Task 1 Knowing the components of vector A calculate rotA and divA.

Slides:



Advertisements
Similar presentations
Electric Flux Density, Gauss’s Law, and Divergence
Advertisements

Continuous Charge Distributions
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.
Fundamentals of Applied Electromagnetics
Chapter 7 – Poisson’s and Laplace Equations
§9-3 Dielectrics Dielectrics:isolator Almost no free charge inside
Physics 2102 Lecture 9 FIRST MIDTERM REVIEW Physics 2102
MAGNETOSTATIC FIELD (STEADY MAGNETIC)
Notes 13 ECE 2317 Applied Electricity and Magnetism Prof. D. Wilton
Lecture 7 Practice Problems
ELECTROSTATICS. Outline Electric Force, Electric fields Electric Flux and Gau  law Electric potential Capacitors and dielectric (Electric storage)
Applied Electricity and Magnetism
Day 4: Electric Field Calculations for Continuous Charge Distributions A Uniform Distribution of Surface charge A Ring of Continuous Charge A Long Line.
Introduction: what do we want to get out of chapter 24?
Wednesday, Sep. 14, PHYS Dr. Andrew Brandt PHYS 1444 – Section 04 Lecture #5 Chapter 21: E-field examples Chapter 22: Gauss’ Law Examples.
Electric Field formulas for several continuous distribution of charge.
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.
3/21/20161 ELECTRICITY AND MAGNETISM Phy 220 Chapter2: Gauss’s Law.
Review on Coulomb’s Law and the electric field definition
EXAMPLES OF SOLUTION OF LAPLACE’s EQUATION NAME: Akshay kiran E.NO.: SUBJECT: EEM GUIDED BY: PROF. SHAILESH SIR.
LINE,SURFACE & VOLUME CHARGES
Chapter 23 Electric Potential & Electric Potential Energy.
Gauss’ Law.
24.2 Gauss’s Law.
Line integral of Electric field: Electric Potential
Applied Electricity and Magnetism
Physics 2102 Lecture 10r: WED04FEB
Line integral of Electric field: Electric Potential
4. Gauss’s law Units: 4.1 Electric flux Uniform electric field
Electric Fields Due to Continuous Charge Distributions
(Gauss's Law and its Applications)
Chapter 23 Electric Potential
Chapter 25 Electric Potential.
Last Time Insulators: Electrons stay close to their own atoms
To be worked at the blackboard in lecture. R
Electric Flux & Gauss Law
Fields & Forces Coulomb’s law Q r q How does q “feel” effect of Q?
Chapter 9 Vector Calculus.
4. Gauss’s law Units: 4.1 Electric flux Uniform electric field
Chapter 23 Electric Potential
Physics 2113 Jonathan Dowling Physics 2113 Lecture 13 EXAM I: REVIEW.
Chapter 23 Electric Potential
Electric flux, Gauss’s law
Electric flux, Gauss’s law
ELECTRIC FIELD ELECTRIC FLUX Lectures 3, 4 & 5 a a R 2R
Chapter 25 Electric Potential.
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Flux Capacitor (Operational)
Electric Fields Electric Flux
That reminds me… must download the test prep HW.
Jan Physics II Get Fuzzy: Jan 20, 2009.
Chapter 23 Electric Potential
Question for the day Can the magnitude of the electric charge be calculated from the strength of the electric field it creates?
Last Lecture This lecture Gauss’s law Using Gauss’s law for:
Electric Flux Density, Gauss’s Law, and Divergence
4. Gauss’s law Units: 4.1 Electric flux Uniform electric field
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
Chapter 23 Gauss’ Law Key contents Electric flux
Electric Flux Density, Gauss’s Law, and Divergence
Chapter 22 Gauss’s Law The Study guide is posted online under the homework section , Your exam is on March 6 Chapter 22 opener. Gauss’s law is an elegant.
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.
Chapter 21, Electric Charge, and electric Field
Chapter 21, Electric Charge, and electric Field
Chapter 1 Test Review.
Fundamentals of Applied Electromagnetics
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.
Electrical Energy and Electric Potential
Presentation transcript:

Task 1 Knowing the components of vector A calculate rotA and divA. Data:

Solution Formula Hence, in our case: Conclusion: Sourced field.

Task 2a Data: a) Check if the field is potential. b) If it is, calculate the potential: using the reference point N(1,1,1).

Hints for b) 1) Use relation between vector field and potential: A=grad , (d/dx=Ax….) 2) Use the formula:

Solution a) Checking : Hence conditions for vector field A to be potential are: Conclusion: vector field A is potential.

Solution b) We are looking for function: satisfying equation: N(1,1,1) is reference point (potential is equall to zero) We are looking for function: satisfying equation: Following conditions are formulated:

STEP 1 Integrating respectively equations 1-3 we obtain: Comparing derivative of this function with condition 2) we obtain:

STEP 2 Comparing derivative of the above function with condition 3) we obtain:

Final STEP

This problem may be solved also by use of formula:

Task 3 Calculate the flux of through S. Use: The flux definition Stoke’s theorem 2 4 x y z

Solution a) Hence the flux: Calculation of vector B Remark:Only z-component of B creates the nonzero flux

Solution b) Hence the flux: 2 4 x y z Stoke’s Theorem 2 4 x y z Hence the flux: Remark:Only tangentional components are integrated.

Task 4 Calculate the flux of through the surface of perpendicular. y Calculate the flux of through the surface of perpendicular. From definition. Gauss theorem. 3 x 2 4 z

Solution a) b) Gauss Theorem

Task 5 Evaluate both sides of the divergence theorem for a vector field: y x within the unit cube centered about origin. z

Solution a) b) The closed-surface integral consists of only two terms that are evaluated at x=-0.5 and x=0.5:

Task 6 Evaluate the flux of vector through the area S shown in figure. Use: The flux definition 2 3 x y z

Task 7 Calculate the electric field intensity and inductance around a point charge. Apply a spherical symmetry of the problem.

Task 8 The bar on dielectric is charged with the charge q. Calculate field intensity and potential at point P.

x h a r P

Task 9 For the capacitor shown in Fig.9_1 calculate tte potential if: a) voltage V between plates is known b) charge q is known +q -q a b x V Fig.9_1

Task 10 The charge q is distributed uniformly on the very thin ring with the radius R. Determine the field intensity E in the point P in the distance a from the ring plate. R dE R dEx a x dE P dq Fig.10_1

Task 11 The charge q is distributed uniformly on the circle plate (radius R). Determine the field intensity E in the point P in the distance a from the plate. R dq dE r b dEx a x dE P dq Fig.11_1