Gauss Law. 2. Gauss s Law: connects charge and field 3. Applications of Gauss s Law

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1 Gauss Law 1. Review on 1) Couomb s Law (charge and force) 2) Eectric Fied (fied and force) 2. Gauss s Law: connects charge and fied 3. Appications of Gauss s Law

2 Couomb s Law and Eectric Fied Couomb s Law: the force between two point charges

3 Couomb s Law and Eectric Fied Couomb s Law: the force between two point charges F e = K e q 1 q 2 r 2 ˆr 12

4 Couomb s Law and Eectric Fied Couomb s Law: the force between two point charges F e = K e q 1 q 2 r 2 ˆr 12 The eectric fied is defined as

5 Couomb s Law and Eectric Fied Couomb s Law: the force between two point charges F e = K e q 1 q 2 r 2 ˆr 12 The eectric fied is defined as E F q 0 and is represented through fied ines.

6 Couomb s Law and Eectric Fied Couomb s Law: the force between two point charges F e = K e q 1 q 2 The eectric fied is defined as E F q 0 r 2 ˆr 12 and is represented through fied ines. The force a charge experiences in an eectric fied is

7 Couomb s Law and Eectric Fied Couomb s Law: the force between two point charges F e = K e q 1 q 2 The eectric fied is defined as E F q 0 r 2 ˆr 12 and is represented through fied ines. The force a charge experiences in an eectric fied is F = q E 0

8 Couomb s Law and Eectric Fied A one page summary of two chapters Couomb s Law: the force between two point charges F e = K e q 1 q 2 The eectric fied is defined as E F q 0 r 2 ˆr 12 and is represented through fied ines. The force a charge experiences in an eectric fied is F = q E 0

9 From Couomb s Law to Gauss s aw Try to cacuate the eectric fied generated by a point charge ç easy an infinitey ong straight wire with eveny distributed charge ç hard a wire oop ç ony at specia ocations a round disk ç ony at specia ocations an infinitey arge pane ç what??? a soid sphere with eveny distributed charge Are there other ways to cacuate eectric fied generated from a charge distribution? Eectric fied is generated by source charges, are there ways to connect eectric fied directy with these source charges? The answer is YES

10 When the eectric fied is perpendicuar to the chosen surface A, the eectric fux is the product of the magnitude of E and the surface area A: The unit of is N. m 2 /C Some preparation: Eectric Fux, a specia case E = E A Compare to a water fux in a tube: Φ W = V 1 A 1 = V 2 A 2 This sign means water fows into the tube, by convention.

11 Eectric Fux when E and A is at an ange θ When the eectric fied ines make an ange θ with the direction (i.e., the norma) of the surface, the fux is cacuated as: = EA = EAcosθ = E A E A And the eectric fied has to be a constant a over the area. Question: when this (eectric fied is constant) is not the case, what do you do the get the fux? = d = E d A Review on math: 1. direction of a surface is defined as the (outwards) norma to that surface. 2. Dot product of two vectors.

12 Eectric Fux, the genera case In a more genera case, ook at a sma area eement Δ = E i Δ A = E i ΔA i cosθ i this becomes Δ = im ΔA i 0 = surface Review on math: Integra over a surface. E i Δ A E d A The surface integra means the integra must be evauated over the surface in question In genera, the vaue of the fux wi depend both on the fied pattern and on the shape of the surface When the surface is cosed, the direction of the surface (i.e. the norma of it) points outwards.

13 fux through a cube in a uniform eectric fied The fied ines that pass through surfaces 1 and 2 perpendicuary and are parae to the other four surfaces For side 1, Δ = E 2 For side 2, Δ = E 2 For the other sides, Δ = 0. Why? Therefore: = 0

14 fux through a sphere with a charge at its center A positive point charge, q, is ocated at the center of a sphere of radius r According to Couomb s Law, the magnitude of the eectric fied everywhere on the surface of the sphere is E = k e r q 2

15 fux through a sphere with a charge at its center The fied ines are directed radiay outwards and are perpendicuar to the surface at every point, so " E d A = " E n da = E da = " E da " = E 4πr 2 Combine these two equations, we have = E 4πr 2 = 1 4πε 0 q r 2 4πr2 = q ε 0

16 The Gaussian Surface and Gauss s Law You choose a cosed surface and ca it a Gaussian Surface. This Gaussian Surface can be any shape. It may or may not encose charges. Gauss s Law (Kar Friedrich Gauss, ) states: The net fux through any cosed surface surrounding a charge q is given by q/ε o and is independent of the shape of that surface. = " E d A = q ε 0

