PHY 303L-Exam2 - midterm 02 VARELA, CHRISTOPHER Due: Oct 18...

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midterm 02 – VARELA, CHRISTOPHER – Due: Oct 18 2007, 11:00 pm 1 E & M - Basic Physical Concepts Electric force and electric Feld Electric force between 2 point charges: | F | = k | q 1 || q 2 | r 2 k =8 . 987551787 × 10 9 Nm 2 /C 2 C 0 = 1 4 π k . 854187817 × 10 - 12 C 2 /N m 2 q p = - q e =1 . 60217733 (49) × 10 - 19 C m p . 672623 (10) × 10 - 27 kg m e =9 . 1093897 (54) × 10 - 31 kg Electric Feld: E E = > F q Point charge: | E | = k | Q | r 2 , E E = E E 1 + E E 2 + ··· ±ield patterns: point charge, dipole, ± 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 ²ux and Gauss’ law ±lux: ΔΦ = E Δ A = E E · ˆ n Δ A Gauss law: Outgoing Flux from S, Φ S = Q enclosed < 0 Steps: to obtain electric ±eld –Inspect E E pattern and construct S –Find Φ s = ± surface E E · d E A = Q encl < 0 , solve for E E Spherical: Φ s =4 π r 2 E Cylindrical: Φ s =2 π r D E Pill box: Φ s = E Δ A , 1 side; = 2 E Δ A , 2 sides Conductor: E E in = 0, E ± surf = 0, E surf = σ surf < 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 = kQ r V = V 1 + V 2 = Energy of a charge-pair: U = kq 1 q 2 r 12 Potential di³erence: | Δ V | = | E Δ s ± | , Δ V = - E E · Δ Es , V B - V A = - ² B A E E · dEs E = - dV dr , E x = - Δ V Δ x ³ ³ ³ fix y,z = - ∂V ∂x , etc. Capacitances Q = CV 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 Ed = < 0 A d Energy: U = ² Q 0 V dq = 1 2 Q 2 C , u = 1 2 C 0 E 2 Dielectrics: C = κC 0 , U κ = 1 2 κ Q 2 C 0 , u κ = 1 2 C 0 κ E 2 κ Spherical capacitor: V = Q 4 π< 0 r 1 - Q 4 0 r 2 Potential energy: U = - E p · E E Current and resistance Current: I = dQ dt = nq v d A Ohm’s law: V = IR , E = ρJ E = V = , J = I A , R = ρ= A Power: P = IV = V 2 R = I 2 R Thermal coe´cient of ρ : α = Δ ρ ρ 0 Δ T Motion of free electrons in an ideal conductor: = v d qE m τ = J nq ρ = m 2 τ Direct current circuits V = Series: V = eq = 1 + 2 + 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 Kirchho²’s Rules –Label currents: i 1 ,i 2 3 ,... –Node equations: i in = i out –Loop equations: ( ²E )+ ( iR )=0” –Natural: “+” for loop-arrow entering - terminal - ” for loop-arrow-parallel to current ³ow RC circuit: if dy dt + 1 RC y = 0, y = y 0 exp( - t ) Charging: E - V c - Ri = 0, 1 c dq dt + R di dt = i c + R dt =0 Discharge: 0= V c - = q c + R dt , i c + R dt Magnetic Feld and magnetic force μ 0 π × 10 - 7 Tm / A Wire: B = μ 0 i 2 Axis of loop: B = μ 0 a 2 i 2( a 2 + x 2 ) 3 / 2 Magnetic force: E F M = i E D × E B qEv × E B Loop-magnet ID: = i E A × E B , = iA ˆ n Circular motion: F = mv 2 r = q v B , T = 1 f = 2 v Lorentz force: E F = q E E + × E B Hall e³ect: V H = F M d q , U = - · E B Sources of E B and magnetism of matter Biot-Savart Law: Δ E B = μ 0 4 π i Δ > = × ˆ r r 2 , B = μ 0 4 π q>v × ˆ r r 2 Δ B = μ 0 4 π i Δ y r 2 sin θ , sin θ = a r , Δ y = r 2 Δ θ a Ampere’s law: M = ± L E B · dEs = μ 0 I encircled Steps: to obtain magnetic ±eld –Inspect E B pattern and construct loop L –Find M and I encl , and solve for E B .
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PHY 303L-Exam2 - midterm 02 VARELA, CHRISTOPHER Due: Oct 18...

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