13Notes:The “Gaussian” surface can have any shape. You choose the surface.The flux is positive if more electric field lines go out of the surface and negative if more go in.The flux from a particular charge qin is the same regardless of the shape or size of the surface.If the surface contains no charge, the flux through it is zero. That means that every field line that enters the surface will also exit the surface.
14More Notes on Gauss’s Law The normal area vector is taken to point from the inside of the closed surface out.Gauss’s Law is also known as Maxwell’s first equation.Gauss’s Law is second way used to find the electric field. What is the first way you learned?
15Problem 1 The Electric Field due to a point charge. Prove E=kq/r2
16What do you think would be your answer for E, if you chose a different Gaussian surface?
17Problem 2: Show E= σ /(2 ε0) for a rectangular metal slab positively charged with a uniform surface charge density σ.
18Problem 3: Field Due to a Spherically Symmetric Charge Distribution of an Insulator a)Outside the sphere( r>a)
19Problem 3: Field Due to a Spherically Symmetric Charge Distribution of an Insulator b)Inside the sphere( r<a)qin < Q and since the volume charge density is uniform:
20Problem 3: Field Due to a Spherically Symmetric Charge Distribution (Insulator)
21Problem 4: Field Due to a Thin Spherical Shell For r > a, the enclosed charge is Q and E = keQ / r2For r < a, the charge inside the surface is 0 and E = 0