17 The Gaussian Surface and Gauss s Law = E d A " = q ε 0 The net eectric fux through a cosed surface that surrounds no charge is zero. The eectric fied due to many charges is the vector sum of the eectric fieds produced by the individua charges the fux through any cosed surface can be expressed as = E d A " = % ' & E i i ( * d A ) " = i ε 0 Gauss s Law connects eectric fied with its source charge q i

18 Gauss s Law Summary Gauss s aw states = " E d A = q in ε 0 q in is the net charge inside the Gaussian surface E represents the eectric fied at any point on the surface E is the tota eectric fied at a point in space and may have contributions from charges both inside and outside of the surface Athough Gauss s aw can, in theory, be soved to find E for any charge configuration, in practice it is imited to a few symmetric situations

19 Recap: Eectric Fux and Gauss Law Eectric fux, definition = EA = EAcosθ = E A = im ΔA i 0 = surface i Gauss aw = E i ΔA i E d A E d A = q in ε 0 q in is the net charge inside the surface

20 Appying Gauss Law To use Gauss aw, you need to choose a Gaussian surface over which the surface integra can be simpified and the eectric fied determined Take advantage of symmetry Remember, the Gaussian surface is a surface you choose, it does not have to coincide with a rea surface

21 Conditions for a Gaussian Surface Try to choose a surface that satisfies one or more of these conditions: The vaue of the eectric fied can be argued from symmetry to be constant over the surface The dot product of E d A can be expressed as a simpe agebraic product EdA when E and d A are parae. The dot product is 0 when and are perpendicuar Sti no cue about how to use Gauss Law? There are ony three types of probems. See exampes in the foowing pages. E d A

22 Probem type I: Fied Due to a Sphericay Symmetric Even Charge Distribution, incuding a point charge. The fied must be different inside (r <a) and outside (r >a) of the sphere. For r >a, seect a sphere as the Gaussian surface, with radius r and concentric to the origina sphere. Because of this symmetry, the eectric fied direction must be radiay aong r, and at a given r, the fied s magnitude is a constant. Can you write down the mathematica expression based on the above reasoning?

23 Probem type I: Fied Due to a Sphericay Symmetric Even Charge Distribution, incuding a point charge. = The fied must be different inside (r <a) and outside (r >a) of the sphere. For r >a, seect a sphere as the Gaussian surface, with radius r and concentric to the origina sphere. Because of this symmetry, the eectric fied direction must be radiay aong r, and at a given r, the fied s magnitude is a constant. E d A " = E da " = E 4πr 2 = q in ε 0 E = 1 q in 4πε 0 r 2 Q E = k e r r 2 = k e Q r 2 E is constant at a given r. Gauss Law As if the charge is a point charge Q

24 Fied inside the sphere For r < a, seect a sphere as the Gaussian surface. A the arguments are the same as for r > a. The ony difference is here q in < Q Find out that q in = Q(r/a) 3 (How?) = E d A = E = 1 q in 4πε 0 r 2 Q E = k e r a 3 = k e E da " = E 4πr 2 = q in ε 0 q in r 2 = k e 1 r 2 r 3 a 3 Q = k e Q a 3 r Increase ineary with r, not with 1/r 2

25 Pot the resuts (assume positive Q) Inside the sphere, E varies ineary with r E 0 as r 0 The fied outside the sphere is equivaent to that of a point charge ocated at the center of the sphere This concudes Gauss Law type I probems.

26 Probem type II: Fied at a Distance from a Straight Line of Charge Seect a cyinder as the Gaussian surface. The cyinder has a radius of r and a ength of E is constant in magnitude and parae to the surface (the direction of a surface is its norma) at every point on the curved part of the surface (the body of the cyinder).

27 Cacuate the fux Because of this ine symmetry, the end view iustrates more ceary that the fied is parae to the curved surface, and constant at a given r, so the fux is = E 2πr The fux through the ends of the cyinder is 0 since the fied is perpendicuar to these surfaces. r

28 Eectric fied from Gauss Law This concudes Gauss Law type II probems. = E d A = E da = q 0 ε 0 E 2πr = λ ε 0 λ E = 2πε 0 r = 2k e λ r One can change the thin wire into a rod as we did in the sphere case and find the eectric fied inside and outside of the rod.

29 Probem type III: Fied Due to an Infinitey Large Pane of Charge Argument about the eectric fied: Because the pane is infinitey arge, any point can be treated as the center point of the pane, so at that point E must be parae to the pane direction (again this is its norma) and must have the same magnitude at a points equidistant from the pane. Choose the Gaussian surface to be a sma cyinder whose axis is parae to the pane s direction (third time, this is the norma of the pane).

