2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum
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1 2. Electostatics D. Rakhesh Singh Kshetimayum 1
2 2.1 Intoduction In this chapte, we will study how to find the electostatic fields fo vaious cases? fo symmetic known chage distibution fo un-symmetic known chage distibution when electic potential, etc. what is the enegy density of electostatic fields? how does electostatic fields behave at a media inteface? We will stat with Coulomb s law and discuss how to find electic fields? What is Coulomb s law? It is an expeimental law 2
3 2.2 Coulomb s law and electic field And it states that the electic foce between two point chages q 1 and q 2 is along the line joining them (epulsive fo same chages and attactive fo opposite chages) diectly popotional to the poduct q 1 and q 2 invesely popotional to the squae of distance between them Mathematically, u q1q u 2 q1q F α ˆ F = k ˆ k = πε 0 F 3
4 2.2 Coulomb s law and electic field Electic field is defined as the foce expeienced by a unit positive chage q kept at that point u u 1 Qq u F 1 Q F = ˆ E = = ˆ (N/C) 2 2 4πε 0 q 4πε 0 Pinciple of Supeposition: The esultant foce on a chage due to collection of chages is equal to the vecto sum of foces due to each chage on that chage Next we will discuss How to find electic field fom Gauss s law? 4 Convenient fo symmetic chage distibution
5 2.3 Electic flux and Gauss s law Electic flux: We can define the flux of the electic field though an aea d s to be given by the scala poduct ψ =. Fo any abitay suface S, the flux is obtained by integating ove all the suface elements u ψ = dψ = D d s S S u d D d s 5
6 ψ 2.3 Electic flux and Gauss s law Gauss s law ψ = S v D ds = Q enclosed Total electical flux coming out of a closed suface S is equal to chage enclosed by the volume defined by the closed suface S iespective of the shape and size of the closed suface 6
7 ψ 2.3 Electic flux and Gauss s law Applying divegence theoem, v ψ = D ds = enclosed = ρ S V ( D) dv = Q dv Since it is tue fo any abitay volume, we may equate the two integands and wite, D = ρ E = ρ ε 0 [Fist law of Maxwell s Equations] V Next we will discuss How to find electic field fom electic potential? Easie since electic potential is a scala quantity 7
8 2.4 Electic potential Suppose we move a potential chage q fom point A to B in an electic field E The wok done in displacing the chage by a distance dl u u dw = - F dl = -q E dl The negative sign shows that the wok is done by an extenal agent. B W = -q u E dl A The potential diffeence between two points A and B is given B by W φ AB = = - u E dl q A 8
9 2.4 Electic potential Electic field as negative of gadient of electic potential: Fo 1-D case, φ x x = - dx ( x) E ( x) Diffeentiate both sides with espect to the uppe limit of integation, i.e., x dφx dx Extending to 3-D case, fom fundamental theoem of gadients, = - E dφ = - E dx x x x x 9
10 2.4 Electic potential dφ = - E dx - E dy - E dz x y z E φ φ φ dφ = dx + dy + dz x y z E = - x φ x = - φ E = - y φ y E = - z φ z Electic field intensity is negative of the gadient of φ 10
11 2.4 Electic potential Maxwell s second equation fo electostatics: Electostatic foce is a consevative foce, i.e., the wok done by the foce in moving a unit chage fom one point to anothe point is independent of the path connecting the two points B A Path B E dl = E dl A 1 Path 2 Q B A A E dl = E dl B 11
12 2.4 Electic potential B A Path A E dl + E dl 0 = B 1 Path 2 E dl = 0 Applying Stoke s theoem, we have, E dl = E ds 0 ( ) = 12 E = 0 [Second law of Maxwell s Equations fo electostatics]
13 2.5 Bounday value poblems fo electostatic fields Basically thee ae thee ways of finding electic field E : Fist method is using Coulomb s law and Gauss s law, when the chage distibution is known Second method is using E = Φ, when the electic potential Φ is known 13
14 2.5 Bounday value poblems fo electostatic fields Thid method In pactical situation, neithe the chage distibution no the electic potential is known Only the electostatic conditions on chage and potential ae known at some boundaies and it is equied to find them thoughout the space 14
15 2.5 Bounday value poblems fo electostatic fields In such cases, we may use Poisson s o Laplace s equations o method of images fo solving bounday value poblems Poisson s and Laplace s equations D = ρ v v E = ρ ε o 15
16 2.5 Bounday value poblems fo electostatic fields Since E = Φ E = Φ = Φ = Poisson s equation Φ = ρ ε 2 v ρ 2 v Fo chage fee condition, Laplace s equation 2 Φ = 0 o ε o 16
17 2.5 Bounday value poblems fo electostatic fields Uniqueness theoem: Solution to Laplace s o Poisson s equations can be obtained in a numbe of ways Fo a given set of bounday conditions, if we can find a solution to 17
18 2.5 Bounday value poblems fo electostatic fields Poisson s o Laplace s equation satisfying those bounday conditions the solution is unique egadless of the method used to obtain the solution 18
19 2.5 Bounday value poblems fo electostatic fields Pocedue fo solving Poisson s o Laplace s equation: Solve the Laplace s o Poisson s equation using eithe diect integation whee is a function of one vaiable Φ 19
