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Transcript
Gauss’s Law and Electric Flux Preview Gauss’s Law relates number of E field lines entering and leaving a surface to the net charge inside the surface. Consider imaginary spheres centered on: a.) +q (blue) b.) -q (red) c.) midpoint (yellow) Number of lines exiting: a.) (blue) 24 b.) -q (red) -24 c.) midpoint (yellow) 0 More on Gauss’s Law later. c a b Electric Flux • Flux: Let’s quantify previous discussion about fieldline “counting” Define: electric flux ΦΕ through the closed surface S r r Φ E ≡ ∫ E • dA S E dA S “S” is surface of the box Electric Flux r r ΦE ≡ ∫ E • dA S •What does this new quantity mean? • The integral r r is over a CLOSED SURFACE • Since E • dA is a SCALAR product, the electric flux is a SCALAR quantity r • The integration vector dA is normal to the surface and points OUT r r of the surface. E • d A is interpreted as the component of E which is NORMAL to the SURFACE • Therefore, the electric flux through a closed surface is the sum of the normal components of the electric field all over the surface. • The sign matters!! Pay attention to the direction of the normal component as it penetrates the surface… is it “out of” or “into” the surface? • “Out of” is “+” “into” is “-” Electric Flux Special case: uniform E perpendicular to plane surface A. ΦE E r r ≡ ∫ E • d A = ∫ E dA cos θ = E ∫ dA = EA S S constant dA s S =1 Another: uniform E at angle to plane surface A. constants ΦE s θ dA r r ≡ ∫ E • d A = ∫ E dA cos θ = E cos θ ∫ dA = EA cos θ S S S E CLICKER QUESTION A flat disk with radius r = 0.1m, is oriented with its normal unit vector relative the a constant E field as shown above. What is the electric flux through the disk area? r 2 r o A) π (0.1) E 2 r o B ) π (0.1) E cos 30 2 r o C ) π (0.1) E sin 30 ( ) ( ) ( ) ( ) D) 2π (0.1) E cos 30 r E ) 2π (0.1) E sin 30o How to think about flux • We will be interested in net flux in or out of a closed surface like this box w z • This is the sum of the flux through each side of the box “A” is surface of the box y x – consider each side separately surface area vector: A r A = area • yˆ = w 2 yˆ • Let E-field point in y-direction r – then E and A are parallel and • Look at this from on top – down the z-axis r r r 2 E • A =| E | w A UI3PF3: 6) A cube is placed in a uniform electric field. Find the flux through the bottom surface of the cube. a) Фbottom < 0 b) Фbottom = 0 c) Фbottom > 0 Exercise UI3A2 2A •Imagine a cube of side a positioned in a region of constant electric field as shown •Which of the following statements about the net electric flux ΦE through the surface of this cube is true? (a) ΦE = 0 2B (b) ΦE ∝ 2a2 a a (c) ΦE ∝ 6a2 • Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown. – Which of the following statements about the net electric flux through the 2 surfaces (Φ2R and ΦR) is true? R 2R (a) ΦR < Φ2R (b) ΦR = Φ2R (c) ΦR > Φ2R Exercise UI3A2 2A •Imagine a cube of side a positioned in a region of constant electric field as shown •Which of the following statements about the net electric flux ΦE through the surface of this cube is true? (a) ΦE = 0 (b) ΦE ∝ 2a2 a a (c) ΦE ∝ 6a2 r r • The electric flux through the surface is defined by: Φ ≡ ∫ E • dA r r • ∫ E • dA is ZERO on the four sides that are parallel to the electric field. r r • E • dA on the bottom face is negative. (dA is out; E is in) ∫r r • ∫ E • dA on the top face is positive. (dA is out; E is out) • Therefore, the total flux through the cube is: r r Φ ≡ ˜ E • dS +Φ + Φtop = 0 − Ea 2 + Ea 2 = 0 A =Φ sides bottom Exercise UI3A2 2B • Consider 2 spheres (of radius R and 2R) drawn around a single charge as shown. – Which of the following statements about the net electric flux through the 2 surfaces (Φ2R and ΦR) is true? (a) ΦR < Φ2R (b) ΦR = Φ2R (c) ΦR > Φ2R R 2R • Look at the lines going out through each circle -- each circle has the same number of lines. • The electric field is different at the two surfaces, because E is proportional to 1 / r 2, but the surface areas are also different. The surface area of a sphere is proportional to r 2. • Since flux = ∫ r r E • d A , the r 2 and 1/r 2 terms will cancel, and the two S circles have the same flux! • There is an easier way. Gauss’ Law states the net flux is proportional to the NET enclosed charge. The NET charge is the SAME in both cases. • But, what is Gauss’ Law ??? --You’ll find out next lecture! c a b r r qenclosed ∫ E • dA = Φ E = ε0 Fundamental Law of Electrostatics • Coulomb’s Law Force between two point charges OR • Gauss’ Law Relationship between Electric Flux and charges Gauss’ Law • Gauss’ Law (a FUNDAMENTAL LAW): The net electric flux through any closed surface is proportional to the charge enclosed by that surface. r r qenclosed ∫ E • dA = Φ E = ε0 • How do we use this equation?? • The above equation is ALWAYS TRUE but it may not be easy to use. • It is very useful in finding E when the physical situation exhibits massive SYMMETRY. UI4PF4: dA 1 2) A positive charge is contained inside a spherical shell. How does the electric flux dФE through the surface element dA change when the charge is moved from position 1 to position 2? a) dФE increases b) dФE decreases c) dФE doesn’t change dA 2 dA 1 3) A positive charge is contained inside a spherical shell. How does the flux ФE through the entire surface change when the charge is moved from position 1 to position 2? a) ФE increases ФE decreases c) ФE doesn’t change b) dA 2 dA 1 3) A positive charge is contained inside a spherical shell. How does the flux ФE through the entire surface change when the charge is moved from position 1 to position 2? a) ФE increases ФE stays the same c) ФE = 0 b) dA 2