10/21/2004 section 4_4 E-field calculations using Coulombs Law blank.doc 1/1 Jim Stiles The Univ. of Kansas Dept. of EECS 4-3 E-field Calculations using Coulomb’s Law Reading Assignment: pp. 93-98 1. Example: The Uniform, Infinite Line Charge 2. Example: The Uniform Disk of Charge 3. Example: An Infinite Charge Plane10/21/2004 The Uniform Infinite Line Charge.doc 1/5 Jim Stiles The Univ. of Kansas Dept. of EECS The Uniform, Infinite Line Charge Consider an infinite line of charge lying along the z-axis. The charge density along this line is a constant value of ρA C/m. Q: What electric field ()rE is produced by this charge distribution? A: Apply Coulomb’s Law! We know that for a line charge distribution that: ()()30rr-rr4r-rCdρπε′′′=′∫EAA r r′ ρA ()rE r-r′10/21/2004 The Uniform Infinite Line Charge.doc 2/5 Jim Stiles The Univ. of Kansas Dept. of EECS Q: Yikes! How do we evaluate this integral? A: Don’t panic! You know how to evaluate this integral. Let’s break up the process into smaller steps. Step 1: Determine d′A The differential element d′A is just the magnitude of the differential line element we studied in chapter 2 (i.e., dd′′=AA). As a result, we can easily integrate over any of the seven contours we discussed in chapter 2. The contour in this problem is one of those! It is a line parallel to the z-axis, defined as x’ =0 and y’ =0. As a result, we use for d′A: ˆzd a dz dz′′′==A Step 2: Determine the limits of integration This is easy! The line charge is infinite. Therefore, we integrate from z′=−∞ to z′=∞. Step 3: Determine the vector r-r′. Since for all charge x’ = 0 and y’ =0, we find: ()()()()r-r x x xxˆˆˆ ˆ ˆ ˆˆˆˆ ˆˆˆ ˆxyz x y zxyz zxy zayaza ayazaayaza zaaya zza′′′′=++−++′=++−′=++−10/21/2004 The Uniform Infinite Line Charge.doc 3/5 Jim Stiles The Univ. of Kansas Dept. of EECS Step 4: Determine the scalar 3r-r′ Since ()222r-rxy zz′′=++−, we find: ()323222r-rxy zz⎡⎤′′=++−⎣⎦ Step 5: Time to integrate ! ()()()()()()()()()()30322220322220322220322220rr-rr4r-rx14x4x44Cxy zxy zxyzdaya zzadzxy zzaya zzadzxy zzayadzxy zzazzdzxy zzρπερπερπερπερπε∞−∞∞−∞∞−∞∞−∞′′′=′′++−′=⎡⎤′++−⎣⎦′++−′=⎡⎤′++−⎣⎦+′=⎡⎤′++−⎣⎦′′−+⎡⎤′++−⎣⎦∫∫∫∫∫EAAAAAAˆˆ ˆˆˆ ˆˆˆˆ()()220220x204x2xyxyayaxyayaxyρπερπε+=+++=+AAˆˆˆˆ10/21/2004 The Uniform Infinite Line Charge.doc 4/5 Jim Stiles The Univ. of Kansas Dept. of EECS This result, however, is best expressed in cylindrical coordinates: 22 2xcossincos sinˆˆ ˆ ˆˆˆxy x yxyaya a axyaaρφρφρφφρ++=++= And with cylindrical base vectors: ()()()()()() ()()22cos sin1cos sin1cos sin1cos sin1cos sin1-cos sin sin cos1cos 0 sin 0ˆˆˆˆ ˆˆˆˆˆ ˆˆˆˆˆ ˆˆˆˆˆˆˆxyxyxyxz yzzzaaaa aaaaa aaaaa aaaaaaaρρρφφφρφρφφφφρρφφρφφρφφρφφ φ φρφφρρ+=⋅+⋅+⋅+⋅+⋅+⋅=+++++=10/21/2004 The Uniform Infinite Line Charge.doc 5/5 Jim Stiles The Univ. of Kansas Dept. of EECS As a result, we can write the electric field produced by an infinite line charge with constant density ρA as: ()0r2ˆaρρπερ=EA Note what this means. Recall unit vector ˆaρis the direction that points away from the z-axis. In other words, the electric field produced by the uniform line charge points away from the line charge, just like the electric field produced by a point charge likewise points away from the charge. It is apparent that the electric field in the static case appears to diverge from the location of the charge. And, this is exactly what Maxwell’s equations (Gauss’s Law) says will happen ! i.e.,: ()()0rrvρε∇⋅ =E Note the magnitude of the electric field is proportional to 1ρ, therefore the electric field diminishes as we get further from the line charge. Note however, the electric field does not diminish as quickly as that generated by a point charge. Recall in that case, the magnitude of the electric field diminishes as 21r.10/21/2004 The Uniform Disk of Charge.doc 1/5 Jim Stiles The Univ. of Kansas Dept. of EECS The Uniform Disk of Charge Consider a disk radius a, centered at the origin, and lying entirely on the z =0 plane. This disk contains surface charge, with density of sρ C/m2. This density is uniform across the disk. Let’s find the electric field generated by this charge disk! From Coulomb’s Law, we know: ()()30rr-rr4r-rsSdsρπε′′′=′∫∫E rr′sρ()rEr-r′10/21/2004 The Uniform Disk of Charge.doc 2/5 Jim Stiles The Univ. of Kansas Dept. of EECS Step 1: Determine ds’ This disk can be described by the equation z’ = 0. That is, every point on the disk has a cordinate value z’ that is equal to zero. This is one of the surfaces we examined in chapter 2. The differential surface element for that surface, you recall, is: zds ds d dρρφ′′′′== Step 2: Determine the limits of integration . Note over the surface of the disk, ρ′ changes from 0 to radius a, and φ′ changes from 0 to 2π. Therefore: 002aρφπ′′<< << Step 3: Determine vector r-r′. We know that z’ = 0 for all charge, therefore we can write: ()()()()()()r-r x x xxx-xˆˆˆ ˆ ˆ ˆˆˆˆ ˆ ˆˆˆˆxyz x y zxyz x yxyzayaza ayazaayaza ayaayyaza′′′′=++−++′′=++−+′′=+−+ Since the primed coordinates in ds’ are expressed in cylindrical coordinates, we convert the coordinates to get:10/21/2004 The Uniform Disk of Charge.doc 3/5 Jim Stiles The Univ. of Kansas Dept. of EECS ()()()()()()r-r x xxx'xcos sinxyz x yxyzxyzayaza ayaayyazaay azaρφ ρφ′′′=++− +′=− +− +′′ ′′=− +− +ˆˆˆ ˆ ˆˆˆˆˆˆˆ Step 4: Determine 3r-r′ We find that: ()()322232r-r x- cos sinyzρφ ρφ⎡⎤′′′′′=+−+⎣⎦ Step 5: Time to integrate ! ()()()()()()30232222000rr-rr4r-rx- cos sin4x- cos sinsSaxyzsdsay azaddyzπρπερφ ρφρρρφπερφ ρφ′′′=′′′ ′′+− +′′′=⎡⎤′′ ′′+− +⎣⎦∫∫∫∫Eˆˆˆ Yikes! What a mess! To simplify our integration let’s determine the electric field ()rE along the z-axis only. In other words, set x = 0 and y = 0.10/21/2004 The Uniform Disk of Charge.doc 4/5 Jim Stiles The Univ. of Kansas Dept. of EECS