20 Bounday value poblems fo electostatic fields o method of sepaation of vaiables if Φ is a function of moe than one vaiable Note that this is not unique since it contains the unknown integation constants Then, apply bounday conditions to detemine a unique solution fo. Φ Once is obtained, We can find electic field and flux density using E = Φ D = ε ε E o Φ
21 2.5 Bounday value poblems fo electostatic fields Method of images: Q ρl ρ V Q ρl ρ V Q ρl ρv (a) Point, line and volume chages ove a pefectly conducting plane and its (b) images and equi-potential suface 21
22 2.5 Bounday value poblems fo electostatic fields commonly used to find electic potential, field and flux density due to chages in pesence of conductos 22
23 2.5 Bounday value poblems fo electostatic fields States that given a chage configuation above an infinite gounded pefect conducting plane may be eplaced by the chage configuation itself, its image and an equipotential suface A suface in which potential is same is known as equipotential suface Fo a point chage the equipotential sufaces ae sphees 23
24 2.5 Bounday value poblems fo electostatic fields In applying image method, two conditions must always be satisfied: The image chages must be located within conducting egion and the image chage must be located such that on conducting suface S, the potential is zeo o constant 24
25 2.5 Bounday value poblems fo electostatic fields Fo instance, Suppose a point chage q is held at a distance d above an infinite gound plane What is the potential above the plane? Note that the image method doesn t give coect potential inside the conducto It gives coect values fo potential above the conducto only 25
26 2.6 Electostatic enegy Assume all chages wee at infinity initially, then, we bing them one by one and fix them in diffeent positions To find the enegy pesent in an assembly of chages, we must fist find the amount of wok necessay to assemble them W = W1 + W2 + W3 = Φ q + q ( Φ + Φ )
27 2.6 Electostatic enegy If the chages wee placed in the evese ode W = W3 + W2 + W1 = 0 + q ( Φ ) + q ( Φ + Φ ) Theefoe, 2 W = q ( Φ + Φ ) + q ( Φ + Φ ) + q ( Φ Φ ) W = 1 ( q Φ + q Φ + q Φ ) In geneal, if thee ae n point chages W 1 2 n = q Φ k = 1 k k 27
28 2.6 Electostatic enegy If instead of point chages, the egion has a continuous chage distibution, the summation becomes integation Fo Line chage 1 W = ρ Φ Fo suface chage Fo volume chage W W 2 L L 1 = 2 ρsφ S ρ 1 = 2 vφ V dl ds dv 28
29 2.6 Electostatic enegy 29 Since we have, D = ρ v W = D Φ dv ( ) Fom vecto analysis, Hence Theefoe, 1 2 v Φ D = D Φ + Φ D ( ) Φ( D) = ΦD D Φ ( ) W D dv D dv ( ) ( ) = Φ Φ V V
30 2.6 Electostatic enegy 30 Applying Divegence theoem on the 1 st integal, we have, W = 1 2 S ΦD ds 1 2 V ( D Φ)dv ΦD emains as 1/ 3 while ds emains as 1/ 2, theefoe the fist integal vaies as 1/, tend to zeo as the suface becomes lage and Hence tends to be infinite W = ( D Φ) dv 1 2 V D E dv = 2 o E dv V V ε
31 2.6 Electostatic enegy The integal E 2 can only incease (the integand being positive) 1 = ρ 2 v Note that the integal and is ove the egion V whee the chage is located, so any lage volume would do just as well W The exta space and volume will not contibute to the integal Since fo those egions = 0 ρ v dv 31
32 2.6 Electostatic enegy the enegy density in electostatic field is 2 dw d d D = = = = 2ε o w 2 D E dv 2 ε o E dv dv dv dv V V 32
33 2.7 Bounday conditions fo electostatic fields Two theoems o Maxwell s fist and second equations in integal fom ae sufficient to find the bounday conditions Bounday conditions fo electic field Let us conside the small ectangula contou PQRSP (see Fig. 2.8 l is chosen such that E 1t and E 2t ae constant along this length 33
34 2.7 Bounday conditions fo electostatic fields σ S S Fig. 2.8 Bounday fo electostatic fields at the inteface of two media 34
35 2.7 Bounday conditions fo electostatic fields Note that h 0 at the bounday inteface and theefoe thee is no contibution fom QR and SP in the above line integal Also note that the diection of the line integal along PQ and RS ae in the opposite diection Q C E dl = E 1t = E 2t 0 = Q P E 1 dl + 1 S R E 2 dl 2 = E 1t l E 2t l The tangential component of electic field vecto is continuous at the inteface 35
36 2.7 Bounday conditions fo electostatic fields Bounday conditions fo electic flux density Let us conside a small cylinde at the inteface Coss section of the cylinde must be such that vecto D is the same Note that h 0 at the bounday inteface theefoe, thee ae no contibution fom the cuved suface of the pillbox in the above suface integal So only the top and bottom sufaces emains in the suface integal 36
37 2.7 Bounday conditions fo electostatic fields D ds = D1 ds1 + D2 ds2 = pillbox top suface bottom suface Q enclosed The nomal is in the upwad diection in the top suface and downwad diection in the bottom suface D S D S = σ S D D = σ 2n 1n 2n 1n the nomal component of electic flux density can only change at the inteface if thee is chage on the inteface, i.e., suface chage is pesent 37
38 2.7 Bounday conditions fo electostatic fields If medium 2 is dielectic and medium 1 is conducto Then in conducto D 1 =0 and hence D 2n =σ o in geneal case, D n =σ 38
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