Conductors in Electrostatic Equilibrium

Slides:



Advertisements
Similar presentations
Gauss’s Law Electric Flux
Advertisements

Announcements Monday guest lecturer: Dr. Fred Salsbury. Solutions now available online. Will strive to post lecture notes before class. May be different.
Physics 24-Winter 2003-L031 Gausss Law Basic Concepts Electric Flux Gausss Law Applications of Gausss Law Conductors in Equilibrium.
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
The electric flux may not be uniform throughout a particular region of space. We can determine the total electric flux by examining a portion of the electric.
Chapter 22: The Electric Field II: Continuous Charge Distributions
Electric Potential AP Physics Montwood High School R. Casao.
Continuous Charge Distributions
Conductors in Electrostatic Equilibrium
© 2012 Pearson Education, Inc. A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2)
C. less, but not zero. D. zero.
1 Chapter Flux Number of objects passing through a surface.
Karl Friedrich Gauss ( ) – German mathematician Ch 24 – Gauss’s Law.
Chapter 24 Gauss’s Law.
Chapter 23 Gauss’ Law.
Chapter 24 Gauss’s Law.
A Charged, Thin Sheet of Insulating Material
Chapter 24 Gauss’s Law.
Physics for Scientists and Engineers II, Summer Semester 2009 Lecture 3: May 22 nd 2009 Physics for Scientists and Engineers II.
Chapter 23 Gauss’s Law.
Nadiah Alanazi Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions.
1 Fall 2004 Physics 3 Tu-Th Section Claudio Campagnari Lecture 9: 21 Oct Web page:
Steps to Applying Gauss’ Law
Electric Forces and Electric Fields
From Chapter 23 – Coulomb’s Law
Gauss’ Law. Class Objectives Introduce the idea of the Gauss’ law as another method to calculate the electric field. Understand that the previous method.
Summer July Lecture 3 Gauss’s Law Chp. 24 Cartoon - Electric field is analogous to gravitational field Opening Demo - Warm-up problem Physlet /webphysics.davidson.edu/physletprob/webphysics.davidson.edu/physletprob.
Chapter 24 Gauss’s Law.
Electric Field Lines - a “map” of the strength of the electric field. The electric field is force per unit charge, so the field lines are sometimes called.
Gauss’s law : introduction
III.A 3, Gauss’ Law.
Fig 24-CO, p.737 Chapter 24: Gauss’s Law قانون جاوس 1- Electric Flux 2- Gauss’s Law 3-Application of Gauss’s law 4- Conductors in Electrostatic Equilibrium.
Physics Lecture 3 Electrostatics Electric field (cont.)
Physics 213 General Physics Lecture 3. 2 Last Meeting: Electric Field, Conductors Today: Gauss’s Law, Electric Energy and Potential.
Definitions Flux—The rate of flow through an area or volume. It can also be viewed as the product of an area and the vector field across the area Electric.
Chapter 21 Gauss’s Law. Electric Field Lines Electric field lines (convenient for visualizing electric field patterns) – lines pointing in the direction.
Electric Flux and Gauss Law
1 Gauss’s Law For r > a Reading: Chapter Gauss’s Law Chapter 28.
Faculty of Engineering Sciences Department of Basic Science 5/26/20161W3.
Chapter 24 Gauss’s Law. Let’s return to the field lines and consider the flux through a surface. The number of lines per unit area is proportional to.
Lecture 2 The Electric Field. Chapter 15.4  15.9 Outline The Concept of an Electric Field Electric Field Lines Electrostatic Equilibrium Electric Flux.
CHAPTER 24 : GAUSS’S LAW 24.1) ELECTRIC FLUX
1 Lecture 3 Gauss’s Law Ch. 23 Physlet ch9_2_gauss/default.html Topics –Electric Flux –Gauss’
ELECTRICITY PHY1013S GAUSS’S LAW Gregor Leigh
Physics 1161 Lecture 4 Electric Flux and Shielding
Introduction: what do we want to get out of chapter 24?
Q22.1 A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses.
Gauss’ Law Chapter 23 Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Chapter 24 Gauss’s Law. Intro Gauss’s Law is an alternative method for determining electric fields. While it stem’s from Coulomb’s law, Gauss’s law is.
Physics 2102 Gauss’ law Physics 2102 Gabriela González Carl Friedrich Gauss
Review on Coulomb’s Law and the electric field definition Coulomb’s Law: the force between two point charges The electric field is defined as The force.
Chapter 17-3b 1. A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the electric field called electric.
18.8 THE ELECTRIC FIELD INSIDE A CONDUCTOR: SHIELDING In conducting materials electric charges move in response to the forces that electric fields exert.
Fig 24-CO, p.737 Chapter 24: Gauss’s Law قانون جاوس 1- Electric Flux 2- Gauss’s Law 3-Application of Gauss’s law 4- Conductors in Electrostatic Equilibrium.
24.2 Gauss’s Law.
Gauss’ Law Symmetry ALWAYS TRUE!
Gauss’s Law ENROLL NO Basic Concepts Electric Flux
Electric Flux & Gauss Law
Gauss’ Law Symmetry ALWAYS TRUE!
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Reading: Chapter 28 For r > a Gauss’s Law.
Flux and Gauss’s Law Spring 2009.
Chapter 21 Gauss’s Law.
C. less, but not zero. D. zero.
Chapter 23 Gauss’s Law.
Chapter 24 - Summary Gauss’s Law.
Norah Ali Al-moneef King Saud university
Chapter 23 Gauss’s Law.
Gauss’s Law: applications
Chapter 23 Gauss’s Law.
Presentation transcript:

