Chapter 21 Electric Potential Topics: Sample question:
Electric potential energy Electric potential Conservation of energy Sample question: Shown is the electric potential measured on the surface of a patient.This potential is caused by electrical signals originating in the beatingheart. Why does the potential have this pattern, and what do thesemeasurements tell us about the hearts condition? Slide 21-1 The Potential Inside a Parallel-Plate Capacitor
Slide 21-25 Reading Quiz The electric potential inside a parallel-plate capacitor
is constant. increases linearly from the negative to the positive plate. decreases linearly from the negative to the positive plate. decreases inversely with distance from the negativeplate. decreases inversely with the square of the distance fromthe negative plate. Answer: B Slide 21-10 Answer The electric potential inside a parallel-plate capacitor
is constant. increases linearly from the negative to the positiveplate. decreases linearly from the negative to the positive plate. decreases inversely with distance from the negativeplate. decreases inversely with the square of the distance fromthe negative plate. Answer: B Slide 21-11 Dielectrics and Capacitors Dielectrics and Capacitors
The molecules in a dielectric tend to become oriented in a way that reduces the external field. This means that the electric field within the dielectric is less than it would be in air, allowing more charge to be stored for the same potential. Dielectric Constant With a dielectric between its plates, the capacitance of a parallel-plate capacitor is increased by a factor of the dielectric constant : Dielectric strength is the maximum field a dielectric can experience without breaking down. Storage of Electric Energy
The energy density, defined as the energy per unit volume, is the same no matter the origin of the electric field: (17-11) The sudden discharge of electric energy can be harmful or fatal. Capacitors can retain their charge indefinitely even when disconnected from a voltage source be careful! Capacitors and Capacitance (Key Equations)
C = |Q| / |Delta V| Property of the conductors and the dielectric Special Case - Parallel Plate Capacitor C = Kappa * Epsilon0*A / d Energy Pee = 1/2 |Q| |Delta V| |Delta V| = Ed Slide 21-16 Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Electricity key concepts (Chs. 20 & 21) - Slide 1
General Concepts - These are always true Electric Force and Field Model Charge Model E-field Definition E-field vectors E-field lines Superposition(note that for forces and fields, we need to work in vector components) Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Electricity key concepts (Chs. 20 & 21) - Slide 2
General Concepts - These are always true Energy, Electric Potential Energy, and Electric Potential Energy Definitions: KE, PEe, Peg, W, Esys, Eth and V Work-Energy Theorem Conservation of Energy Work by Conservative force = -- change of PE Electric Potential Energy and Electric Potential Energy Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Electricity - concepts (Chs 20 & 21)
General Concepts - These are always true Electric Force and Field Model Charge Model E-field Definition E-field vectors E-field lines Superposition Energy, Electric Potential Energy, and Electric Potential Energy Definitions: KE, PEe, Peg, W, Esys, Eth and V Work-Energy Theorem Conservation of Energy Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Electricity - General key concepts (Chs 20 & 21)
Charge Model Electric forces can be attractive or repulsive Objects with the same sign of charge repel each other Objects with the opposite sign of charge attract each other Neutral objects are polarized by charged objects which createsattractive forces between them There are two kinds of charges, positive (protons) and negative (electrons). In solids, electrons are charge carriers (protons are 2000 time moremassive). A charged object has a deficit of electrons (+) or a surplus of electrons (-). Neutral objects have equal numbers of + and charges Fe gets weaker with distance: Fe 1/r2 Fe between charged tapes are > Fe between charged tapes & neutral objects Rubbing causes some objects to be charged by charge separation Charge can be transferred by contact, conduction, and induction Visualization => charge diagrams Slide 21-16 Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Nature of Electric Field Vectors
Test charge is a small positive charge to sample the E-Field Charge of test charge is small compared to source charges (source charges are the charges that generate the field) E-field vectors E-field is the force per charge E = Fe / q E-field vectors points away from + charges E-field vectors point towards - charges E-field for point charges gets weaker as distance from source point charges increases For a point chargeE = Fe / q = [k Q q / r2] / q = k Q / r2 Electric Force Fe = qE Nature of Electric Field Lines
