Electromagnetism Physics 15b

Transcription

1 Electromagnetism Physics 15b Lecture #2 Guass s Law Electric Field and Flux Purcell Administravia Online sectioning due Wednesday (tudy Card Day) Go to Do both discussion sections and lab sections Giovanni zevi@physics.harvard.edu for questions Participate in Learning Gains tudy by Prof. Mazur Take two online questionnaires and get 2 extra points toward your course grade CEM = Conceptual urvey of Electricity and Magnetism Once each at the beginning and end of the semester Go to You will have to register first by clicking Not registered? Jessica Watkins watkinsj@seas.harvard.edu or Doug van Wieren dvw@deas.harvard.edu for questions 1

2 What We Did Last Time Electric charge = ource and recipient of electric forces Positive and negative; conserved; and quantized Coulomb s Law F 2 = q q q q2 ˆr 2 r 21 F r F Inverse-square law same as gravity Multiple charges uperposition Principle Electrostatic force is conservative Electrostatic energy depends only on the positions of the charges For 2-body: U = q 1q 2 r 12 For N-body: U = 1 2 N j =1 k j F j = q j q k r jk k j F jk Today s Goals Introduce electric field E First (and the easiest) of the E&M fields Representation by field lines Define flux Φ How much E flows through an area Introduce Gauss s Law Useful tool for many E&M problems Apply Gauss s Law to a few examples pherical charge distribution Infinite sheet of charge Discuss energy in the electric field if I get that far J.C.F. Gauss,

3 Electric Field A charge q 0 is placed near {q 1, q 2,, q N } Force on q 0 N q 0 q N j q F = jˆr 0 j ˆr 2 0 j = q 0 2 j=1 r 0 j j=1 r 0 j What all the other charges are and where they are Most of the interesting part has nothing to do with the value of q 0 But the position of q 0 is important Let s call it r r 0 j = r r j r 0 j = r r j ˆr 0 j = r r j r r j q j (r r j ) F = q 0 r r j 3 j A function of r for a given configuration of {q 1, q 2,, q N } Electric Field Imagine we have a fixed configuration of {q 1, q 2,, q N } and we can move q 0 around to test what happens to it E(r) is the electric field q j (r r j ) F(r) = q 0 E(r) = q 0 r r j 3 It is a vector field, i.e. a vector function of the position r We ve introduced a middle-man for the electric force Charge creates E field E field creates force on charge End result is unchanged (of course) Dimension of E is force/charge Unit of E is dyne/esu j 3

4 Electric Field E field created by a single charge q at the origin is E(r) = q r 2 ˆr E points outward One can express E field graphically by small arrows But it s busy and tedious to draw Faraday came up with an easier way to draw E field lines Long, continuous arrows originating from the charge represent the direction What about the size of E? Field Line Density E field weakens with distance Inverse-square law E = q r 2 Field lines become less crowded with distance Consider a sphere at distance r Number of field lines N is constant Density = N = N 4πr 2 Density of field lines is proportional to the E field magnitude if we choose the number of lines N to be proportional to q 4

5 Field Line Rules You can often draw field lines without calculation Just follow a few rules Field lines start from a positive charge and end in a negative charge Exception: it s OK to come from/go to infinitely far away Field lines cannot split, merge, or cross each other The number of field lines coming out of a charge is proportional to the amount of the charge The last rule ensures proportionality between the field line density and the electric field Practice #

6 Practice # Charge Distribution Electric field created by multiple charges is E = Electric charge may be distributed over objects, e.g. Along a length of wire (1-d) On the surface of a plate (2-d) Inside the volume of a ball (3-d) For continuous charge distributions, we replace the sum with an integral E = i q i r i 2 ˆr i uperposition Principle dq r ˆr 2 Integral over line/surface/volume 6

7 Uniform Linear Charge Charge Q uniformly distributed on a thin rod of length l What is the electric field E at point A, which is distance r away from the center of the rod? et up is everything de 1. Define x-y coordinates A 2. Pick a small piece of the rod and call it dx 3. Calculate the field due to this piece 4. Integrate along the rod The charge on the dx piece is Linear charge density (esu/cm) Field from this piece is de = Q dx r 2 + x 2 Uniform Linear Charge Note that the field from the green piece is same size and flipped in x Only the y-component will remain after adding up de y = Q Integrate this! E = 2 2 dx r 2 + x 2 Q r r 2 + x 2 r (r 2 + x 2 ) dx ŷ 3 2 Q = ŷ r r de Check the direction, dimension, r-dependence, large-r limit 7

8 Concept of Flux Consider a flow of water The water velocity is described by a Immerse a tiny wire loop Area of the loop is a Define the loop area vector a as being perpendicular to the loop, and the magnitude a equals the area of the loop It represents the size and the orientation of the loop If the loop is small, the shape is irrelevant Q: how much water will flow through the loop? v a Generalize for a loop of not-so-small size loop v da Call this water flux through the loop Field Flux Imagine a surface in an electric field E It could be any shape, any size, any angle We can define the flux of the electric field through Φ E da Density of the field lines is proportional to E o Φ represents how many field lines goes through 8

