Electric Flux Phys 122 Lecture 4
y Last class: Electric Dipole +Q a a -Q r x Case of special interest: (antennas, molecules) r >> a For pts along x-axis: For pts along y-axis: For r >>a, For r >>a,
y +Q a a -Q Last class: Electric Dipole Summary r Case of special interest: x (antennas, molecules) r >> a Along x-axis Along y-axis Along arbitrary angle
Electric Field Distribution Summary Dipole ~ 1 / R 3 Point Charge ~ 1 / R 2 Infinite Line of Charge ~ 1 / R
Outline Electric field acting on charges Defining Electric Flux A measure of flow of the electric field through a given area. A brief introduction to Gauss Law Some of your Concerns Looking forward to more hand written walkthroughs of the word problems. One of the slides said that the electric field is everywhere and at every direction as it is represented by the vector field. So how could it be denser and less dense when it is represented as field lines? The idea of flux was extremely confusing. I do not understand all of the derivations of N and epsilon naught that they went through. What is an insulator vs what is a conductor? How do they affect charges differently? Why is Gauss's law useful practically?
New Stuff: A moving charge Recall from mechanics and force on q in E Acceleration of a charge is then: F = ma F = qe a = (q/m)e Tipler Problem: A particle beam of protons, each with a KE = 3.25 10-15 J, is stopped by a uniform electric field in 2 m. A proton has: m p = 1.673 10-27 kg q p = 1.602 10-19 C What is E? 1. Given the kinetic energy KE = ½ mv 2 2. 1-D kinematics, uniform a; stops in distance x: 2ax = v 2 3. Substitute: ax = KE/m 4. Recall from above that a = (q/m)e 5. Therefore qex = KE 6. Or, E = KE/(qx) 7. You know all these, so only now calculate: E = 10143 N/C
Clicker An electron is fired upward at speed v 0 toward the top of the page as shown. A uniform electric field points to the right. Which trajectory best represents its motion: (A) (B) (C) (D) (E) Vertical speed unchanged; no vertical force Horizontal force: F x = qe x = ma Recall constant acceleration parabolic trajectory in x Which way? Electron is NEGATIVE charge so sign of F x is opposite to sign of E
Clicker: Torque on a dipole An electric dipole consists of two equal and opposite charges, fixed a distance 2a apart. It is placed in a uniform electric field as shown. It will A. Not move at all B. Translate horizontally C. Translate vertically D. Start to rotate clockwise E. Start to rotate counter-clockwise Consider the Forces on the charges: equal in magnitude opposite in direction Dipole will not translate Displaced compared to center of mass non-zero torque Dipole will initially rotate Clockwise
Torque on a dipole Electric dipole moment: q = charge = vector from to + charge Units: C-m Torque on dipole in uniform field Aligns dipole with field Potential energy of dipole is usually 0 Aside: UW Hg-199 EDM experiment Test of new physics; most sensitive in the world
Fundamental Law of Electrostatics Coulomb's Law Force between two point charges OR Gauss' Law Relationship between Electric Fields and charges Completely equivalent for static fields. For time-dependent only Gauss correct
Electric Dipole Field Lines Consider imaginary spheres centered on : a) +q (green) b) -q (red) c) midpoint (yellow) c a All lines leave a All lines enter b Equal number leaves or enters sphere c b
Quantify Electric Flux Define: electric flux through the closed surface S The integral is over the entire CLOSED SURFACE The result (the net electric flux) is a SCALAR quantity depends on the component of E which is NORMAL to the SURFACE (normal means it points OUT, ds is positive ) Net Electric flux is the sum of the normal components of the electric field all over the closed surface. The direction of the normal component of E as it penetrates the surface can point IN or it can point OUT. IN is NEGATIVE OUT is POSITIVE
Direction Matters: What might be in this box???? E E da da da E da E E For a closed surface, da points outward E E da E S E da 0 S
CheckPoint: Electric Flux and Field Lines An infinitely long charged rod has uniform charge density of λ, and passes through a cylinder (gray). The cylinder in case 2 has twice the radius and half the length compared to the cylinder in case radius 1. Compare the magnitude of the flux, Φ, through the surface of the cylinder in both cases. Take s to be the radius 1 2 2 1 2 (A) (B) 1 1/2 2 (C) none (D)
CheckPoint: Electric Flux and Field Lines Definition of Flux: E da Case 1 E1 2 s 0 surface E constant on barrel of cylinder E perpendicular to barrel surface (E parallel to da) E da EA barrel 1 barrel A1 2sL L 0 A E 2 1 2 2 1 2 (A) (B) Case 2 2 0 (2s) (2 (2s)) L / 2 2sL 2 2 1 1/2 2 (C) ( L / 2) 0 none (D) RESULT: GAUSS LAW proportional to charge enclosed!
Density of lines: clearly E A < E C, E A < E B E C largest because closer to +q and also at C fields add up.
Q gets closer to da E A gets larger. Still true that flux over whole sphere does not change.
Clicker Imagine a cube of side a positioned in a region of constant electric field as shown. Which of the following statements about the electric flux E through the surface of this cube is true? a a (a) E = (b) E 2a (c) E 6a The electric flux through the surface is defined by: is ZERO on the four sides that are parallel to the electric field. on the bottom face is negative. (ds is out; E is in) on the top face is positive. (ds is out; E is out) Therefore, the total flux through the cube
Clicker Consider 2 spheres (of radius R and 2R) drawn around an electric dipole Which of the following statements about the net electric flux through the 2 surfaces ( 2R and R ) is true? (a) R < 2R (b) R = 2R (c) R > 2R R 2R Calculating flux here is a challenge. no ready symmetry Look at the inner circle only Number of lines going out is equal to the number coming in. It means the net flux is zero Look at outer circle. Less clear. Lines not extended in drawing. Imagine them. Same would occur. The net flux is zero. Easier way: Gauss Law states the net flux is proportional to the NET enclosed charge. The NET charge is ZERO in both cases. We need to know about Gauss Law!!
Gauss' Law Gauss' Law (a FUNDAMENTAL Law): The net electric flux through any closed surface is proportional to the charge enclosed by that surface. Always TRUE but not always easy to use Useful when situation exhibits SYMMETRY Choose a closed surface such that the integral is trivial Direction: chose a surface so that E is parallel or perpendicular to each piece of the surface Magnitude: chose surface where E has the same value at all points on the surface when E is perpendicular to the surface. Therefore: then, bring E outside of the integral
Geometry and Surface Integrals If E is constant over a surface, and normal to it everywhere, we can take E outside the integral, leaving only a surface area z x a b c y z R R L
Next Time we ll use this powerful tool to solve many problems Stay Tuned