phys2exam2 - exam 02 – ANDERSON ZACH – Due Mar 6 2008...

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Unformatted text preview: exam 02 – ANDERSON, ZACH – Due: Mar 6 2008, 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 Nm 2 /C 2 ǫ = 1 4 π k = 8 . 854187817 × 10 − 12 C 2 /Nm 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 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 , solve for vector E Spherical: Φ s = 4 π r 2 E Cylindrical: Φ s = 2 π r ℓE 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 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 = − d V 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 A d Energy: U = integraltext Q V dq = 1 2 Q 2 C , u = 1 2 ǫ E 2 Dielectrics: C = κC , U κ = 1 2 κ Q 2 C , u κ = 1 2 ǫ κE 2 κ Spherical capacitor: V = Q 4 π ǫ r 1 − Q 4 π ǫ r 2 Potential energy: U = − vector p · vector E Current and resistance Current: I = d Q dt = nq v d A Ohm’s law: V = I R , E = ρJ E = V ℓ , J = I A , R = ρℓ A Power: P = I V = V 2 R = I 2 R Thermal coefficient of ρ : α = Δ ρ ρ Δ T Motion of free electrons in an ideal conductor: aτ = v d → q E m τ = J n q → ρ = m n q 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 ,......
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This test prep was uploaded on 04/10/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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phys2exam2 - exam 02 – ANDERSON ZACH – Due Mar 6 2008...

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