PARADIGMS IN PHYSICS STATIC VECTOR FIELDS
PARADIGMS IN PHYSICS STATIC VECTOR FIELDS PH 422
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This 11 page Class Notes was uploaded by Mylene Russel on Monday October 19, 2015. The Class Notes belongs to PH 422 at Oregon State University taught by Staff in Fall. Since its upload, it has received 7 views. For similar materials see /class/224539/ph-422-oregon-state-university in Physics 2 at Oregon State University.
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Date Created: 10/19/15
PH 422 Day 2 1 The Electric eld of a uniformly charged plane Recall that a the electric eld of a uniform disk is given along the axis by Elt 27m 2 z z 2 7 i xzi2 22 R2 where of course Z2 i1 depending on the sign of 2 In the limit as R a 00 one gets the electric eld of a uniformly charged plane which is just 47TEO 02 sgnz 2760 which is valid everywhere as any point can be thought of as being on the axis But the calculation leading to the rst expression above was somewhat involved ls there a better way An in nite plane is highly symmetric Not only is every point like every other at any point all directions in the plane are equivalent This rotational symmetry means that the electric eld must be orthogonal to the plane 7 otherwise you could face a different direction repeat the computation and get a different answer There is a subtlety here in that the electric eld must in fact point in opposite directions on opposite sides of the plane A similar argument using the translational symmetry shows that the electric eld can only depend on the distance from the plane The electric eld must therefore be of the form EE2 assuming that the z direction is orthogonal to the plane Now recall Gauss7 Law which relates the ux of the electric eld through any closed surface to the charge enclosed by the surface that is Ed32 60 Choose a closed surface which uses the symmetry A rectangular box is one possibility two of whose faces are parallel to the plane and equidistant from it The ux through the sides of this box is zero since the normal vector to these sides is parallel to the plane but E is perpendicular to the plane so that E dA 0 What about the top The electric eld is perpendicular to the top7 but constant in magnitude Thus7 E dA E2A top where z is the distance from the plane to the top of the box where z gt 0 By syrnrnetry7 the ux through the bottom of the box is the same as that though the top Point out that this implies that E7z Finally7 the charge enclosed by the box is just the charge density 039 times the area of the part of the plane inside the box7 which is again A lnserting all of this into Gauss7 Law7 we obtain A 2E2A L 60 so that 039 E 7 Z 260 which turns out to be independent of z for z gt 0 PH 422 Day 3 1 Electric Field Lines You probably already know the following facts about electric eld lines o The density of eld lines is proportional to the strength of the electric eld there 0 Field lines only start at positive charges and end at negative charges 0 Field lines never cross These rules can all be explained using Gauss7 Law since the ux of the electric eld can be interpreted as counting the number of eld lines which cross the surface 2 Divergence Consider a small closed box with sides parallel to the coordinate planes What is the ux of E out of the box Consider rst the vertical contribution namely the ux up through the top plus the ux down through the bottom These two sides each have area elernent dA d2 dy but the outward norrnal vectors point in opposite directions so we get 2E d2 E2d2 1ddy drdy topbottom E42 d2 7 d2 dy E42 12 7 E42 12 BE 82 d2 dy d2 d2 dy d2 Repeating this argument using the remaining pairs of faces it follows that the total ux out of the box is total uxZEdA d x y 2 box Since this is proportional to the volume of the box7 it approaches zero as the box shrinks down to a point The interesting quantity is therefore the ratio of the ux to volume This is the divergence At any point P7 we therefore de ne the divergence of a vector eld E written to be the ux of E per unit volume leaving a small box around P In other words7 the divergence is the limit as the box collapses around P of the ratio of the ux of the electric eld out of the box to the volume of the box Thus7 the divergence of E at P is the ux per unit volume through a small box around P7 Which is given in rectangular coordinates by a ux BEE aEy BEZ V39Eunitvolume 8m 8 y 82 You may have seen this formula before7 but remember that it is merely the rectangular coordinate expression for the divergence of E the divergence is de ned as ux per unit volume Similar computations can be used to determine expressions for the divergence in other coordinate systems PH 422 Day 13 1 Electrostatic Energy How much work is done in assembling a collection of point charges Work is force times distance7 which in this context takes the form WFd7qEdqAV Moving the rst charge requires no work 7 since there is no electric eld The second charge needs to be moved in the Coulomb eld of the rst7 the third in the eld of the rst two7 and so on Continuing in this manner7 we see that the work done in assembling the charges is 1 12 1 12 W 7 7 47reogmirjl 87reogmirjl The advantage of the second expression7 in which each term is double counted7 is that it can be rewritten in the form W klVi where V is the potential at the location E of the ith charge due to all the other charges This expression in turn generalizes naturally to a continuous charge distribution but see the discussion in Gri iths about some subtleties in this limit7 for which it becomes 1 W deT