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Chapter 22 Gauss’ Law Mathematics Reminder u = aiˆ + bĵ v = ciˆ + dĵ u × v º ac + bd = u v cos f where f is the angle between the vectors Math: Area Vector Gauss’ Law F= qenclosed e0 F : Electric flux through a closed surface qenclosed : Total charge enclosed by the closed surface But how do you find F? Through which closed surface? Why it makes sense All capture the same number of field lines. Flux by Integration dA º n̂dA where n̂ is the unit normal vector. Gauss' Law: qenclosed F= e0 Relates the outside to the inside F= Flux through the area of the Gaussian surface q e0 Charge inside the Gaussian surface A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it. Compared to the electric flux through surface #1, the flux through surface #2 is +q Gaussian surface #1 A. greater. B. the same. C. less, but not zero. D. zero. Gaussian surface #2 E. not enough information given to decide Two point charges, +q (in red) and –q (in blue), are arranged as shown. Through which closed surface(s) is the net electric flux equal to zero? A. surface A B. surface B C. surface C D. surface D E. both surface C and surface D Area vector of a cube Convention: Area vector of a close surface points outward. Sign of Flux • Positive when E field is going out • Negative when E field is going in F>0 F<0 Sign of Flux (Example) F=0 Example (constant E) Six surfaces: Ftotal = F1 + F2 + F 3 + F 4 + F 5 + F 6 = -EA + EA + 0 + 0 + 0 + 0 Þ Ftotal = 0 qenclosed but Ftotal = e0 Þ qenclosed = 0 is difficult to find in general … so we need to be smart. + E here is different than E there. You are allowed to change the shape of the Gaussian surface to make the surface integral easy. Spherical Gaussian Surface Simplify the integral E parallels to dA everywhere Þ E × dA = E cos(0 )dA = EdA. By spherical symmetry: E is constant everywhere on the sphere Therefore the flux becomes: F= ò E × dA = ò EdA = E ò dA = EA Outside Uniform Sphere of Charge F= (Electric field) ´ (Area) Area of sphere = 4p r q e0 q 2 Þ F = (E)(4p r ) 2 (E)(4p r ) = 2 ÞE= q e0 q 4pe 0 r 2 Gaussian surface r x What to write in the exam In the exam, just a few words of explanation is enough in simplifying the flux integral: Þ (E)(4p r ) = 2 ÞE= q 4pe 0 r 2 q e0 Must include diagrams When answering questions on Gauss’ law, you MUST include a diagram to indicate the Gaussian surface you picked. Without defining the Gaussian surface with the diagram, the flux cannot be defined, and your integral will have no meaning. You will lose half your points without a diagram. Linear fly density L=4m There are N = 8 flies on the rod. The number of flies per unit length is: N 8 flies l= = = 2 flies / m. L 4m How many flies n are on l = 2m of the rod? n l= l Þ n = ll = (2 flies / m)(2m) = 4 flies. Linear charge density, +++++++++++++++++++++++++++++++++++++ L=4m There are Q = 8C on the rod. The charge per unit length is: Q 8C l= = = 2C / m. L 4m How much charge q is on l = 2m of the rod? q l= l q = ll = (2C / m)(2m) = 4C. Surface charge density Q = 10C on a disc of radius R = 4m. What is the charge per unit area? ++++++ ++++++++++++++++++ ++++++ Q Q 10C 2 s= = = = 0.2C / m A p R 2 p (4m)2 How much charge q is in a circle of r = 2m? (red circle) q s= 2 pr Þ q = sp r 2 = (0.2C / m 2 )(p )(2m)2 = 2.5C Volume charge density Q q r= = V v R Þ q = rv, where q is the charge on a volume v, Q is the total charge on the whole volume V. Unit: C / m 3 Charge Density Summary 1D (l is linear charge density, C / m) : q = l ´ length 2D (s is surface charge density, C / m 2 ) : q = s ´ area 3D ( r is volume charge density, C / m 3 ) : q = r ´ volume [But watch out for regions where the charge density drops to zero!] A Line of Charge l : Linear charge density (charge per unit length) Unit: C/m Simplify the integral Ftotal = F1 + F2 + F 3 = ò S1 E × dA + ò E × dA + ò E × dA S2 S1 S3 By cylindrical symmetry: E points radially outward Therefore the flux on S1 and S3 are zero: F1 = ò E × dA = ò F = ò E × dA = ò S1 3 S3 F1 = F 3 = 0 S3 S1 E cos 90 dA = 0 S3 E cos 90 dA = 0 S2 Simplify the integral E parallels to dA everywhere on S2 Þ E × dA = E cos(0 )dA = EdA. S3 S1 S2 By cylindrical symmetry: E is constant everywhere on S2 Therefore the flux on the curved surface becomes: F2 = ò E × dA = S2 ò S2 EdA = E ò dA = EA2 S2 Ftotal = F1 + F2 + F 3 = 0 + EA2 + 0 = EA2 F= (Electric field) ´ (Area) Area of cylinder A2 = 2p rl Þ F = (E)(2p rl) q e0 q = ll Charge enclosed E(2p rl) = ll / e 0 ; l : Linear charge density l ÞE= 2 e 0p r An Plane Sheet of Charge A Plane Sheet of Charge F= (Electric field) ´ (Area) q q = sA e 0 Charge enclosed σ is the charge per area F = EA + EA = 2EA sA s 2EA = ÞE= e0 2e 0 area A r x Gaussian surface (cylinder, pillbox) Two Parallel Plates E = σ/ ε0 Derivation with Gauss Law + + + + + + + + + + + + + + + + F = EA, q = s A E where s is charge per unit area q F= e0 sA Þ EA = e0 s ÞE= e0 No E inside a conductor A Solid Conducting Sphere Outside: qenclosed = Q E(4p r ) = 2 Þ Eout = Q Q Q e0 Q 4pe 0 r Inside: qenclosed = 0 Þ Ein = 0 2 Useful fact E field is always zero inside a conductor at equilibrium (no movement of charges). Since F=qE, if E is non-zero, the electric force will push the charges around. Therefore the only way equilibrium can be established is if F=0, which implies E=0. An Uniformly Charged Sphere Total charge: Q Radius: R We can compute the volume charge density: Q r= 4 3 pR 3 R Outside (r > R) Total charge inside the Guassian surface of radius r: qenclosed = Q F = E(4p r 2 ) Þ E(4p r ) = 2 Þ Eout = Q e0 Q 4pe 0 r 2 R Inside (r < R) r= Total charge inside the Guassian surface of radius r: qenclosed 4 3 = r pr 3 4 3 r pr Þ E(4p r 2 ) = 3 e0 rr Qr ÞE= = 3e 0 3e ( 4 p R 3 ) 0 3 Q r Þ Ein = 4pe 0 R 3 R Q 4 3 pR 3 R