Homework 2

Homework 2 - homework 02 KANUNGO ANIL 1 E M Basic Physical...

• Homework Help
• 5

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

homework 02 – KANUNGO, ANIL 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 rℓE 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 κE 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 = nq 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: = 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 ,...

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

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