Copyright © 2009 Pearson Education, Inc. Chapter 21 Electric Charge and Electric Field
Copyright © 2009 Pearson Education, Inc Electric Dipoles An electric dipole consists of two charges Q, equal in magnitude and opposite in sign, separated by a distance. The dipole moment, p = Q, points from the negative to the positive charge.
Copyright © 2009 Pearson Education, Inc Electric Dipoles An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque: So how was I able to move the 2x4 using the electric field of the rod? (cf. Chapt. 10)
Copyright © 2009 Pearson Education, Inc Electric Dipoles Recall: The electric field created by a dipole is the sum of the fields created by the two charges; far from the dipole, the field shows a 1/r 3 dependence:
Copyright © 2009 Pearson Education, Inc. Along p axis
Copyright © 2009 Pearson Education, Inc Electric Dipoles Example: The dipole moment of a water molecule is 6.3 x C·m. A sample contains water molecules, with their dipole moments all oriented in the direction of an electric field of 2.5 x 10 5 N/C. How much work is required to rotate the dipoles from this orientation ( = 0 ) to one in which all moments are perpendicular to the field ( = 90 )?
Copyright © 2009 Pearson Education, Inc. Solution:
Copyright © 2009 Pearson Education, Inc. Photocopy machine: drum is charged positively image is focused on drum only black areas stay charged and therefore attract toner particles image is transferred to paper and sealed by heat Photocopy Machines and Computer Printers Use Electrostatics
Copyright © 2009 Pearson Education, Inc Photocopy Machines and Computer Printers Use Electrostatics
Copyright © 2009 Pearson Education, Inc. Laser printer is similar, except a computer controls the laser intensity to form the image on the drum Photocopy Machines and Computer Printers Use Electrostatics
Copyright © 2009 Pearson Education, Inc. Two kinds of electric charge – positive and negative. Charge is conserved. Charge on electron: e = x C. Conductors: electrons free to move. Insulators: nonconductors. Summary of Chapter 21
Copyright © 2009 Pearson Education, Inc. Charge is quantized in units of e. Objects can be charged by conduction or induction. Coulomb’s law: Electric field is force per unit charge: Summary of Chapter 21
Copyright © 2009 Pearson Education, Inc. Electric field of a point charge: Electric field can be represented by electric field lines. Static electric field inside conductor is zero; surface field is perpendicular to surface. Summary of Chapter 21
Copyright © 2009 Pearson Education, Inc. Chapter 22 Gauss’s Law
Copyright © 2009 Pearson Education, Inc. Electric Flux Gauss’s Law Applications of Gauss’s Law Experimental Basis of Gauss’s and Coulomb’s Laws Units of Chapter 22
Copyright © 2009 Pearson Education, Inc. Electric flux: Electric flux through an area is proportional to the total number of field lines crossing the area Electric Flux
Copyright © 2009 Pearson Education, Inc Electric Flux Example 22-1: Electric flux. Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°.
Copyright © 2009 Pearson Education, Inc. Flux through a closed surface: 22-1 Electric Flux
Copyright © 2009 Pearson Education, Inc Gauss’s Law
Copyright © 2009 Pearson Education, Inc Gauss’s Law
Copyright © 2009 Pearson Education, Inc. The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law: This can be used to find the electric field in situations with a high degree of symmetry Gauss’s Law
Copyright © 2009 Pearson Education, Inc Gauss’s Law For a point charge, Therefore, Solving for E gives the result we expect from Coulomb’s law: Spherical symmetry
Copyright © 2009 Pearson Education, Inc Gauss’s Law Using Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives: Looking at the arbitrarily shaped surface A 2, we see that the same flux passes through it as passes through A 1. Therefore, this result should be valid for any closed surface.
Copyright © 2009 Pearson Education, Inc Gauss’s Law Finally, if a gaussian surface encloses several point charges, the superposition principle shows that: Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.
Copyright © 2009 Pearson Education, Inc Gauss’s Law The figure shows five charged clumps of plastic and an electrically neutral coin as well as a Gaussian surface S. What is the flux through S if the charges are: ?
Copyright © 2009 Pearson Education, Inc Gauss’s Law A charge of 1.8 nC is placed at the center of a cube 3 cm on an edge. What is the electric flux through each face?
Copyright © 2009 Pearson Education, Inc Gauss’s Law A charge of 1.8 nC is placed at the center of a cube 3 cm on an edge. What is the electric flux through each face?
Copyright © 2009 Pearson Education, Inc Applications of Gauss’s Law Example 22-3: Spherical conductor. A thin spherical shell of radius r 0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? Spherical symmetry
Copyright © 2009 Pearson Education, Inc Applications of Gauss’s Law Example 22-4: Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r 0. Determine the electric field (a) outside the sphere ( r > r 0 ) and (b) inside the sphere ( r < r 0 ). Spherical symmetry
Copyright © 2009 Pearson Education, Inc Applications of Gauss’s Law Example 22-5: Nonuniformly charged solid sphere. Suppose the charge density of a solid sphere is given by ρ E = α r 2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r 0. (b) Find the electric field as a function of r inside the sphere. Spherical symmetry
Copyright © 2009 Pearson Education, Inc Applications of Gauss’s Law Example 22-6: Long uniform line of charge. A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends. Cylindrical symmetry
Copyright © 2009 Pearson Education, Inc Applications of Gauss’s Law Example 22-7: Infinite plane of charge. Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA ) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane. Planar, Cylindrical, & Mirror Symmetry
Copyright © 2009 Pearson Education, Inc Applications of Gauss’s Law Example 22-8: Electric field near any conducting surface. Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by E = σ / ε 0 where σ is the surface charge density on the conductor’s surface at that point. Approximate Planar & Cylindrical Symmetry
Copyright © 2009 Pearson Education, Inc Applications of Gauss’s Law Conceptual Example 22-9: Conductor with charge inside a cavity. Suppose a conductor carries a net charge + Q and contains a cavity, inside of which resides a point charge + q. What can you say about the charges on the inner and outer surfaces of the conductor?
Copyright © 2009 Pearson Education, Inc Applications of Gauss’s Law Procedure for Gauss’s law problems: 1. Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). 2. Draw the surface. 3. Use the symmetry to find the direction of E. 4. Evaluate the flux by integrating. 5. Calculate the enclosed charge. 6. Solve for the field.
Copyright © 2009 Pearson Education, Inc Experimental Basis of Gauss’s and Coulomb’s Laws In the experiment shown, Gauss’s law predicts that the charge on the ball flows onto the surface of the cylinder when they are in contact. This can be tested by measuring the charge on the ball after it is removed – it should be zero.
Copyright © 2009 Pearson Education, Inc. Electric flux: Gauss’s law: Gauss’s law can be used to calculate the field in situations with a high degree of symmetry. Gauss’s law applies in all situations. Follows from Coulomb’s Law Summary of Chapter 22
Copyright © 2009 Pearson Education, Inc. Questions?