ENGI 4430 Gauss & Stokes Theorems; Potentials Page 10.01

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1 ENGI 443 Gauss & tokes heorems; Potentials Page.. Gauss Divergence heorem Let be a piecewise-smooth closed surface enclosing a volume in vector field. hen the net flux of F out of is F d F d, N 3 and let F be a where F N is the component of F normal to the surface. But the divergence of F is a flux density, or an outflow per unit volume at a point. Integrating div F over the entire enclosed volume must match the net flux out through the boundary of the volume. Gauss divergence theorem then follows: F d F d Example. (Example 9.8 repeated) Find the total flux of the vector field F z k ˆ through the simple closed surface x y z a b c Use Gauss Divergence heorem: F is differentiable everywhere in F d div F d 3, so Gauss divergence theorem is valid. ˆ zk z div F x y z 4 abc div F d d the volume of the ellipsoid! 3 herefore F d 4abc 3 In fact, the flux of that surface. F z k ˆ through any simple closed surface is just the volume enclosed by

2 ENGI 443 Gauss & tokes heorems; Potentials Page. Example. Archimedes Principle Gauss divergence theorem may be used to derive Archimedes principle for the buoyant force on a body totally immersed in a fluid of constant density (independent of depth). Examine an elementary section of the surface of the immersed body, at a depth z below the surface of the fluid: he pressure at any depth z is the weight of fluid per unit area from the column of fluid above that area. z is the height of the column (note z < ). the mass is Az he volume of the column is Az the weight is Az g and the pressure is the weight per unit area, so pressure = p = gz he normal vector N to is directed outward, but the hydrostatic force on the surface (due to the pressure p) acts inward. he element of hydrostatic force on is pressure area direction g z N ˆ g z N ˆ he element of buoyant force on of ˆk (vertically upwards): is the component of the hydrostatic force in the direction g z Nˆ k ˆ Define F gzk ˆ and d N ˆ d. umming over all such elements, the total buoyant force on the immersed object is

3 ENGI 443 Gauss & tokes heorems; Potentials Page.3 Example. Archimedes Principle (continued) g z ˆ k d g z kˆ Nˆ d F d F d (by the Gauss Divergence heorem) x But g z ˆ k g z g z g y z z gz z provided that z g, which is a reasonable assumption for incompressible fluids over depths shallow enough for the acceleration g due to gravity to be nearly constant. total buoyant force g d g = weight of fluid displaced by the immersed body herefore the total buoyant force on an object fully immersed in a fluid equals the weight of the fluid displaced by the immersed object (Archimedes principle).

4 ENGI 443 Gauss & tokes heorems; Potentials Page.4 Gauss Law A point charge q at the origin O generates an electric field q q E r r ˆ 3 4 r 4 r If is a smooth simple closed surface not enclosing the charge, then the total flux through is E d E d (Gauss divergence theorem) But Example 6.4 showed that r r 3. r E d. herefore here is no net outflow of electric flux through any closed surface not enclosing the source of the electrostatic field. If does enclose the charge, then one cannot use Gauss divergence theorem, because E is undefined at the origin. Remedy: Construct a surface identical to except for a small hole cut where a narrow tube connects it to another surface, a sphere of radius a centre O and entirely inside. Let * (which is a simple closed surface), then O is outside *! Applying Gauss divergence theorem to *, E d * * E d E d E d E d

5 ENGI 443 Gauss & tokes heorems; Potentials Page.5 Gauss Law (continued) As the tube approaches zero thickness, EN on opposite walls of the tube tend to become equal in magnitude and opposite in direction, cancelling each other perfectly in the limit of zero thickness all over the tube. E d and therefore E d E d But is a sphere, centre O, radius a. Using parameters, on the sphere, r sin cos a sin sin cos r r Finding, as before leads to Nasin r. But the outward normal to actually points towards O. N a sin r on the sphere q and E r everywhere on 3. 4 a Also r arˆ r r a q sin sin 3 sin q q E N r a r r r 4 a 4 a 4 Recall that d Nd N N d d N d d q E d E N d d sin d d 4

6 ENGI 443 Gauss & tokes heorems; Potentials Page.6 Gauss Law (continued) q cos q q 4 4 q E d E d (which is independent of the radius of the sphere ) a O, But, as he surface looks more and more like the surface as the tube collapses to a line and the sphere collapses into a point at the origin. Gauss law then follows. Gauss law for the net flux through any smooth simple closed surface, in the presence of a point charge q at the origin, then follows: E d q if encloses otherwise O

7 ENGI 443 Gauss & tokes heorems; Potentials Page.7 Example.3 Poisson s Equation he exact location of the enclosed charge is immaterial, provided it is somewhere inside the volume enclosed by the surface. he charge therefore does not need to be a concentrated point charge, but can be spread out within the enclosed volume. Let the charge density be (x, y, z), then the total charge enclosed by is q d Gauss law E d q Apply Gauss divergence theorem to the left hand side, substitute for q on the right hand side and assume that the permittivity is constant throughout the volume: E d d E d his identity will hold for all volumes only if the integrand is zero everywhere. Poisson s equation then follows: E E and his reduces to Laplace s equation when.

