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PHYSICS 51 lecture 24.1. Plan for the day. All about capacitance Calculating capacitance from geometries Plates Spheres (skip) Capacitor at Circuit element Adding capacitors in parallel and series Calculating charge and voltage on a cpacitor Quiz. Homework # 5. Homework #5
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Plan for the day All about capacitance • Calculating capacitance from geometries • Plates • Spheres (skip) • Capacitor at Circuit element • Adding capacitors in parallel and series • Calculating charge and voltage on a cpacitor • Quiz
Homework # 5 • Homework #5 • Ch. 24: #17, #20, #31, #72 • Due next Tuesday, Feb 25
Circuit Elements Capacitors: Ch 24 Store charges/energy in Electric field Block DC currents Pass AC currents Resistors: Ch 25 Pass current (AC and DC) Dissipate energy into heat Electromotive force (battery) Ch 25 Store energy Provide a constant voltage source Inductors: Ch 30 Store energy in magnetic field Block AC currents Pass DC currents
Capacitance and Capacitors • We will calculate capacitance for a few simple geometries • Use definition of capacitance and background with calculating fields and potential • Most useful and important is parallel plates • Add dielectrics inside capacitor to change capacitance • We will also treat a capacitor as a simple circuit element • Stores charge • How to add capacitors • Stored energy in a capacitor
Capacitor • An element that stores charge when a voltage is applied • Q = CV (definition) • C is the capacitance • Unit is farad or C/V or C2/Nm or C2/J • Capacitance is a function of geometry and material
Parallel Plate Capacitor C = 0A/d 0 = 8.85 x 10-12 F/m
Example of Capacitance A capacitor consists of two square, flat conductors each 1.0 centimeters on a side, The separation is 0.10 mm What is the capacitance? If connected to a 12V battery what is the charge on the capacitor? c) What is the Electric Field between the plates? (assume air)
Capacitors A capacitor can be of any two conductors isolated Parallel plates are just a easy way to think of the problem General method to calculate a) Assume a charge on the conductors b) Calculate the voltage difference due to the charge ` c) Ratio Q/V is the capacitance
Capacitor as a circuit element • Typical value of a capacitor • A few mF • Adding capacitors in series and parallel • Calculating charge and voltage on a capacitor
Parallel Capacitors (a) Two capacitors in parallel, (b) the equivalent circuit. Across C1 and C2 Voltage same Charge splits Ceq = C1 + C2
Series Capacitors (a) Two capacitors in series (b) the equivalent capacitor. Across C1 and C2 Charge same Voltage splits 1 / Ceq = 1 / C1 + 1 / C2
Capacitor practice C1 B A C2 C1 = 4 mF, C2 = 2 mF What is total capacitance from A to B
Capacitor practice C3 C2 C1 A B C1 = 4 mF, C2 = 2 mF, C3 = 2 mF What is total capacitance from A to B
Capacitor practice C2 C1 A C3 C4 B C1 = 4 mF, C2 = 4 mF, C3 = 2 mF, C4 = 2 mF What is total capacitance from A to B
Typical Capacitance Problem: 1 • Find the equivalent capacitance • What is the voltage across capacitor C2 C1 = 4.0 mF, C2 = C3 = 1.0mF, VAB = 3.0 V C2 C1 B A C3
Typical Capacitance Problem: 2 Given 9V across the elements what is the charge on C1 C1 = C2 = 1.5 nF, C3 = C4 = 3.0 nF, VAB = 9.0 V C1 C3 C4 B A C3
Energy Storage in Capacitors • When we charge a capacitor we have stored energy in the capacitor • U = Q2/2C • Using C=Q/V can rewrite the above as: • U = CV2/2 or QV/2
Energy Density in Capacitor Here's another way to think of the energy stored in a charged capacitor: If we consider the space between the plates to contain the energy (equal to 1/2 C V2) we can calculate an energy DENSITY (Joules per volume). The volume between the plates is area x plate separation, or A d. Then the energy density u isu = 1/2 C V2 / A d = eo E2 / 2 Recall C = eo A / d and V =E d.
Homework # 5 • Homework #5 • Ch. 24: #14, #17, #22, #30, #66 • Due next Tuesday, Sept 24, 2013
Capacitance Problem: What if numbers are not so convenient 10.0V across the elements, what is the charge on C1 C1 = 1.0 nF, C2 = 4.0 nF, C3 = 2.0 nF, C4 = 7.0 nF, VAB = 10.0 V C1 C3 C4 B A C2
Capacitor multiple choice A string of 4 identical capacitors are connected in parallel to a 10.0 V ideal source. If a fifth cap is added in parallel, the charge on the first cap. 1) Decreases 2) Increases 3) Says the same
Capacitor multiple choice 2 A string of 4 identical capacitors are connected in series to a 10.0 V ideal source. If a fifth cap is added in series, the charge on the first cap. 1) Decreases 2) Increases 3) Says the same
Energy Storage in Capacitors • When we charge a capacitor we have stored energy in the capacitor • U = Q2/2C • Using C=Q/V can rewrite the above as: • U = CV2/2 or QV/2
Energy Example What energy was stored in C1 of previous example ?
Cap energy multiple choice Three identical caps are connected to an ideal battery. Will the system store more energy in the caps are connected in parallel or series? • Parallel • Series
Energy Density Energy density: u = eo E2 / 2 This is an important result because it tells us that empty space contains energy if there is an electric field (E) in the "empty" space. If we can get an electric field to travel (or propagate) we can send or transmit energy and information through empty space!!! C 2007 J. F. Becker
Capacitors • In real capacitors we place a material between the metal plates because: • It keeps the metal plates from shorting • It increases the capacitance • It increase the “breakdown voltage”
Polarization in a metal A neutral sphere B in the electric field of a charged sphere A is attracted to the charged sphere because of polarization.
Polarization in a dielectric material “Polarization” of a dielectric in an electric field E gives rise to thin layers of bound charges on the dielectric’s surfaces, creating surface charge densities +siand –si.
Dielectric in a capacitor The charges induced on the surface of the dielectric reduce the electric field.
Dielectric Constant • If the field in the capacitor is reduced then the voltage (V=Ed) must also be reduced • If for the same charge the voltage is reduced then the capacitance must be increase • C=Q/V (definition of capacitance) • The ratio of the amount the capacitance changes is the definition of the dielectric constant • K = C/C0 (K>1)
Capacitors with a Dielectric • C = KC0 • Capacitance increases • V = V0/K • For a given charge the voltage decreases • E = E0/K • For a given charge the electric field decrease
Capacitors and Breakdown C0 = 0A/d and C = C0K To get high capacitance we want a very small distance between plates Imagine if D was just one monolayer of material (about 10-10m) The electric field between the plates is E = V/d So 10V is a field of 1011 V/m Breakdown (Dielectric strength) is the maximum field that A material can support without a short (spark)
Example of filled Capacitor • A capacitor consists metal foil with an area of 10 m2 • (each side). The foil is separated by a 10 micron film of • Polyester. • What is the capacitance of this element? • What is the maximum voltage it can handle?