1 1 W02D1 Electric Dipoles and Continuous Charge Distributions Online Registration If you are in a M/W/F class for 8.02, you will need to register for the course “8.02r-MW Electricity and Magnetism (Monday and Wednesday)”. If you are in a Tuesday/Thursday/Friday class for 8.02, register for “8.02r-TTh: Electricity and Magnetism (Tuesday and Thursday)”. The following link will get you to either course and the web site will require certificates: https://lms.mitx.mit.edu/ Announcements Math Review Week Two Tues from 9-11 pm in 26-152 PS 1 Handwritten due Week Two Tuesday at 9 pm in boxes outside 32-082 or 26-152 PS 1 Online due Week Two Wednesday at 8:30 am Finalize Group Formations Make sure your clicker is registered 32 4 Outline Review: Electric Field Main Problem of Electrostatics: Electric Field Lines Electric Dipoles Force and Torque on Dipole Continuous Charge Distributions 5 Review: Electric Field due to Source Charges The electric field at a point P due to a charged object (source) with charge qs is the force acting on a test point-like charged object with charge qt at that point P, divided by the charge qt : Es(P) ≡Fst(P)qt Es(P) = keqsrst2ˆrs(P)Units: N/C 6 Types of Source Charges Positively charged point-like object Negatively charged point-like object Discrete number of charged point-like objects Example: Electric Dipole Continuous Distribution of charged objects Examples: Charged wire Charged disc Charged sphere Very large charged plane3 7 Concept Question: Electric Field Two charged objects are placed on a line as shown below. The magnitude of the negative charge on the right is greater than the magnitude of the positive charge on the left, . Other than at infinity, where is the electric field zero? 1. Between the two charged objects. 2. To the right of the charged object on the right. 3. To the left of the charged object on the left. 4. The electric field is nowhere zero. 5. Not enough info – need to know which is positive. qR> qL8 Force on Charged Object in Electric Field due to Source Charges Force on a charged object with charge q at a point P in an electric field due to a source is: Es Fq(P) = mqaq(P)qEs(P) = mqaq(P)9 Newton’s Second Law Force on a charged object with charge q at a point P in an electric field due to a source is: Fq(P) = mqaq(P)qEs(P) = mqaq(P) Es4 10 Electric Field Lines 1. Direction of field at any point is tangent to field line at that point 2. Field lines point away from positive charges and terminate on negative charges 3. Field lines never cross each other 11 Concept Question: Field Lines Electric field lines show: 1. Directions of forces that exist in space at all times. 2. Directions in which positive charges on those lines will accelerate. 3. Paths that charges will follow. 4. More than one of the above. 5. I don’t know. 12 Electric Dipole5 13 Nature Likes to Make Dipoles http://youtu.be/EMj10YIjkaY ! 14 Demonstration: Electric field Lines D16 15 Electric Dipole Two equal but opposite charges +q and –q, separated by a distance 2a p ≡ q2aˆj = 2qaˆjpoints from negative to positive charge prq -q 2a Dipole moment two charges prDipole moment for neutral charge distribution, N point charges p ≡ qii=1i= N∑ri6 16 Electric Field Created by Dipole Week 01 Friday Problem Solving Ex= keqΔxr+3−Δxr−3⎛⎝⎜⎞⎠⎟= keqxx2+ ( y − a)2⎡⎣⎤⎦3/2−xx2+ ( y + a)2⎡⎣⎤⎦3/2⎛⎝⎜⎜⎞⎠⎟⎟ Ey= keqΔy+r+3−Δy−r−3⎛⎝⎜⎞⎠⎟= keqy − ax2+ ( y − a)2⎡⎣⎤⎦3/2−y + ax2+ ( y + a)2⎡⎣⎤⎦3/2⎛⎝⎜⎜⎞⎠⎟⎟ ˆrr2=rr3=Δxr3ˆi +Δyr3ˆj17 Point Dipole Approximation r >> a Ex(r,θ) →3p4πε0r3sinθcosθ Ey(r,θ) →p4πε0r33cos2θ− 1( )Finite Dipole Point Dipole You can show… For distances 18 Concept Question: Dipole Field As you move to large distances r away from a dipole, the electric field will fall-off as: 1. 1/r2, just like a point charge 2. More rapidly than 1/r2 3. More slowly than 1/r27 19 Demonstration: Dipole in a Van de Graaff Generator D22 20 Dipole in Uniform Field E = Eˆi p = 2qa(cosθˆi + sinθˆj) Fnet=F++F−= qE + (−q)E = 0Total Net Force: Torque on Dipole: tends to align with the electric field p τ =r ×F = (2a)(qE)sin(θ) =p ×E τ= rF+sin(θ) = pE sin(θ)Charges fixed at ends of rod 21 Torque on Dipole Total Field (dipole + background) shows torque: • Field lines transmit tension • Connection between dipole field and constant field “pulls” dipole into alignment8 22 Concept Question: Dipole in Non-Uniform Field A dipole sits in a non-uniform electric field E E Due to the electric field this dipole will feel: 1. force but no torque 2. no force but a torque 3. both a force and a torque 4. neither a force nor a torque 23 Continuous Charge Distributions 24 ( )?P =ErV Continuous Charge Distributions Q = Δqii∑Break distribution into parts: ΔE = keΔqr2ˆrE field at P due to Superposition: E = ΔE∑→ dE∫ → dqV∫∫∫ → dE = kedqr2ˆr Δq9 25 Continuous Sources: Charge Density Length L=LwL Area = A = wLRLdLdQλ=dAdQσ=dVdQρ= ρ= Q / V (uniform) Volume = V =πR2L σ= Q / A (uniform) λ= Q / L (uniform)26 Group Problem: Charge Densities A solid cylinder, of length L and radius R, is uniformly charged with total charge Q. (a) What is the volume charge density ρ? (b) What is the linear charge density λ? (c) What is the relationship between these two densities ρ and λ? 27 Examples of Continuous Sources: Finite Line of Charge LQ=λLength L=LdLdQλ=E field on perpendicular bisector10 28 Examples of Continuous Sources: Finite Line of Charge LQ=λLength L=LdLdQλ=E field off axis 29 Examples of Continuous Sources: Finite Line of Charge LQ=λLength L=LdLdQλ=Grass seeds of total E field 30 Concept Question Electric Field of a Rod A rod of length L lies along the x-axis with its left end at the origin. The rod has a uniform charge density λ. Which of the following expressions best describes the electric field at the point P 1.E(P) = −keλd′x(′x + d )3′x = 0′x = L∫ˆi 2.E(P) = keλd′x(′x + d )3′x = 0′x = L∫ˆi 3.E(P) = −keλd′x(′x + d )2′x = 0′x =
View Full Document
Unlocking...