E-Field lines start on + charges and end on -- charges Larger charges will have more field lines going out/coming in Density of Field lines is a measure of field strength the higher the density the stronger the field The E-field vector at a point in space is tangent to the field line at that point.If there is no field line, extrapolate Chapter 21 Key Equations (Physics 151)
Key Energy Equations from Physics 151 Definition of Work Where Work- Energy Theorem (only valid when particle model applies) Work done by a conservative force (Fg, Fs, & Fe) Also work done by conservative forceis path independent Conservation of Energy Equation Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Chapter 21 Key Equations (2)
Key Energy Equations from Physics 152 Electric Potential Energy for 2 point charges (zero potential energy when charges an infinite distance apart) Potential Energy for a uniform infinite plate For one plate, zero potential energy is at infinity For two plates, zero potential energy is at one plate or inbetween the two plates Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Chapter 21 Key Equations (3)
Key Points about Electric Potential Electric Potential is the Electric Potential Energy per Charge Electric Potential increases as you approach positive source charges and decreases as you approach negative source charges (source charges are the charges generating the electric field) A line where V= 0 V is an equipotential line (The electric force does zero work on a test charge that moves on an equipotential line and PEe= 0 J) Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Assembling a square of charges
Slide 21-16 Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Analyzing a square of charges
Energy to Assemble Wme = PEE = PEEf - PEEi (PEEi = 0 J) PEEf = =+ Energy to move(Move 2q from Corner to Center) Wme = PEE = PEEf - PEEi = - Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Problem 21.51 A point charge and a point charge are 15.0 apart on the x-axis. Part A. What is the electric potential at the point on the x-axis where the electric field is zero?Do not consider x = + or - infinity. Part B. What is the magnitude of the electric field at the point between the charges on the x-axis where the electric potential is zero? Slide 21-16 Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Problem 21.65 Two 2.2-diameter disks spaced 1.9 apart form a parallel-plate capacitor. The electric field between the disks is 4.6105V/m. Part A.What is the voltage across the capacitor? Part B.How much charge is on each disk Part C.An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.1107m/s. What was the electron's speed as it left the negative plate? Slide 21-16 Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Problem 21.69 A proton is fired from far away toward the nucleus of an iron atom. Iron is element number 26, and the diameter of the nucleus is 9.0 fm. (1 fm = 1e-15 m.) Assume the nucleus remains at rest. What initial speed does the proton need to just reach the surface of the nucleus? Slide 21-16 Copyright 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Example Problem A parallel-plate capacitor is held at a potential difference of 250 V.A proton is fired toward a small hole in the negative plate with a speed of 3.0 x 105 m/s. What is its speed when it emerges through the hole in the positive plate? (Hint: The electric potential outside of a parallel-plate capacitor is zero). Slide 21-26 Example Problem What is Q2? Slide 21-35 A Conductor in Electrostatic Equilibrium
Slide 21-27 Reading Quiz The electric field
is always perpendicular to an equipotential surface. is always tangent to an equipotential surface. always bisects an equipotential surface. makes an angle to an equipotential surface that dependson the amount of charge. Answer: A Slide 21-12 Answer The electric field
is always perpendicular to an equipotential surface. is always tangent to an equipotential surface. always bisects an equipotential surface. makes an angle to an equipotential surface that dependson the amount of charge. Answer: A Slide 21-13 Graphical Representations of Electric Potential
Slide 21-13 The Potential Inside a Parallel-Plate Capacitor
Slide 21-25 Electric Potential of a Point Charge
Slide 21-27 Reading Quiz The electric potential inside a parallel-plate capacitor
is constant. increases linearly from the negative to the positive plate. decreases linearly from the negative to the positive plate. decreases inversely with distance from the negativeplate. decreases inversely with the square of the distance fromthe negative plate. Answer: B Slide 21-10 Answer The electric potential inside a parallel-plate capacitor
is constant. increases linearly from the negative to the positiveplate. decreases linearly from the negative to the positive plate. decreases inversely with distance from the negativeplate. decreases inversely with the square of the distance fromthe negative plate. Answer: B Slide 21-11