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Introduction to Electromagnetism - 7

University of Ontario Institute of Technology (UOIT)
Uploaded: 7 years ago
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Category: Electromagnetic Theory
Type: Lecture Notes
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Filename:   Introduction to Electromagnetism - 7.ppt (1.01 MB)
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Physics for Bioscience (II)
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Announcements Assignment deadline has been moved to February 9 Today: More on conductors in electrostatic equilibrium Applications using Gauss’s law Potential Energy Quiz Properties of a Conductor in Electrostatic Equilibrium The electric field is zero everywhere inside the conductor If an isolated conductor carries a charge, the charge resides on its surface The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of ?/?o On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest Property 1: Einside = 0 Consider a conducting slab in an external field E If the field inside the conductor were not zero, free electrons in the conductor would experience an electrical force These electrons would accelerate These electrons would not be in equilibrium Therefore, there cannot be a field inside the conductor Property 2: Charge Resides on the Surface Choose a gaussian surface inside but close to the actual surface The electric field inside is zero (prop. 1) There is no net flux through the gaussian surface Because the gaussian surface can be as close to the actual surface as desired, there can be no charge inside the surface Property 3: Field’s Magnitude and Direction Choose a cylinder as the gaussian surface The field must be perpendicular to the surface If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be in equilibrium Conductors in Equilibrium, example The field lines are perpendicular to both conductors There are no field lines inside the cylinder Electric Potential Instructor Franco Gaspari PHY 1040U (Physics for the biosciences) Introduction to Electromagnetism and Optics Lecture 7 January 30, 2007 Infinite Charged Conducting Plate + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Surface 1 Surface 2 a b c We have seen before a conductor in a pre-existing electric field. Let us consider now a charged (+) conductor. When a metal plate is given a net charge, the charge distributes itself over the entire outer surface of the plate. If the plate is of uniform thickness and infinitely large, the charge per unit area will be uniform and the same on both surfaces. We treat the problem as two sheets of charge. By symmetry, E is perpendicular to the plate and uniform. Let us consider 3 points, on the left (a), inside the conductor (b) and on the right (c) Shell Theorems A shell of uniform charge at the surface attracts or repels a charged particle that is outside the shell as if all the charge of the shell were concentrated at the centre of the shell. 2. A shell of uniform charge exerts no electrostatic force on a charged particle that is located inside. Apply Gauss’s Law That is, for r>a, the electric field at r is the same as if all the charge q were at the center. A solid conducting sphere of radius a and carrying a charge 2Q surrounded by a conducting spherical shell carrying a charge Q. Key point: the electric field inside a conductor is zero! Using Gauss’s Law find the charge at the 3 surfaces Qa=2Q Qb=-2Q Qc+Qb=Q ? Qc=+3Q Gaussian surface within the conducting shell 42. A solid copper sphere of radius 15.0 cm carries a charge of 40.0 nC. Find the electric field (a) 12.0 cm, (b) 17.0 cm, and (c) 75.0 cm from the center of the sphere. (d) What If? How would your answers change if the sphere were hollow? 51. A hollow conducting sphere is surrounded by a larger concentric spherical conducting shell. The inner sphere has charge –Q , and the outer shell has net charge +3Q. The charges are in electrostatic equilibrium. Using Gauss’s law, find the charges and the electric fields everywhere. 55. A solid insulating sphere of radius a carries a net positive charge 3Q , uniformly distributed throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c, and having a net charge –Q , as shown in Figure P24.55. (a) Construct a spherical gaussian surface of radius r > c and find the net charge enclosed by this surface. (b) What is the direction of the electric field at r > c? (c) Find the electric field at r > c. (d) Find the electric field in the region with radius r where c > r > b. (e) Construct a spherical gaussian surface of radius r, where c > r > b, and find the net charge enclosed by this surface. (f) Construct a spherical gaussian surface of radius r, where b > r > a, and find the net charge enclosed by this surface. (g) Find the electric field in the region b > r > a. (h) Construct a spherical gaussian surface of radius r < a, and find an expression for the net charge enclosed by this surface, as a function of r. Note that the charge inside this surface is less than 3Q. (i) Find the electric field in the region r < a. ( j) Determine the charge on the inner surface of the conducting shell. (k) Determine the charge on the outer surface of the conducting shell. (l) Make a plot of the magnitude of the electric field versus r. Using Coulomb’s Law or Gauss’s Law (in case of symmetric charge distribution) allows us to obtain E The similarities between Newton’s Law and Coulomb’s Law suggest that it is possible to associate a Potential Energy (scalar) to the electrostatic field. This would allow us to examine the movement of charges within the electrostatic field in terms of work by the field and work against the field. Review of some fundamentals of mechanics A B If a body goes from B to A under the influence of a conservative force, the change in Potential Energy is: A force is a conservative force if the work done by the force is independent of the path of the particle. We define (arbitrarily) a condition of zero potential, that is a point R where UR = 0.Therefore we can define a property at any point P, called potential energy, as: UB-UR = UB-0 = UB Conservative forces: Gravity, Coulomb Dissipative forces: friction, etc. Work: s is the length of the path of the body subject to the force. A B It can be proven that the work done along path 1 and 2 is the same. 1 2 It follows that if the object traveled from B to A along 1 and back to B along 2 When we deal with a conservative force, i.e., when the only relevant parameters are the initial and final position, we can define a potential energy function U, as the amount of energy decrease after a work done by a force F.

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