9 Electric Flux Unit of Φ is (dyne/esu) cm 2 This equals to esu, because esu = i.e., same as the unit of charge (???) dyne cm Φ E da ign of flux depends on the direction of da That is, you must define which side of is positive = 1cm 2 da Φ = 1 esu da Φ = +1 esu E = 1dyne/esu Which Way is da? Defined unambiguously only for a closed surface i.e., a surface that wraps around a volume completely At any point on the surface, da is perpendicular to the surface By convention, da points outward from of the surface In other words: flux is positive (negative) if the net flow is coming out of (going into) the volume 9

10 Flux Through a phere Consider a sphere of radius r around a charge q da always points outward, i.e. parallel to r Coulomb: Φ = E = q r 2 ˆr E da = sphere sphere = q r 2 4πr 2 = 4πq q r ˆr 2 daˆr ( ) r da This is hardly surprising Φ should be proportional to the number of field lines coming out of the charge, which is proportional to q We just didn t know that the constant was 4π Non-pherical urface Now take an arbitrary closed surface around charge q Define da at r from the charge q da θ Infinitesimal flux dφ through da is uppose da covers solid angle dω around q Integrate Φ = E da = dφ = E da = q r 2 cosθda cosθda = r 2 dω q r r 2 dω = q dω = 4πq 2 4π 4π The flux does not depend on the shape of the surface r 10

11 Gauss s Law Now consider multiple charges contained inside surface Total E field is a sum of fields from each charge Φ = E j da = E j da = 4πq j = 4π q j j Net flux through a closed surface is given by the net charge inside the surface by Φ = E da = 4πq inside The law connects charge and field in yet another way Coulomb s law did it one way they are equivalent j j j pherical Charge Distribution Problem: Calculate the electric field (everywhere in space) due to a uniformly-charged sphere olid sphere of radius R Volume charge density: olution by Coulomb s Law We know the E due to a point charge dq by Coulomb s Law Integrate over the volume of the sphere Repeat for inside (r < R) and outside (r > R) of the sphere Of course you can do it, but not fun R 11

12 Apply Gauss s Law Consider an imaginary sphere of radius r Concentric with the charged sphere ymmetry of the problem tells us that E field is radial in direction, and its magnitude is constant over Think about it Apply Gauss s Law to Φ = The integral is easy: E(r) = Q r 2 E da = 4πQ E(r)da = E(r) 4πr 2 This is for r > R R r E Apply Gauss s Law Now make smaller than the charged sphere The symmetry argument still holds Apply Gauss s Law Φ = Charge inside is E da = 4πq inside q inside = r 3 Integral is the same as before: E(r) = Qr R 3 R Q 3 E(r)da = E(r) 4πr 2 This is for r < R r R 12

13 Am I Done? One point remains before we can get full credit We were asked to determine the electric field a vector We need to specify the direction! Complete solution: E Q Q r ˆr for r R R 2 E = 2 Qr R ˆr for r < R 3 R E field outside a charged sphere = E field for a point charge NB: the same holds for gravity r Checklist for E&M Problems Read the problem Look for symmetries: Which coordinate system works best? What cancels out? Which way the vectors should point? Look for ways to avoid integration Turn the math crank Write down the complete solution e.g. magnitudes and directions for all the different regions Read the problem again Did you answer what is asked? Box the solution: your TF will be grateful 13

14 Infinite heet of Charge Problem: Calculate the electric field at a distance z from a positively charged infinite plane urface charge density: σ = charge area Use Gauss again Which surface to use? What symmetry do we have? z E Consider a cylinder Area A and height 2z E field must be vertical How do we know that? Infinite heet of Charge Total flux Φ total = Φ top + Φ side + Φ bottom ide is parallel to E E da = 0 No flux Top and bottom are symmetric ame flux Charge inside the cylinder is q inside = Aσ Using Gauss Φ total = 2Φ top = 2AE(z) Don t forget the direction! E = E(z) = 2πσ +2πσẑ for z > 0 2πσẑ for z < 0 The result is worth remembering: Infinite sheet of charge produces uniform E field of 2πσ above and below z E 14

15 Pair of Charged heets Place two oppositely-charged large sheets in parallel Consider an area A of them E fields from the two sheets overlap and add up Between the sheets: E = 4πσ Cancel each other outside Two sheets also attract each other (obviously) Top sheet feels E bottom = 2πσẑ The force on area A of the top sheet is F = σ AE bottom = 2πσ 2 Aẑ = E 2 z 8π Aẑ E top +σ σ E bottom Pair of Charged heets Imagine we move the top sheet upward by a distance d We must do work W = Fd = E 2 8π Ad The energy of the system increases by W Q: Where exactly is this energy? Note that the volume of the space between the sheets increased by Ad This is also where E field exists z E top +σ σ pace with E holds energy with a volume density Total electrostatic energy of a system is U = E 2 8π dv u = E 2 E bottom 8π 15

16 ummary Defined electric field by F = qe Can be expressed by field lines Defined flux of electric field Φ = E da Note the sign convention: positive if coming out Gauss s Law E da = 4πq Useful for solving E fields with symmetries pherical charge distribution and infinite sheet Infinite sheet generates uniform E field of 2πσ Energy density of electric field u = E 2 8π U = E 2 8π dv 16