{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# midterm2 - midterm 02 LIN KEVIN Due Mar 6 2008 9:00 pm 1 E...

This preview shows pages 1–2. Sign up to view the full content.

midterm 02 – LIN, KEVIN – Due: Mar 6 2008, 9:00 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 0 = 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 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 = 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 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 = d Q 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 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 , . . . –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 d y dt + 1 R C y = 0, y = y 0 exp( t R C ) Charging: E − V c R i = 0, 1 c d q dt + R d i dt = i c + R d i dt = 0 Discharge: 0 = V c R i = q c + R d q dt , i c + R d i dt = 0 Magnetic field and magnetic force μ 0 = 4 π × 10 7 Tm / 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 = m v 2 r = q v B , T = 1 f = 2 π r v

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern