Electromagnetic Field Theory: A Problem Solving Approach Part 12. Electromagnetic field theory is often the least popular course in the electrical engineering curriculum. Heavy reliance on vector and integral calculus can obscure physical phenomena so that the student becomes bogged down in the mathematics and loses sight of the applications. This book instills problem solving confidence by teaching through the use of a large number of worked problems. To keep the subject exciting, many of these problems are based on physical processes, devices, and models. This text is an introductory treatment on the junior level for a two-semester electrical engineering. | The Electric Potential 85 The minus sign in front of the integral is necessary because the quantity W represents the work we must exert on the test charge in opposition to the coulombic force between charges. The dot product in 1 tells us that it takes no work to move the test charge perpendicular to the electric field which in this case is along spheres of constant radius. Such surfaces are called equipotential surfaces. Nonzero work is necessary to move q to a different radius for which dl drir. Then the work of 1 depends only on the starting and ending positions ra and rb of the path and not on the shape of the path itself 4ir oJra r Ne can convince ourselves that the sign is correct by examining the case when rb is bigger than ra and the charges q and qt are of opposite sign and so attract each other. To separate the charges further requires us to do work on qt so that W is positive in 2 . If q and qt are the same sign the repulsive coulomb force would tend to separate the charges further and perform work on qt. For force equilibrium we would have to exert a force opposite to the direction of motion so that W is negative. If the path is closed so that we begin and end at the same point with ra rb the net work required for the motion is zero. If the charges are of the opposite sign it requires positive work to separate them but on the return equal but opposite work is performed on us as the charges attract each other. If there was a distribution of charges with net field E the work in moving the test charge against the total field E is just the sum of the works necessary to move the test charge against the field from each charge alone. Over a closed path this work remains zero VT -itE-cll O E-dl 0 3 which requires that the line integral of the electric field around the closed path also be zero. 2-5-2 The Electric Field and Stokes Theorem Using Stokes theorem of Section we can convert the line integral of the electric field to a surface integral of the 86 .
