Physics 11b Lecture #3. Electric Flux Gauss s Law

Transcription

1 Physics 11b Lecture #3 lectric Flux Gauss s Law

2 What We Did Last Time Introduced electric field by Field lines and the rules From a positive charge to a negative charge No splitting, merging, or crossing Number of field lines amount of charge Density field Was about to define electric flux We continue from there F Q = Q () r

3 Today s Goals Define electric flux Φ Start from the number of field lines Introduce Gauss s Law Useful tool for many &M problems Apply Gauss s Law to a few examples Spherical charge distribution Infinite sheet of charge Discuss basic rules of conductors Derive them using Gauss s Law

4 Concept of Flux Consider a flow of water The water velocity is described by v( xyz,, ) vxˆ vyˆ vzˆ ( v, v, v) x y z x y z Immerse a tiny wire loop Area of the loop is A Unit vector perpendicular to the loop is Define the loop area vector as Aˆ n It represents the size and the orientation of the loop The loop is so small that the shape is irrelevant Q: how much water will flow through the loop? A ˆn ˆn v Let s call it flux Φ v

5 Water Flux It depends on how the loop is oriented w.r.t. the flow Assuming constant velocity and a flat loop: If A v Φ v = If A v Φ v = Av If A and v makes angle θ Φ v = Av cos θ ˆn v ˆn v ˆn v θ Generalize: Definition of flux Φ v = v da Integrate over the area of the loop

6 Field Flux Imagine a small area in an electric field Let s call the area A It could be any shape, any angle How many field lines run through this area? Remember: density of field lines So we are talking about flux of Math is identical to the water flow case Φ = da

7 lectric Flux We now define electric flux as S is a surface area, da = da n Φ = d A Φ is proportional to the number of field lines going thru S Unit of Φ is Newton m 2 /Coulomb (N m 2 /C) NB: sign of flux depends on the direction of n That is, you must define which side of S is positive 2 S = 1m S n Φ = 1 Nm 2 /C n =1N C Φ = 1 Nm 2 /C

8 Which Way is da? Defined unambiguously only for a closed surface i.e., a surface that wraps around a volume completely At any point in space, da is perpendicular to the surface It points towards the outside of the surface ˆn In other words: flux Φ (or Φ v ) is positive if the net flow is coming out of the volume

9 Simple Flux xample 1 Consider a sphere of radius r around a charge q n always points outward We know and n are parallel 1 q We also know = 2 4πε r 2 Φ = d A = 4π r = sphere q ε q r n This is hardly surprising Φ should be proportional to the number of field lines coming out of charge q, which should be proportional to q We just didn t know that the constant was 1/ε

10 Simple Flux xample 2 Consider a sphere in a uniform field Take polar coordinates (θ, φ) relative to the direction of Φ = 2 cosθr sin d d = = π 2 2 r cos sin θ θ φ π θ θdθ r θ n Again hardly surprising Incoming on one side is balanced by outgoing We know field lines never disappear without a charge

11 Gauss s Law Net flux through a closed surface is given by the net charge inside the surface by This seems natural, from what we ve found so far Formal proof is found in textbook 24.5 The law connects charge and field in yet another way Coulomb s law did it one way and they are consistent It s more useful than it looks Let s investigate Φ = d A = qin ε

12 Spherical Charge Distribution Problem: Calculate the electric field (everywhere in space) due to a uniformlycharged sphere ρ = = Q Q y 3 Solid sphere of radius R Constant charge density V 4π 3 R R x Solution #1 I know the due to a point charge dq by Coulomb s Law I know how to integrate

13 Apply Gauss s Law Solution #2 Why would I ever do an integral is somebody (Gauss) already did it for me? First, consider a sphere S 1 outside R Apply Gauss s Law to S 1 Φ = d A = S Q ε 1 Because of symmetry, should be same size and parallel to da everywhere on S 1 d A = 4π r S 1 2 Combine = R S 1 r Q 4πε r 2

14 Apply Gauss s Law First, consider a sphere S 2 inside R Apply Gauss s Law to S 2 Φ = d A = S 1 qin ε Difference: qin is only a fraction of Q that is inside S 2 r R S 2 q r = Q R 3 in 3 Ratio of volumes = (ratio of radii) 3 Same symmetry argument gives d A = 4π r S 2 2 Combine = Qr 4πε R 3

15 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! Which way is? Problem is spherically symmetric must point radially (in or out) Complete solution: r r R S 1 Q rˆ for r 2 4πε r = Qr rˆ for r < 3 4πε R R R R r

16 Checklist for &M Problems Read the problem Look for symmetries: Which coordinate system works best? What cancels out? Which way the vectors should be? Look for ways to avoid integration Turn the math crank Write down the complete solution (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

17 Infinite Sheet of Charge Problem: Calculate the electric field at a distance z from a positively charged infinite plane charge Surface charge density: σ = area Use Gauss again Which surface to use? What symmetry do we have? Consider the cylinder Area A and height z field must be vertical How do we know that? z

18 Infinite Sheet of Charge Now the total flux Φ total = Φ top Φ side Φ bottom Side is parallel to No flux Top and bottom are symmetric Same flux Φ total = 2Φ top = 2A in q = σ A Gauss 2A σ A = ε = σ 2ε z Don t forget the direction! σ zˆ for z > 2ε = σ zˆ for z < 2ε The result is worth remembering: Infinite sheet of charge produces uniform field of σ/2ε above and below it

19 Conductor A conductor conducts electric current Conductive metal contains freely-movable electrons We won t deal with current until later What we are doing is electrostatics = no moving charges When there is no charge movement, the conductor is in electrostatic equilibrium Immediately obvious: there shouldn t be any field in the conductor in electrostatic equilibrium Otherwise the free electrons would be moving There are less obvious but interesting rules

20 Charge On Conductor Free electrons are negative, but the atoms in the conductors are charge neutral So the core of the atom must be a positive ion Ions are fixed, electrons move Their charges usually cancel What if we took away (or add) some electrons? There will be net positive (or negative) charge Where does that go? Answer: on the surface of the conductor

21 Charge Inside Conductor Suppose there is a net charge q somewhere inside a conductor in electrostatic equilibrium Imagine a sphere around the charge and apply Gauss q Φ = d A S = ε q S For the integral to be non-zero, must be non-zero somewhere on the surface of S But is zero inside the conductor! Rule #2: there is no net charge anywhere inside a conductor in electrostatic equilibrium

22 Surface Charge and Charge can only be on the surface Let s call the charge density σ = Q/area NB: σ may not be constant for the whole surface Apply Gauss on a cylinder sticking out of the metal surface Φ total = Φ top Φ side Φ bottom A = Φ total =Φ top = A qin = σ A = σ ε Rule #3: field just outside a conductor in electrostatic equilibrium is perpendicular to the surface and the magnitude is σ/ε

23 Summary Defined electric flux Φ = d A number of field lines through the surface Gauss s Law Φ = qin ε Very useful for solving field problems Applied it on spherical charge distribution and infinite sheet Infinite sheet generates uniform field = 2ε Also used on conductors in electrostatic equilibrium Charge (if any) lives on the surface S σ