: Parallel Plate Capacitor • Calculate the capacitance. We assume + - charge densities on each plate , with potential difference V : Q A C  ++++ V d -----  Need Q:  Need V: from definition:  Use Gauss’ Law to find E (in next slide) 2/20/2008 8
9. Recall the formula for the electric field of two infinite sheets: + - • Field outside the sheets is zero   + - E=0 + - E=0 Gaussian surface encloses zero net charge + - + - • Field inside sheets is not zero: A + - + - • Gaussian surface encloses non-zero net charge + -  Q A   + -   S AEinside E d E  + -  A - 0 + - (Note that here we consider a capacitor in vacuum) + - E 2/20/2008 9
10.  Q E   A 0 0  b Q A  Q V a  dl   d b V E Ed C  0 a A0 V d Remark: • The capacitance of this capacitor depends only on its shape and size • This formula is true for parallel-plate capacitor (shape), and C depends on A, d (size) (for another shape one has other formula). • When the space between the metal plates is filled with a dielectric material, the capacitance increases by a factor k (see the previous chapter) (Recall: k - dielectric constant; ε- permitivity of the dielectric) In order to increase C d (limitatively), and one must increase A, ε . 2/20/2008 10
11. Example 2: Cylindrical Capacitor • Calculate the capacitance: • Assume +Q, -Q on surface of cylinders with r potential difference V. a b • Gaussian surface is cylinder of radius r (a L < r < b) and length L   Q Q • Apply Gauss' Law:   S 2  E d rLE  E  2  Lr 0 0 If we assume that inner cylinder has +Q, then the potential V is positive if we take the zero of potential to be defined at r = b:   a Q 2 0 L   b C   a b Q Q V   dl     E   Edr  dr  ln  V b  ln   b b a 2  0 rL 20 L    a a  2/20/2008 11
12. Summary on Capacitance • A capacitor is an object with two spatially separated conducting surfaces. • The definition of the capacitance of such an object is: Q C  V • The capacitance depends on the geometry : -Q -Q a A r +Q ++++ a d +Q ----- b b L Parallel Plates Cylindrical Spherical A L ab C C C 2/20/2008 d ln( b / a ) b a 12
13. 2.3 Capacitors in Parallel: a a  Q1 Q2 Q VC C V C 1 2 -Q1 -Q2 -Q b b • Find “equivalent” capacitance C in the sense that no measurement at a, b could distinguish the above two situations. • The voltage across the two is the same….  Q1 Q 2 C Parallel Combination: V   Q 2 Q1 2 C1 C 2 C1 Q Q Q Q (C  ) C Equivalent Capacitor: C  1 2  1 1 2 V V C1V 2/20/2008  C12 C C 13
14. 2.4 Capacitors in Series: +Q -Q C2 +Q -Q a b a b C1 +Q -Q C • Find “equivalent” capacitance C in the sense that no measurement at a, b could distinguish the above two situations. • The charge on C1 must be the same as the charge on C 2 since applying a potential difference across ab cannot produce a net charge on the inner plates of C1 and C2 assume there is no net charge on node between C1 and C2 Q RHS: V ab   1 1 1 C   Q Q C C1 C 2 LHS: Vab  1  2   V V C1 C2 2/20/2008 14
15. Examples: Combinations of Capacitors a C3  a b C1 C2 C b • How do we start?? • Recognize C3 is in series with the parallel combination on C1 and C 2. i.e., C 3 ( C1  2 ) C  1 1 1 C   C C3 C1  2 C C1  2  3 C C 2/20/2008 15
16. §3.Energy of a Capacitor and electric field energy: 3.1 Energy of a charged capacitor: • How much energy is stored in a charged capacitor? – Calculate the work provided (usually by a battery) to charge a capacitor to +/- Q: Calculate incremental work dW needed to add charge dq to capacitor at voltage V (there is a trick here!): - +  q dW  (q)   dq V dq    C • The total work W to charge to Q is then given by: Q 1 1 Q2 W    qdq Look at this! C0 2 C Two ways to write W 1 • In terms of the voltage V: W  CV 2 2/20/2008 2 16
17. 3.2 Energy storage in capacitors: Where is the Energy Stored? • Energy is stored in the electric field itself. Think of the energy needed to charge the capacitor as being the energy needed to create the field. • To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor, where -Q -------- -- ---- 1 Q2 1 Q2 U  2 C 2 ( A / d ) ++++++++ +++++++ +Q 0 This is the energy density, u, of the • The electric field is given by: electric field….  Q  E  1  A U  E2 Ad 0 0 0 2 • The energy density u in the field is given by: U U 1 2 J u   0E Units: volume Ad 2 m3 2/20/2008 17
18. Note 1: The expression for the energy density of the electrostatic field 1 u  E 20 2 is general and is not restricted to the special case of the constant field in a parallel plate capacitor. Note 2: For the electric field in a dielectric (In a small volume surrounding the considered point we can consider the electric field as constant, and apply the formulas of the electric field inside paralell-plates capacitor). 2/20/2008 18
19. §4. Electric current, resistance and electromotive force 4.1 The definition of current and current density:  Charges, e.g. free electrons, exists in conductors with a density, ne (ne approx 1029 m-3 )  “Somehow” put that charge in motion:  effective picture -- all charge moves with a velocity, ve  real picture -- a lot of “random motion” of charges with a small average equal to ve We need a quantity which can characterize flows of moving charges. Definition of current: The rate at which charge flows  unit of I : 2/20/2008 19
20. l=vt Definition of current density: In metal wires, the electrons are the carriers Cross section of charge. We have the following equations: area A Volume = A.l (N – number of electrons which pass through the cross-section A during t ). where n is free electron density The formula gives relation between current density, electron density and 2/20/2008 velocity of electrons. 20 CÓ THỂ BẠN MUỐN DOWNLOAD