Now use Gauss7 Law7 to replace p by 60 times the divergence of the electric eld We have lt EmTv da 30ltE2d7VEdZi where we have integrated by parts7 as justi ed below 1 2 Integration by parts The product rule for the divergence is 006 Ww w Integrating both sides yields vfew6f dTfv d7 Now use the Divergence Theorem to rewrite the rst terrn7 leading to f dzi6f dTfv d7 Which can be rearranged to f d7f d i f d7 Which is the desired integration by parts PH 422 Day 1 1 Flux Recall that V7Ed This is the integral of E along a curve What sort of integral can you take of E over a surface Along a curve there is a natural direction namely that tangent to the curve The line integral above adds up the tangential component of E along the curve The only natural direction associated with a surface on the other hand is the direction perpendicular to it The natural integral to compute over a surface adds up the normal component of This V is called the aw of E through the given surface in the given direction uxEhdAEdzi Be warned that some authors use two integral signs rather than one and other letters are often used in place of A 2 Gauss Law Recall Gauss7 Law which says that a a 1 E 39 dA 6 Qinside 0 box Consider the example of a point charge inside a sphere where 1 if and dAr2sin6d6d f Thus7 the ux is given by a a 1 EdA i dfi 47TEO r2 sphere sphere l 2 gtlt surface area 47TEO r2 2 60 as claimed Note that there is no dependence on r 3 Example Flux through a cube Gauss7 Law does not depend on the shape of the surface being used So lets replace the sphere in the previous example with a cube Suppose the charge is at the origin7 and the length of each side of the cube is 2 What is the ux through one face We have 1 1 1 q q 1 1 dzdy W77 d f W UK 7171 47TEO r2 z y 47TEO 1 1 z y 1 This integral is doable with the help of integral tables or with Maple The worksheet flux mws can be used to compute the answer Does it agree with the previous computation for the sphere Now suppose the charge is not at the origin Maple can still do the integrals nurnerically try some examples Finally7 suppose the charge is on a face7 or an edge What answer do you expect Check and see PH 422 Day 4 1 The Divergence Theorem The total ux of the electric eld out through a small rectangular box is uxZEdA vEdV box But any closed box can be lled With such boxes Furthermore7 the ux out of such a box is just the sum of the ux out of each of the smaller boxes7 since the net ux through any common side Will be zero because adjacent boxes have opposite notions of out of Thus7 the total ux out of any closed box is given by EdA Edv box inside This is the Divergence Theorem 2 Differential form of Gauss Law Recall that Gauss7 Law says that a a l E 39 dA Qinside box 0 But the enclosed charge is just Qinside pdV so we have Putting this all together7 the Divergence Theorem tells us that a a 1 v E dV pdV 6 inside Obox 1 for any closed box This means that the integrands themselves must be equal that is vE 60 This is the di erential form of Gauss7 Law and is one of Maxwell7s Equations It states that the divergence of the electric eld at any point is just a measure of the charge density there 3 The Divergence in Curvilinear Coordinates The rst expression above relating ux to divergence is really just a de nition of the divergence of the electric eld from this point of view the Divergence Theorem is a tautology Our computations yesterday used this geometric de nition to derive an expression for in rectangular coordinates namely ux unit volume 8m By 82 Similar computations to those done yesterday in rectangular coordinates can be done using boxes adapted to other coordinate systems Not surpris ingly this introduces some additional factors of r and sin 6 For instance consider a radial vector eld of the form V EE where 39 is the unit vector in the radial direction The electric eld of a point charge would have this form as do the spherical examples some groups considered in a previous activity What is the ux of E through a small box around an arbitrary point P whose sides are surfaces with one of the spherical coordinates held constant Only the two sides which are parts of spheres contribute and each such contribution takes the form is 121 ETrzsinGdeg An argument similar to the one used in rectangular coordinates leads to E 121 0219 sin dr d6 d Ti2r2E dV where it is important to note that the factor of r2 must also be di erentiated It now nally follows that This is just part of the full expression for the divergence in spherical coor dinates since we started with the very special vector eld E These formulas are so important that they appear on the inside front cover of Gri iths 4 The Divergence of a Coulomb Field The electric eld of a point charge at the origin is given by a 1 A E qr 47r 0 r 2 We can take the divergence of this eld using the expression derived above for the divergence of a radial vector eld which yields a 1 8 1 q 1 8g E 2E 0 v r2 87 T gt 47r 0 r2 r2 87 On the other hand the ux of this electric eld through a sphere centered at the origin is a a 1 1 EdA ldA 14m21 4713960 r2 47r 0 r2 60 sphere sphere V in agreement with Gauss7 Law The Divergence Theorem then tells us that vEdV Ed217 0 inside sphere even though v E 0 Whats going on A bit of thought yields a clue E isn7t de ned at r 0 neither is its divergence So we have a function which vanishes almost everywhere whose integral isn7t zero This should remind you of the Dirac delta function However were in 3 dimensions here so that the correct conclusion is a 1 q 163m gamegt69 47reor 2 60 or equivalently p WW which should not be surprising