Review Calculate E, V, and F Units: E (c/m 2 ), V (c/m=V), and F (c 2 /m 2 =N). Superposition Principle 1. Point charges Equation of motion F=ma, S=v 0.

Slides:



Advertisements
Similar presentations
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.
Advertisements

EE3321 ELECTROMAGENTIC FIELD THEORY
Charge and Electric Flux. Electric Flux A closed surface around an enclosed charge has an electric flux that is outward on inward through the surface.
Phy 213: General Physics III Chapter 23: Gauss’ Law Lecture Notes.
Physics 2113 Lecture 12: WED 11 FEB Gauss’ Law III Physics 2113 Jonathan Dowling Carl Friedrich Gauss 1777 – 1855 Flux Capacitor (Operational)
22-4 Coulomb’s Law q1 Unit q2 Only apply to point charges r
Chapter 24 Gauss’s Law.
Chapter 23 Summer 1996, Near the University of Arizona Chapter 23 Electric Fields.
Norah Ali Al-moneef King Saud university
E The net electric flux through a closed cylindrical surface is zero.
General Physics 2, Lec 6, By/ T.A. Eleyan
Electric Potential with Integration Potential Difference in a Variable E-field If E varies, we integrate the potential difference for a small displacement.
Apply Gauss’s law 2. Choose Gaussian surface so that E can be taken out of integration: explore the symmetry of E Symmetry of EGaussian Surface Spherical.
Physics 2102 Lecture 9 FIRST MIDTERM REVIEW Physics 2102
Outline. Show that the electric field strength can be calculated from the pd.
a b c Gauss’ Law … made easy To solve the above equation for E, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL. (1)
A b c Gauss' Law.
EXERCISES Try roughly plotting the potential along the axis for some of the pairs Exercises on sheet similar to this.
Physics.
Chapter 22 Gauss’s Law.
Gauss’s Law The electric flux through a closed surface is proportional to the charge enclosed The electric flux through a closed surface is proportional.
Last Lecture Gauss’s law Using Gauss’s law for: spherical symmetry This lecture Using Gauss’s law for: line symmetry plane symmetry Conductors in electric.
Electric fields Gauss’ law
Electricity and Magnetism Review 1: Units 1-6
Gauss’sLaw 1 P05 - The first Maxwell Equation A very useful computational technique This is important!
Chapter 22 Gauss’s Law Chapter 22 opener. Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s.
Previous Lectures: Introduced to Coulomb’s law Learnt the superposition principle Showed how to calculate the electric field resulting from a series of.
Wednesday, Feb. 1, 2012PHYS , Spring 2012 Dr. Jaehoon Yu 1 PHYS 1444 – Section 004 Lecture #5 Wednesday, Feb. 1, 2012 Dr. Jaehoon Yu Chapter 22.
2). Gauss’ Law and Applications Coulomb’s Law: force on charge i due to charge j is F ij is force on i due to presence of j and acts along line of centres.
EMLAB 1 Chapter 3. Gauss’ law, Divergence. EMLAB 2 Displacement flux : Faraday’s Experiment charged sphere (+Q) insulator metal Two concentric.
Application of Gauss’ Law to calculate Electric field:
Copyright © 2009 Pearson Education, Inc. Chapter 22 Gauss’s Law.
Flux Capacitor (Schematic)
Wednesday, Sep. 14, PHYS Dr. Andrew Brandt PHYS 1444 – Section 04 Lecture #5 Chapter 21: E-field examples Chapter 22: Gauss’ Law Examples.
A b c. Choose either or And E constant over surface is just the area of the Gaussian surface over which we are integrating. Gauss’ Law This equation can.
Physics 2102 Gauss’ law Physics 2102 Gabriela González Carl Friedrich Gauss
Conductor, insulator and ground. Force between two point charges:
Concepts 1. Coulomb ’ s law – direction and magnitude of force 2. Superposition – add forces (fields) vectorially Field lines & Flux 3. Gauss ’ law – application.
Electricity and Magnetism
Physics 2113 Lecture 13: WED 23 SEP EXAM I: REVIEW Physics 2113 Jonathan Dowling.
Physics 2113 Lecture 10: WED 16 SEP Gauss’ Law III Physics 2113 Jonathan Dowling Carl Friedrich Gauss 1777 – 1855 Flux Capacitor (Operational)
3/21/20161 ELECTRICITY AND MAGNETISM Phy 220 Chapter2: Gauss’s Law.
Review on Coulomb’s Law and the electric field definition
Electric Forces and Fields AP Physics C. Electrostatic Forces (F) (measured in Newtons) q1q1 q2q2 k = 9 x 10 9 N*m 2 /C 2 This is known as “Coulomb’s.
Medan Listrik Statis I (Hk.Couloumb, Hk. Gauss, Potensial Listrik) Sukiswo
Energy in Electric Field1 Energy in the Electric Field © Frits F.M. de Mul.
Example: use Gauss’ Law to calculate the electric field due to an infinite sheet of charge, with surface charge density . This is easy using Gauss’ Law.
Capacitors: parallel plate, cylindrical, spherical.
Physics 212 Lecture 4 Gauss’ Law.
Gauss’s Law Chapter 24.
Permittivity of free space (or electric constant)
Problem-Solving Guide for Gauss’s Law
E The net electric flux through a closed cylindrical surface is zero.
Physics 2113 Jonathan Dowling Physics 2113 Lecture 13 EXAM I: REVIEW.
ELECTRIC FIELD ELECTRIC FLUX Lectures 3, 4 & 5 a a R 2R
ELECTRIC FIELD ELECTRIC FLUX Lectures 3, 4 & 5 a a R 2R
The Electric Field Chapter 23.
Flux Capacitor (Operational)
3. Basic Principles of Electrostatics
Gauss’s Law Chapter 24.
1/2/2019.
Exercises on sheet similar to this
Task 1 Knowing the components of vector A calculate rotA and divA.
Last Lecture This lecture Gauss’s law Using Gauss’s law for:
Physics 2102 Lecture 05: TUE 02 FEB
Task 1 Knowing the components of vector A calculate rotA and divA.
Physics 2102 Lecture: 07 WED 28 JAN
Electric Fields From Continuous Distributions of Charge
Example 24-2: flux through a cube of a uniform electric field
Applying Gauss’s Law Gauss’s law is useful only when the electric field is constant on a given surface 1. Select Gauss surface In this case a cylindrical.
Presentation transcript:

