Lessons 6 and 7: Ch. 23 (3-6) – Applying Gauss’s Law 1 Gauss’s Law finds E~ 2 Gaussian Surfaces 3 Charge Enclosed

Gauss’s Law is a relatively simple rela- To make Φ something simple, we are With Φ out of the way we just have to tion between charge and electric field that going to choose an imaginary surface, a get qenc remembering that it might still mathematically says charge CREATES Gaussian Surface, from a very short list: be a charge distribution! Note that the electric field. In certain cases with sym- Box, Cylinder, Sphere. “enc” part means enclosed, so JUST charge found inside the chosen surface. metry , the equation can solve electric The right surface makes the math easy. field without doing messy integrals. If we can choose a surface where E is Depending on the problem, we might constant and the angle between E~ and A~ still need λ, σ and ρ from last chapter. The procedure to use Gauss’s Law is: makes the dot product irrelevant – then We just need to multiply by the right 1. Declare your approach by writing: the integral gets simple. length, area or volume to turn the right H qenc Φ = E~ · dA~ = H density into charge. 0 Like this: E~ · dA~ = EA cos θ 2. Sketch the electric field and it Each surface has a total area though might show a symmetry. considering the direction of E~ we may 4 Conductors in 3. Choose a “Gaussian Surface” to need to consider each face separately: Electrostatic use considering that symmetry 2 • Total ABox = 6l Equilibrium from: Box, Cylinder, Sphere. 2 • Total ACylinder = 2πr + 2πrh The right choice has Charges in conductors move freely but 2 after a long time they will have moved H E~ · dA~ = ± H EdA or 0 (flux • Total ASphere = 4πr along or through surface) and E into a balance distribution we can We have chosen the right surface when explore with Gauss’s Law. constant over the surface so for each face we can argue that E is H H EdA = E dA = EA constant (all spots on the surface are • Charges move until they stop – the net force on each is 0. 4. Determine q (May involve equal distances from the charge) and the enc ~ ~ calculation if a distribution) angle between E and A has a simple • E=0 inside such a conductor. value where cos(θ) is -1,0, or 1. If 5. Solve for E~ using Φ = qenc/0 cos(θ)=0, then E doesn’t have to be • Gauss tells us that any gaussian constant. surface must then have q = 0 NEVER integrate a Gauss’s law integral enc too! with limits and everything. We are It is likely in a real world problem that looking to see if symmetry can give us a no usable Gaussian Surface exists. But • One implication is that in a simple way to find the electric field when we can find it the math is pretty conductor its net charge is found without more work. easy. on the outside.

1 Exercise 1: You will derive one or C. Electric field near a thin D. Electric field near long thin wire more of the following major results: insulating sheet of charge ~ λ • E = 2π r rˆ • Both sides around sheet are 0 A. Electric field near a conductor • λ is charge/length of wire air/vacuum. • r is radial distance from wire Prove that Gauss’s Law gives the electric • σ = charge/area on surface • rˆ points away from a + wire, • E~ = σ rˆ in air/vacuum. field we expect from Coulomb’s law for 20 towards a - one. the case of being a distance r away from • Unlike conductor equation, E field an isolated point particle with charge q goes 2 ways (hence the 2). • Units of E are N/C or V/m B. Electric field near a conductor • One side of surface is metal, other is air/vacuum. • σ = charge/area on surface • E = 0 INSIDE metal • E~ = σ rˆ in air/vacuum. 0 • Units of E are N/C or V/m

E. Shell Theorem (inside) Use steps similar Gauss’s law to prove that E~ inside a uniform spherical shell of charge is zero. F. Shell Theorem (outside) Repeat exercise A under the assumption that instead of a point charge the charge is in a tiny ball of radius R and volume charge density ρ. Your Gaussian surface will be for a distance r where r > R.

2 Exercise 2: (Charged isolated Exercise 3: (Conductor with cavity) Exercise 4: (Uniform sphere of charge) conductor) Can there be excess charge on the cavity Suppose we have a uniform solid sphere wall? of charge with radius R and total charge q. (a) Find E(r) for r ≥ R.

(b) What is the volume charge density ρ? (a) How do we know E~ = 0 inside the copper? Hint: the system is in (a) What is the net flux through the equilibrium, meaning there is no Gaussian surface surrounding the (c) Find E(r) for r ≤ R. flow of charge. cavity?

(b) If E~ = 0, what must be the net (b) So then what must be the net charge enclosed by the Gaussian charge inside the Gaussian surface? surface?

(c) How do the answers change if there (c) If there is any net excess charge on is a charged particle inside the the copper, where must it be? cavity?

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