Conductors in Electrostatic Equilibrium AP Physics C Montwood High School R. Casao

A good electrical conductor contains electrons that are not bound to any atom and are free to move within the conductor. If the electric field E is not zero in some area, the electrons there feel a force F = E·q and start to move. The electrons adjust their positions until the force on every electron is zero. When there is no net motion of charges within a conductor, the conductor is in electrostatic equilibrium.

Properties of a conductor in electrostatic equilibrium: The electric field is zero everywhere inside the material. Any excess charge on an isolated conductor must be entirely on its surface. The electric field just outside a charged conductor is perpendicular to the conductor’s surface and has magnitude 2··k· (or s/eo). On an irregularly shaped conductor, charge tends to accumulate at locations where the radius of curvature is the smallest (at sharp points).

Gaussian surface is labeled A Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

Generally draw a Gaussian surface to satisfy two conditions: Condition 1: electric field constant over the surface by symmetry. Condition 2: the dot product of EdA = E·A·cos q because E and A are parallel (q = 0°). Proof that electric field just outside a charged conductor is 2··k· (or s/eo).

Consider a section of the surface small enough to neglect any curvature and take the section to be flat. Imagine a tiny cylindrical Gaussian surface to be embedded in the section. One end is fully inside the conductor, the other end is fully outside the conductor and the cylinder is perpendicular to the conductor’s surface.

E at and just outside the conductor’s surface must be perpendicular to the surface. If not, a component of E would exist in the direction of the conductor’s surface that would exert forces on the surface charges and cause them to move. There is no flux thru the internal end of the cylinder because no electric field lines pass through the inner surface of the Gaussian cylinder because the electric field E inside the conductor is zero.

There is no flux thru the curved surface of the cylinder because E = 0 N/C within the conductor and externally E is parallel to the curved portion of the Gaussian cylinder. The only flux thru the Gaussian cylinder occurs thru the external end of the cylinder where E is perpendicular to A. The charge enclosed (qin) by the Gaussian surface lies on the conductor’s surface in an area A; s = Q/A so qin = s·A.