8 ENGI 443 Gauss & tokes heorems; Potentials Page.8 tokes heorem Let F be a vector field acting parallel to the xy-plane. Represent its Cartesian components by F f ˆ ˆ i f j f f. hen ˆ i x f ˆ ˆ f f F j f F k y x y ˆ k z Green s theorem can then be expressed in the form F dr F kˆ da C D Now let us twist the simple closed curve C and its enclosed surface out of the xy-plane, so that the unit normal vector ˆk is replaced by a more general normal vector N. If the surface (that is bounded in 3 by the simple closed curve C) can be represented by z f x, y, then a normal vector at any point on is z z N x y Coherent orientation: C is oriented coherently with respect to if, as one travels along C with N pointing from one s feet to one s head, is always on one s left side. he resulting generalization of Green s theorem is tokes theorem: F dr F N d curl F d C his can be extended further, to a non-flat surface with a non-constant normal vector N.

9 ENGI 443 Gauss & tokes heorems; Potentials Page.9 Example.4 Find the circulation of xy F xyz xz e around C : the unit square in the xz-plane. Because of the right-hand rule, the positive orientation around the square is OGHJ (the y axis is directed into the page). In the xz plane y = F xz Computing the line integral around the four sides of the square: d OG : r r t dt t and d r F F dt G F dr dt t O In a similar way (in Problem et 9 Question 5), it can be shown that H J O F dr, F dr and F dr G H J F dr C OR use tokes theorem:

10 ENGI 443 Gauss & tokes heorems; Potentials Page. Example.4 (continued) Note that we must find F first before setting y = : F y F y ˆ i x yz x xy x e x ˆ xy F j xz x y y e y z xz ˆ xy k e z On D F x x z xz z x da ˆ j da F da z x da da F da F dr D C Note that this vector field F is not conservative, because F. Domain A region of 3 is a domain if and only if ) For all points P o in, there exists a sphere, centre P o, all of whose interior points are inside ; and ) For all points P o and P in, there exists a piecewise smooth curve C, entirely in, from o P to P. A domain is simply connected if it has no holes.

11 ENGI 443 Gauss & tokes heorems; Potentials Page. Example.5 Are these regions simply-connected domains? he interior of a sphere. he interior of a torus. he first octant. YE NO YE On a simply-connected domain the following statements are either all true or all false: F is conservative. F F P P C F dr - independent of the path between the two points. end F dr C C start Example.6 Find a potential function x, y, z for the vector field x y z F. First, check that a potential function exists at all: ˆ i x x curl ˆ F F j y y ˆ k z z herefore F is conservative on 3 and a potential function exists.

12 ENGI 443 Gauss & tokes heorems; Potentials Page. Example.6 (continued) F x y z x x x g y, z g y g y, z y h z y y x y h z dh z hz z c z dz x y z c We have a free choice for the value of the arbitrary constant c. Choose c =, then x, y, z x y z r Often we can spot the potential function, then check that it has the correct partial derivatives.

13 ENGI 443 Gauss & tokes heorems; Potentials Page.3 Example.7 y y Find a potential function for F e ˆi xe z ˆj yz k ˆ that has the value at the origin. We seek a function x, y, z such that y y e, xe z and yz. x y z y Immediately we can spot that xe z y Checking this candidate solution, we find y y xe z y e x and y xe z y yz z suggests xe z y. y y,, x y z xe z y c, where c is an arbitrary constant.,, c c x, y, z xe z y herefore y

14 ENGI 443 Gauss & tokes heorems; Potentials Page.4 Maxwell s Equations (not examinable in this course) We have seen how Gauss and tokes theorems have led to Poisson s equation, relating the electric intensity vector E to the electric charge density : E Where the permittivity is constant, the corresponding equation for the electric flux density D is one of Maxwell s equations: D. Another of Maxwell s equations follows from the absence of isolated magnetic charges (no magnetic monopoles): H B, where H is the magnetic intensity and B is the magnetic flux density. Faraday s law, connecting electric intensity with the rate of change of magnetic flux density, is E dr t B d. Applying tokes theorem to the left side produces C E B t Ampère s circuital law, I H dl, leads to H J Jd, where C the current density is J E v, is the conductivity, is the volume charge D density; and the displacement charge density is J d t he fourth Maxwell equation is H D J t he four Maxwell s equations together allow the derivation of the equations of propagating electromagnetic waves. [End of Chapter ]