4Divergenceandcurl. D ds = ρdv. S

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1 4Divergenceandcurl Epressing the total charge Q V contained in a volume V as a 3D volume integral of charge density ρ(r), wecanwritegauss s law eamined during the last few lectures in the general form D d = ρdv. V This equation asserts that the flu of displacement D = ɛ o E over any closed surface equals the net electrical charge contained in the enclosed volume V onlythechargesincludedwithinv affect the flu of D over surface, with charges outside surface making no net contribution to the surface integral D d. Gauss s law stated above holds true everywhere in space over all surfaces and their enclosed volumes V,largeandsmall. z (, y, z + z) y (, y + y, z) Application of Gauss s law to a small volume V = y z surrounded by a cubic surface of si faces, leads, in the limit of vanishing, y,and z,tothedifferentialformofgauss slawepressed in terms of a divergence operation to be reviewed net: (, y, z) ( +, y, z) Given a sufficiently small volume V = y z, wecanassume that ρdv ρ y z. V 1

2 Again under the same assumption D d (D 2 D 1 ) y z+(d y 4 D y 3 ) z+(d z 6 D z 5 ) y with reference to displacement vector components like D 2 shown on cubic surfaces depicted in the margin. Gauss s law demands the equality of the two epressions above, namely (after dividing both sides by y z) D 2 D 1 + D y 4 D y 3 y + D z 6 D z 5 z ρ, in the limit of vanishing, y, and z. Inthatlimit,weobtain D + D y y + D z z = ρ, which is known as differential form of Gauss s law. z y Amorecompactwayofwritingthisresultis (, y, z) 3 D = ρ, where the operator (, y, z ), known as del, isappliedonthedisplacementvector D =(D,D y,d z ) 2

3 z following the usual dot product rules, ecept that the product of and D, for instance, is treated as a partial derivative D.Intheleftsideabove D = D + D y y + D z z (divergence of D) is known as divergence of D. Eample 1: Find the divergence of D =ˆ5 +ŷ12 C/m 2 olution: In this case Therefore, divergence of D is D =5, D y =12, and D z =0. D = D + D y y + D z z = (5)+ y (12) + z (0) = = 5 C m 3. Note that the divergence of vector D is a scalar quantity which is the volumetric charge density in space as a consequence of Gauss s law (in differential form)

4 z ρ 1 < 0 ρ 2 > 0 Eample 2: Find the divergence E of electric field vector ˆ ρ 1(+W 1 ) ɛ o, for W 1 <<0 E = ˆ ρ 2( W 2 ) ɛ o, for 0 <<W 2 0, otherwise, from Eample 4, last lecture (see margin figures). olution: In this case E y = E z =0,andthereforethedivergenceofE is E = E = ρ 1(+W 1 ) ɛ o, ρ 1 ɛ o, for W 1 <<0 ρ 2 ( W 2 ) ρ ɛ o, = 2 ɛ o, for 0 <<W 2, 0, 0, otherwise, which provides us with ρ(r)/ɛ o of Eample 4 from last lecture (in accordance with Gauss s law). ρ 1 W 1 2ɛ o W 1 W 1 - E + E 1() E 2 () W 2 W 2 ρ 2 W 2 2ɛ o ummarizing the results so far, Gauss s law can be epressed in integral as well as differential forms given by D d = ρdv D = ρ. V W 1 E () ρ 1W 1 ɛ o W 2 The equivalence of integral and differential forms implies that (after integrating the differential form of the equation on the right 4

5 over volume V on both sides) D d = V D dv which you may recall as the divergence theorem from MATH 241. Divergence thm. Note that according to divergence theorem, we can interpret divergence as flu per unit volume. We can also think of divergence as a special type of a derivative applied to vector functions which produces non-zero scalar results (at each point in space) when the vector function has components which change in the direction they point. Asecondtypeofvectorderivativeknownascurl which we review net complements the divergence in the sense that these two types of vector derivatives collectively contain maimal information about vector fields that they operate on: Given their curl and divergences, vector fields can be uniquely reconstructed in regions V of 3D space provided they are known at the bounding surface of region V,howeverlarge(eveninfinite) and V may be this is known as Helmholtz theorem (proof outlined in Lecture 7). The curl of a vector field E = E(, y, z) is defined, in terms of the del 5

6 operator, like a cross product E (, y, z ) (E,E y,e z )= ˆ ŷ ẑ y z E E y E z (curl of E) = ˆ( E z y E y z ) ŷ( E z E z )+ẑ( E y E y ). Eample 3: Find the curl of the vector field E =ˆ cos y +ŷ1 olution: The curl is E = ˆ ŷ ẑ y cosy 1 0 z = ˆ( y 0 1) ŷ( z 0 cos y)+ẑ( z 1 y = ˆ0 ŷ0+ẑ(0 + sin y) =ẑ sin y which is another vector field. cos y) y The diagram in the margin depicts E = ˆ cos y +ŷ1 as a vector map superposed upon a density plot of E = ẑ sin y = sin y indicating the strength of the curl vector E (light color corresponds large magnitude)

7 It is apparent that curl E is stronger in those regions where E is rapidly varying in directions orthogonal to the direction of E itself. As the above eample demonstrates the curl of a vector field is in general another vector field. The only eception is if the curl is identically 0 at all positions r =(, y, z)! In that case, i.e., if E =0,vectorfieldE is said to be curl-free. IMPORTANT FACT: All static electric fields E, obtained from Coulomb s law, and satisfying Gauss s law D = ρ with static charge densities ρ = ρ(r), are also found to be curl-free without eception. The proof of curl-free nature of static electric fields can be given by first showing that Coulomb field of a static charge is curl-free, and then making use of the superposition principle along with the fact that the curl of a sum must be the sum of curls like differentiation, taking curl is a linear operation. You should try to show that E =0with the Coulomb field of a point charge Q located at the origin. 7

8 The calculation is slightly more complicated than the following eample (although similar in many ways) where we show that the static electric field of an infinite line charge is curl-free. Eample 4: Recall that the static field of a line charge λ distributed on the z-ais is λ E(, y, z) =ˆr 2πɛ o r, L λ z r E r = y φ λ 2πɛ o r where r 2 = 2 + y 2 and ˆr =ˆ cos φ +ŷ sin φ =( r, y r, 0). how that field E satisfies the condition E =0. olution: Clearly, we can epress vector E as E = λ ( y 2πɛ o r 2, r2, 0). ince the components r and y 2 r of the vector are independent of z, thecorresponding curl can be epanded as ˆ ŷ ẑ E = y z E E y E z 2 = λ ˆ ŷ ẑ 2πɛ o y z y = λ ẑ( y 2πɛ o r 2 y r 2). r 2 r 2 0 But, y r 2 y r = y 1 2 r 1 2 y r = y 2 2y =0, 2 r 4 r 4 so E =0as requested. 8