241-22_lec-4

# 241-22_lec-4 - Physics 241 Lecture 4 Y E Kim September 2...

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Physics 241 Lecture 4 Y. E. Kim September 2, 2010 September 3, 2010 University Physics, Chapter 22 1

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September 3, 2010 University Physics, Chapter 22 3 Review - Point Charge The electric field created by a point charge, as a function of position r, 2 0 1 ˆ () 4 q E r r r  The force exerted by an electric field on a point charge q located at position r F qE r … direction: tangent to the field line

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September 3, 2010 University Physics, Chapter 22 4 Review Electric Dipole 2 equal but opposite charges, -q and +q Dipole moment (direction: - to +) On the axis, far from the dipole, p qd E p 2  0 r 3
September 3, 2010 University Physics, Chapter 22 5 Forces on an Electric Dipole Place an electric dipole in a uniform electric field The dipole is composed of two equal but opposite charges, +q and -q located a distance d apart The electric field is constant and points in the upward direction The electric dipole moment makes an angle with the electric field The net force on the dipole is zero! The dipole experiences a torque about its center

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September 3, 2010 University Physics, Chapter 22 6 Torque on an Electric Dipole The torque is given by Using the dipole moment : Torque vector: torque F 1 moment arm F 2 moment arm  qE d 2 sin qE d 2 sin qEd sin pE sin moment arm sin 2 d p 
September 3, 2010 University Physics, Chapter 22 7 Gauss’ Law Objective: So far, we have considered point charges How can we treat more complicated distributions, e.g., the field of a charged wire, a charged sphere, or a charged ring? Two methods: Method #1: Divide the distribution into infinitesimal elements dE and integrate to get the full electric field Method #2: If there is some special symmetry of the distribution, use Gauss’ Law to derive the field Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed by the surface. First we need to define the concept of flux

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September 3, 2010 University Physics, Chapter 22 8 Water Flux Let’s imagine that we put a ring with area A perpendicular to a stream of water flowing with velocity v The product of area times velocity, Av, gives the volume of water passing through the ring per unit time The units are m 3 /s If we tilt the ring at an angle , then the projected area is Acos , and the volume of water per unit time flowing through the ring is Av cos .
September 3, 2010

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## This note was uploaded on 07/15/2011 for the course PHYS 241 taught by Professor Wei during the Summer '08 term at Purdue.

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241-22_lec-4 - Physics 241 Lecture 4 Y E Kim September 2...

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