September 6, 2013
Physiology (function, how stuff happens) vs. Anatomy (structure)
Teleological (why) vs. mechanistic (how)
Casual Chains: _ -> _ -> _ -> etc.
Tissues: group of cells with similar
Bio 242 Exam 1 Study Guide
Describe the organization of life (Cell->tissue->etc) and describe the characteristics
Cell: smallest unit of life
Tissues: organized group of cells with a similar structure and function
Organs: organized in such a way
3Mechanist Approach: To answer how something occurs using causal chains, or
cause and effect sequences.
Teleological Approach: To answer questions as to why something occurs.
Levels of Organization
Basic unit of life is the cell (also the smallest unit of
Particle Projected Perpendicular to Uniform Electric Field
A charged particle (m = 3kg, q = 1C) is launched at t0 = 0 with initial speed v0 = 2m/s in an electric field of magnitude E = 6 106 N/C as shown.
y m q 1m 1m
(a) Find the position of the pa
Electric Dipole Potential
Use spherical coordinates: V = V (r, ) independent of azimuthal coordinate . , r- - r+ (-q) q = kq Superposition principle: V = V+ + V- = k + r+ r- r- r+ Large distances (r L): r- - r+ L cos , r- r+ r2 V (r, ) k qL cos
Action and Reaction due to Coulomb Interaction
Two particles with masses m1 , m2 and charges q1 , q2 are released from rest a distance r apart. We consider the following four distinct configurations: (a) m1 = 1kg, m2 = 1kg, q1 = 1C, q2 = 1C (b) m1 =
Particles Accelerated by Uniform Electric Field
A uniform electric field E = 0.75 103 N/C exists in the box. (a) A charged particle of mass m1 = 1.9 10-9 kg is released from rest at x = 3cm, y = 0. It exits the box at x = 3cm, y = 6cm after a time
Electric Potential of Conducting Spheres (1)
A conducting sphere of radius r1 = 2m is surrounded by a concentric conducting spherical shell of radii r2 = 4m and r3 = 6m. The graph shows the electric field E(r). (a) Find the charges q1 , q2 , q3 on th
Electric Field of a Point Charge
exerts force locally
exerts force over distance
(1) Electric field E generated by point charge q: E = k
q r ^ r2
(2) Force F1 exerted by field E on poi
When any closed-circuit loop is traversed, the algebraic sum of the changes in electric potential must be zero.
At any junction in a circuit, the sum of the incoming currents must equal the sum of the outg
Conducting cylinder of radius a and length L surrounded concentrically by conducting cylindrical shell of inner radius b and equal length. Assumption: L b.
: charge per unit length (magnitude) on each cylinder Q = L: magnitu
Charged Particle in Crossed Electric and Magnetic Fields (1)
Release particle from rest. Force: F = q(E + v B) dvx = -qvy B dt dvy = qvx B + qE (2) Fy = m dt Ansatz: vx (t) = wx cos(0 t) + ux , (1) Fx = m qB dvx =- vy dt m qB dvy qE = vx + dt
Method for dermining whether charge carriers are positively or negatively charged. Magnetic field B pulls charge carriers to one side of conducting strip. Accumulation of charge carriers on that side and depletion on opposite side produ
Capacitor and Capacitance
Two oppositely charged conductors separated by an insulator. The charges +Q and -Q on conductors generate an electric field E and a potential difference V (voltage). Only one conductor may be present.
Ampre's Law: Magnetic Field Inside a Toroid
Apply Ampre's law, I B d = 0 IC , to the circular Amperian loop shown.
Magnetic field inside: directed tangentially with magnitude depending on R only. Magnetic field outside: negligibly weak. Number o
Magnetic Force Application (3)
The dashed rectangle marks a region of uniform magnetic field B pointing out of the plane. Find the direction of the magnetic force acting on each loop with a ccw current I.
4 2 I 1 I 3 I B
NW W SW
NE E SE
Calculating the Resistance of a Wire
Uniform cross section
Length of wire: L Area of cross section: A Resistivity of material: Current density: J = Current: I = JA Voltage: V = EL Resistance: R E [A/m2 ]
Variable cross section
Resistors Connected in Parallel
Find the equivalent resistance of two resistors connected in parallel. Current through resistors: I1 + I2 = I Voltage across resistors: V1 = V2 = V Equivalent resistance: 1 1 1 = + R R1 R2 I I1 1 I2 = + R V V1 V2
Power Dissipation in Resistor
Consider a resistor in the form of a uniform wire. Voltage between ends: V Va - Vb = E(xb - xa ) Displaced charge: dq = Idt
dq E xa xb
Work done by electric field E on displaced charge dq: WE = F (xb -
Resistor Problem (1)
A heating element is made of a wire with a cross-sectional area A = 2.60 10 -6 m2 and a resistivity = 5.00 10-7 m. (a) If the element dissipates 5000W when operating at a voltage V 1 = 75.0V, what is its length L1 , its resist
Projectile Motion in Electric Field
electrostatic force: Fx = 0
Fy = -eE e E -a me vy (t) = v0 sin - at y(t) = v0 [sin ]t - 1 2 at 2
equation of motion: F = me a acceleration: ax = 0 velocity: vx (t) = v0 cos position: x(t) = v0 [cos ]t ay
Electric Field of Point Charges in Plane (1)
Determine magnitude of E1 and E1 and identify directions in plane: E1 = k|q1 | k|q2 | = 7.99N/C, E2 = = 4.32N/C. (3m)2 (5m)2
E E1 E2 3 2 1 q = +8nC
Determine x- and y-components of E1 and E1 and of th
Electric Field on Line Connecting Point Charges (1)
Consider the x-component of the electric field. Electric field at point P1 : E = E1 + E2 = Electric field at point P2 : E = E1 + E2 = kq1 kq2 - = 7.99N/C - 108N/C = -100N/C. (3m)2 (1m)2 kq2 kq1 +
Calculating E from Gauss' Law: Point Charge
Consider a positive point charge Q. Use a Gaussian sphere of radius R centered at the location of Q. Surface area of sphere: A = 4R2 . Electric flux through Gaussian surface: E = I E dA = E(4R2 ).
Vector Field and Electric Field Lines
The electric field is a vector field: ^ E(r) = E(x, y, z) = Ex (x, y, z)^ + Ey (x, y, z)^ + Ez (x, y, z)k i j Electric field lines: graphical representation of vector field. Properties of electric field lines
Electric Flux: Application (1)
Consider a rectangular sheet oriented perpendicular to the yz plane as shown and positioned in a uniform electric field E = (2^ j)N/C.
z 2m 3m 4m x
(a) Find the area A of the sheet. (b) Find the angle between A and E.
The net electric flux E through any closed surface is equal to the net charge Q in inside divided by the permittivity constant 0 : I Qin Qin i.e. E = with 0 = 8.85410-12 C2 N-1 m-2 EdA = 4kQin =
The closed surface can be real or ficti
Gaussian surface problem (1)
The electric fluxes through the Gaussian surfaces S A and SB are E E
= 3C/ 0 , respectively.
SA q1 q2
Find the electric charges q1 and q2 .
Gaussian surface problem (2)
Calculating E from Gauss' Law: Strategy
Design the Gaussian surface such that it reflects the symmetry of the problem at hand. Use concentric Gaussian spheres in problems with spherically symmetric charge distributions. The electric field is perpend