30 E is perpendicuar to the curved surface direction, so the fux through this surface is 0, because cos(90 o ) = 0. E is parae to the ends, so the fux through each end of the cyinder is EA and the tota fux is 2EA Find out the fux

31 Eectric fied from Gauss Law This concudes Gauss Law type III probems. The tota charge in the surface is Appying Gauss s aw σ A = 2EA = σ A E = σ ε 0 2ε 0 Note, this does not depend on r, the distance from the point of interest to the charged pane. Why? Therefore, the fied is uniform everywhere One can aso change the pane (without thickness) into a pate with thickness d, and foow the discussion in exampe I.

32 To summarize: the 3 types of Gauss Law probems Q E = k e a r $ # & ˆr, when r < a " 3 % Q $ E = # k e & ˆr, when r a " % r 2 E = λ 2πε 0 r = 2k e λ r E = σ 2ε 0

33 Other appications for Gauss Law: Eectrostatic Equiibrium Definition: When there is no net motion of charge within a conductor, the conductor is said to be in eectrostatic equiibrium When in eectrostatic equiibrium, the properties: The eectric fied is zero everywhere inside the conductor, whether the conductor is soid or hoow ç Faraday Cage If an isoated conductor carries a charge, the charge resides on its surface The eectric fied just outside a charged conductor is perpendicuar to the surface and has a magnitude of " #, s is the surface charge density at that point On an irreguary shaped conductor, the surface charge density is inversey proportiona to the radius at that oca surface, so s is greatest at ocations where the radius of curvature is the smaest ç Lightning Rod

34 More discussions about eectrostatic equiibrium properties. Property 1: for a conductor, Fied inside = 0 Consider a neutra conducting sab, when there is no externa fied, charges are distributed throughout the conductor, experience no force and are in eectrostatic equiibrium. When there is an externa fied E This externa fied wi exert a force on the charges inside the conductor and redistribute them in such a way that the interna eectric fied generated by these redistributed charges cance the externa fied so that the net fied inside the conductor is zero to prevent further motion of charges. Hence the conductor reaches again eectrostatic equiibrium This redistribution takes about secons and can be considered instantaneous E induced

35 Property 2: For a charged conductor, charge resides ony on the surface, and the fied inside the conductor is sti zero. Charges (have to be the same sign, why?) repe and move away from each other unti they reach the surface and can no onger move out: charge resides ony on the surface because of Couomb s Law. Choose a Gaussian surface inside but cose to the actua surface Since there is no net charge inside this Gaussian surface, there is no net fux through it. Because the Gaussian surface can be any where inside the voume and as cose to the actua surface as desired, the eectric fied inside this voume is zero anywhere

36 Property 3: Fied s Magnitude and Direction on the surface Direction: Choose a cyinder as the gaussian surface The fied must be parae to the surface (again this is its norma) If there were an ange ( q ¹ 0), then there were a component from E E ^ and tangent to the surface that woud move charges aong the surface. Then the conductor woud not be in equiibrium (no charge motions)

37 Property 3: Fied s Magnitude and Direction on the surface Magnitude: Choose a Gaussian surface as an infinitesima cyinder with its axis parae to the conductor surface, as shown in the figure. The net fux through the Gaussian surface is that ony through the fat face outside the conductor The fied here is parae to the surface The fied on a other surfaces of the Gaussian cyinder is either perpendicuar to that surface, or zero. Appying Gauss s aw, we have = EA = σ A ε 0 E = σ ε 0

38 Another exampe: Eectric fied generated by a conducting sphere and a conducting she Charge and dimensions as marked Anayze: System has spherica symmetry, Gauss Law probem type I. Eectric fied inside conductors is zero There are two other ranges, a<r<b and b<r that need to be considered Arguments for eectric fied Simiar to the sphere exampe, because the spherica symmetry, the eectrica fied in these two ranges a<r<b and b<r is ony a function of r, and goes aong the radius.

39 Construct Gaussian surface and cacuate the fux, and use Gauss Law to get the eectric fied E = 0 when r<a, and b<r<c Construct a Gaussian sphere with its center coincides with the center of the inner sphere When a<r<b: The fux = E πr Appy Gauss Law = Q/ε 0 E = 1 Q 4πε 0 r = k Q 2 e r or Q E = k 2 e r r 2 When b<r The fux = E πr Appy Gauss Law = (-2Q+Q)/ε 0 E = 1 2Q +Q 4πε 0 r 2 Q = k e r or Q E = k 2 e r 2 r

40 Two more exampe probems +σ -σ Exampe 1: find the eectric fied between and outside of two parae pates that are infinitey arge and oppositey charged with equa surface density of σ y Exampe 2: a soid sphere of radius 2r with a charge density ρ has a cavity of radius r. Find the eectrica fied aong the y-axis from 0 to 2r. 2r ρ x

41 Reading materia and Homework assignment Pease watch this video (about 50 minutes): Pease check wieypus webpage for homework assignment.