Review Calculate E, V, and F Units: E (c/m 2 ), V (c/m=V), and F (c 2 /m 2 =N). Superposition Principle 1. Point charges Equation of motion F=ma, S=v 0 t+1/2at 2 Energy Conservation qV=  K (Kinetic energy change)

2. Continuous Charge Distribution g. Check the units. Problem solving strategies: a. Setup a coordinate; Label positions of charge location, point of interests, etc. b. Choose a small element (line segment in 1D, small area in 2D, and small volume in 3D). Avoid choosing special points such as center, end points, midpoint, etc. c. Consider the chosen small element as a point charge and use the formula for point charge to write down dE, dV, or dF, which all depends on charges (dq) carried by the small element. d. Express dq in terms of total charges or density times the small length ( dl), area (  dA) or volume (  dV). e. The total E, V, or F is the sum of all these small element. (integral). f. Properly determine the limits and evaluate the integration.

3. Continuous Charge distribution with symmetries Use Gauss' law. Common geometries are sphere, cylinder, and sheet. Problem solving strategies: a. Choose a point of interest (where you want to calculate E) b. Choose an appropriate Gaussian Surface that contains that point on the surface. 1) E || dA, or E  dA, or combination of two. 2) E has to have the same magnitude on the Gaussian surface. c. Evaluate the integration. Typically you can take E out of integration if you properly choose a Gaussian surface. Consequently, you just need to evaluate the area of the Gaussian surface. Symmetry of EGaussian Surface Spherical (point & spherical charges)Sphere Cylindrical (line charge)Cylinder Planar (sheet of charge)Box or cylinder

4. Calculating Potential Notice that states i and f could be any two points. For examples, i and f could two plates of a capacitor, one sphere to another sphere, etc. Last hint: Don't forget energy conservation principle.