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# FWLec4 - A F Peterson Notes on Electromagnetic Fields Waves...

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 9/04 Fields & Waves Note #4 The Electric Field Objectives : Introduce the electric field associated with a point charge and illustrate its computation. Generalize to the electric field of a linear distribution of charge density and a planar distribution of charge density. The force field associated with a distribution of electric charge is one of the fundamental electromagnetic effects. The normalized force field, known as the electric field , is of primary interest. In this Note, the electric field associated with several simple charge distributions is determined. The concept of superposition is used to obtain the electric field due to continuous distributions of charge density. Coulomb’s Law In physics, students encounter Coulomb’s Law F QQ R R = 1 2 0 2 4 pe ˆ (4.1) describing the force between two point charges. This is an experimental result that was independently obtained by Coulomb, Franklin, and other scientists in the late 1700s. In equation (4.1), F denotes the force vector (N), Q denotes a point charge in Coulombs (C), R is the displacement vector pointing from the “source” charge to the “observer” charge, as defined in Note #1, and e 0 is a constant of proportionality known as the permittivity of free space. This parameter has the numerical value e 0 12 2 8 854 10 = ¥ - . C Nm 2 (4.2) We note that the units of e 0 are equivalent to C Nm F m 2 2 = (4.3) where Farads (F) are the standard electrical units of capacitance. Equation (4.2) is often approximated by e p 0 9 1 36 10 @ ¥ F m (4.4) The application of Coulomb’s Law is easily illustrated by example.

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 9/04 Example: A point charge Q 1 = 15 m C is located at ( x =0, y =1, z =2), while a second charge Q 2 = –200 m C is placed at ( x =3, y =2, z =1), as illustrated in Figure 1. Find the force on charge Q 1 . Solution: From the given construction, we can think of charge Q 2 as the “source” of the force field and charge Q 1 as the “observer” where the force is measured. Thus the vector R points from ( x =3, y =2, z =1) to ( x =0, y =1, z =2) and has the form R x y z x y z = - + - + - = - - + ( ) ˆ ( ) ˆ ( ) ˆ ˆ ˆ ˆ 0 3 1 2 2 1 3 (4.5) It follows that R = + + = 9 1 1 11 (4.6) ˆ ˆ ˆ ˆ R R R x y z = = - + - + 3 11 1 11 1 11 (4.7) Then, from (4.1) we obtain F QQ R R R R R = = ¥ - ¥ ¥ Ê Ë Á ˆ ¯ ˜ = - = - - - 1 2 0 2 6 6 9 2 4 15 10 200 10 4 1 36 10 11 27 11 2 45 pe p p ˆ ( )( ) ( ) ˆ ˆ . ˆ (N) (4.8) In words, the force is 2.45 Newtons in the - ˆ R direction. By substituting the explicit value for ˆ R , we could equivalently express the answer as F x y z = + - 2 22 0 74 0 74 . ˆ . ˆ . ˆ (N) (4.9) In this situation, the charges attract each other (like charges repel, unlike charges attract), which is consistent with the force pointing in the - ˆ R direction (toward the “source”).

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 9/04 The Electric Field Coulomb’s Law describes situations where two charges interact. In many situations, it would be more convenient to remove the effect of the “observer” charge, in order to describe the force in an observer-independent manner. This is easily accomplished by normalizing the observer charge to 1.0 (C).
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