University:
University of Michigan
Course:
PHYSICS 250 | Fundamental Physics for the Life Sciences II
Academic year:

2020
Views:

181

Pages:

1
Author:

Marilyn James

Final Exam Equation sheet Powers of tenm = 10−3 Constants k = 9 × 109 N.m2 /C 2 me = 9.11 × 10−31 kg µ = 10−6 n = 10−9 p = 10−12 M = 106 ǫo = 8.85 × 10−12 C 2 /N.m2 g = 9.8 m/s2 e = 1.6 × 10−19 C mp = 1.67 × 10−27 kg µo = 4π × 10−7 T.m/A Electric Force and Electric Field q1 q2 • Magnitude of electric force between two charges q1 and q2 a distance r apart: F = k 2 r q • Magnitude of electric field at a distance r from a point charge q: E = k2 r Electric Flux ~ = qin E · dA ǫo ~ Gauss’s law: φE = Electric Potential and Energy Consevation I • At a distance r from a point charge q: V = k r V • For a uniform electric field: E = d q1 q2 , ∆U = q ∆V • Potential Energy: U = k r • For an isolated system: ∆U + ∆K = 0 Capacitance Q , • C= ∆V Energy stored: UE = • Parallel-plate: C = ǫo A , d Current and Resistance ρL ∆Q R= , Iavg = ∆t A Cross product q Q2 1 1 C ∆V 2 = Q ∆V = 2 2C 2 Parallel-plate energy density: uE = Current density: J = I A , 1 ǫE 2 o 2 Power: P = I ∆V = I 2 R = ∆V 2 R • Magnitude of cross product of two vectors:A| ~× ~B| = AB sin θ • ˆi × ˆ i = 0, ˆi × ˆj = ˆk, ˆj × ˆk = ˆi, ˆk × ˆi = ˆj Magnetic Field B • Magnetic force on charge q moving in a magnetic field: FB =~ q ~v × ~ ~ I ~L × ~B • Magnetic force on a current-carrying wire in a magnetic field: FB = • For a charge q moving in a uniform B, ~where B~ ⊥ ~v, radius of its circular path: R = • Velocity selector: v = mv qB E B µo I • Magnetic field at a distance ”a” from an infinitely long straight wire: B = 2πa µo I • Magnetic field due to a curved wire segment of radius ”a” at the center: B = θ 4πa • Magnetic force per unit length between two parallel wires carrying currents I1 and I2 , F B µo I 1 I 2 = seprated by a distance ”a”: 2πa l I • Ampère’s law: B · d~s = µo Iin ~