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PHY 303L-Exam1 - midterm 01 VARELA CHRISTOPHER Due 11:00 pm...

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midterm 01 – VARELA, CHRISTOPHER – Due: Sep 20 2007, 11:00 pm 1 E & M - Basic Physical Concepts Electric force and electric field Electric force between 2 point charges: | F | = k | q 1 | | q 2 | r 2 k = 8 . 987551787 × 10 9 N m 2 /C 2 ǫ 0 = 1 4 π k = 8 . 854187817 × 10 12 C 2 /N m 2 q p = q e = 1 . 60217733 (49) × 10 19 C m p = 1 . 672623 (10) × 10 27 kg m e = 9 . 1093897 (54) × 10 31 kg Electric field: vector E = vector F q Point charge: | E | = k | Q | r 2 , vector E = vector E 1 + vector E 2 + · · · Field patterns: point charge, dipole, bardbl plates, rod, spheres, cylinders, . . . Charge distributions: Linear charge density: λ = Δ Q Δ x Area charge density: σ A = Δ Q Δ A Surface charge density: σ surf = Δ Q surf Δ A Volume charge density: ρ = Δ Q Δ V Electric flux and Gauss’ law Flux: ΔΦ = E Δ A = vector E · ˆ n Δ A Gauss law: Outgoing Flux from S, Φ S = Q enclosed ǫ 0 Steps: to obtain electric field –Inspect vector E pattern and construct S –Find Φ s = contintegraltext surface vector E · d vector A = Q encl ǫ 0 , solve for vector E Spherical: Φ s = 4 π r 2 E Cylindrical: Φ s = 2 π r ℓ E Pill box: Φ s = E Δ A , 1 side; = 2 E Δ A , 2 sides Conductor: vector E in = 0, E bardbl surf = 0, E surf = σ surf ǫ 0 Potential Potential energy: Δ U = q Δ V 1 eV 1 . 6 × 10 19 J Positive charge moves from high V to low V Point charge: V = k Q r V = V 1 + V 2 = . . . Energy of a charge-pair: U = k q 1 q 2 r 12 Potential difference: | Δ V | = | E Δ s bardbl | , Δ V = vector E · Δ vectors , V B V A = integraltext B A vector E · dvectors E = dV dr , E x = Δ V Δ x vextendsingle vextendsingle vextendsingle fix y,z = ∂V ∂x , etc. Capacitances Q = C V Series: V = Q C eq = Q C 1 + Q C 2 + Q C 3 + · · · , Q = Q i Parallel: Q = C eq V = C 1 V + C 2 V + · · · , V = V i Parallel plate-capacitor: C = Q V = Q E d = ǫ 0 A d Energy: U = integraltext Q 0 V dq = 1 2 Q 2 C , u = 1 2 ǫ 0 E 2 Dielectrics: C = κC 0 , U κ = 1 2 κ Q 2 C 0 , u κ = 1 2 ǫ 0 κ E 2 κ Spherical capacitor: V = Q 4 π ǫ 0 r 1 Q 4 π ǫ 0 r 2 Potential energy: U = vector p · vector E Current and resistance Current: I = dQ dt = n q v d A Ohm’s law: V = I R , E = ρJ E = V , J = I A , R = ρℓ A Power: P = I V = V 2 R = I 2 R Thermal coefficient of ρ : α = Δ ρ ρ 0 Δ T Motion of free electrons in an ideal conductor: a τ = v d q E m τ = J nq ρ = m nq 2 τ Direct current circuits V = I R Series: V = I R eq = I R 1 + I R 2 + I R 3 + · · · , I = I i Parallel: I = V R eq = V R 1 + V R 2 + V R 3 + · · · , V = V i Steps: in application of Kirchhoff’s Rules –Label currents: i 1 , i 2 , i 3 , . . . –Node equations: i in = i out –Loop equations: ( ±E ) + ( iR )=0” –Natural: “+” for loop-arrow entering terminal ” for loop-arrow-parallel to current flow RC circuit: if dy dt + 1 RC y = 0, y = y 0 exp( t RC ) Charging: E − V c R i = 0, 1 c dq dt + R di dt = i c + R di dt = 0 Discharge: 0 = V c R i = q c + R dq dt , i c + R di dt = 0 Magnetic field and magnetic force μ 0 = 4 π × 10 7 T m / A Wire: B = μ 0 i 2 π r Axis of loop: B = μ 0 a 2 i 2( a 2 + x 2 ) 3 / 2 Magnetic force: vector F M = i vector × vector B q vectorv × vector B Loop-magnet ID: vector τ = i vector A × vector B , vectorμ = i A ˆ n Circular motion: F = mv 2 r = q v B , T = 1 f = 2 π r v
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