The magnitude of E just outside a conductor is proportional to the surface charge density at that location on the conductor. If the charge on the conductor is positive, E is directed away from the conductor. If the charge on the conductor is negative, E is directed toward the conductor.

Spherical Gaussian surfaces around (a) positive and (b) negative point charge.

Off-Centered Charge Distribution Consider a –5 mC point charge located closer to one side of an electrically neutral shell.

A charge of +5 mC will lie on the inner wall of the shell; the E inside the shell must be 0 N/C; qin = -5 mC + 5 mC = 0 C. If the point charge were centered inside the shell, the positive charge would be uniformly distributed along the inner wall. Since the point charge is off-center, the distribution of positive charge tends to collect on the section of the wall nearest the -5 mC. Because the shell is electrically neutral, its inner wall can only have a charge of +5 mC if electrons with a total charge of -5 mC leave the inner wall and move to the outer wall. There they spread out uniformly because the shell is spherical and because the distribution of positive charge on the inner wall cannot produce an electric field in the shell to affect the distribution of charge on the outer wall.

Example 24.7 A Sphere Inside a Spherical Shell

Example 24.7 A Sphere Inside a Spherical Shell Consider a solid conducting sphere of radius a surrounded by a conducting spherical shell of inner radius b and outer radius c. The net charge on the solid conducting sphere is +2·Q. The electric field E from the solid conducting sphere radiates outward from the sphere of radius a and would be uniform. The net charge on the conducting spherical shell is –Q. The electric field E from the conducting spherical shell radiates inward and would be uniform.

The entire system is in electrostatic equilibrium. The spherical nature of the objects satisfies condition 1 (electric field E is constant over the surface by symmetry). Region 1: the charge 2·Q will lie on the outer surface of the solid conducting sphere of radius a. Draw a spherical Gaussian surface of radius r < a.

No electric field E can exist within the conductor. The negative charges from the outer conducting spherical shell attracts the positive charge on the solid conducting sphere to the surface. The repulsion of the positive charges on the solid conducting sphere also contributes to the positioning of the positive charge on the surface of the solid conducting sphere. No charge lies within the solid sphere, therefore, qin = 0 C.

Region 2: the charge 2·Q will lie on the outer surface of the solid conducting sphere of radius a. Draw a spherical Gaussian surface of radius r  a.

Charge located inside the spherical Gaussian surface of radius r is 2·Q; therefore, qin = 2·Q Spherical Gaussian surface satisfies condition 1: electric field constant over the surface by symmetry. Spherical Gaussian surface satisfies condition 2: the dot product of EdA = E·A·cos  because E and A are parallel (q = 0°).

Qnet = Qinner surface + Qouter surface Region 3: inside the conducting spherical shell which is in electrostatic equilibrium: The negative charges on this conductor are located on the inner surface of the conducting spherical shell because they are attracted to the positive charge located on the solid conducting sphere. The charge on the inner surface of the spherical shell must be -2·Q to balance the +2·Q charge located on the solid sphere. Therefore, a charge of +Q must reside on the outer surface of the conducting shell, bringing the net charge to –Q; (-2·Q +Q = -Q). Qnet = Qinner surface + Qouter surface

The distribution of negative charge on the inner surface and the positive charge on the outer surface of the shell means that there is no charge located within the spherical shell itself because all the charge is located on the surface. Draw a spherical Gaussian surface of radius b  r < c. The spherical Gaussian surface satisfies condition 1 and condition 2; the electric field E can be considered to be constant over the surface by symmetry and the dot product of EdA = E·A·cos  because E and A are parallel (q = 0°).

Region 4: lies outside the conducting spherical shell. A spherical Gaussian surface can be drawn to enclose the conducting spherical shell and the solid conducting sphere. The radius of the spherical Gaussian surface is r  c. The spherical Gaussian surface satisfies condition 1 and condition 2; the electric field E can be considered to be constant over the surface by symmetry and the dot product of EdA = E·A·cos  because E and A are parallel (q = 0°).