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AP Physics C Mrs. Coyle Electric Flux, F: The number of electric (flux) field lines which pass through a given cross-sectional area A. Which has the largest electric flux? Does flux depend on area? Answer: A Which has the largest electric flux? Does flux depend on direction? Answer:1 The area vector A is perpendicular to the surface A and has a magnitude equal to the area A. Electric Flux F= E dA = EdA cos q Flux is a dot product and therefore a scalar Units: N 2 [F ]= m C [E]= N C [A]= area m2 q: angle formed between E and the normal to the surface. When E is perpendicular to the loop? F= EA Why? What if you have a closed surface? For a closed surface: Gauss’s Law: the total electric flux through a closed surface is proportional to the enclosed charge. qin F= surface E dA = 0 Permittivity of free space: ε0 = 8.8542 x 10-12 C2 / (N m2) Remember: Coulomb’s Law constant, k 1 k= = 9 x10 Nm / C 9 2 2 4 0 Note: the dot product qin F= surface EdA cos q = 0 Notes: 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. More 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? Problem 1 The Electric Field due to a point charge. Prove E=kq/r2 What do you think would be your answer for E, if you chose a different Gaussian surface? Problem 2: Show E= σ /(2 ε0) for a rectangular metal slab positively charged with a uniform surface charge density σ. Problem 3: Field Due to a Spherically Symmetric Charge Distribution of an Insulator a)Outside the sphere( r>a) qin F E = E dA = EdA = εo EA = E 4πr 2 Q Q E= = ke 2 4πεo r 2 r Problem 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: Q qin r3 = qin = Q 3 4 a 3 4 r 3 a qin F E = E dA = EdA = εo qin Q E= = ke 3 r 4πεo r 2 a Problem 3: Field Due to a Spherically Symmetric Charge Distribution (Insulator) Outside the sphere Q E = ke 2 r Inside the sphere Q E = ke 3 r a Problem 4: Field Due to a Thin Spherical Shell For r > a, the enclosed charge is Q and E = keQ / r2 For r < a, the charge inside the surface is 0 and E = 0 Problem 5: Field due to a Line of Charge Problem 5: Field Due to a Line of Charge End view The flux through the ends of the gaussian cylinder is 0. Why? Problem 5: Field Due to a Line of Charge qin F E = E dA = EdA = εo λ E 2πr = εo λ λ E= = 2ke 2πεo r r