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Announcements. Electric Field from a Ring of Charge. Consider symmetry: all perpendicular fields cancel, only parallel components (cos q ). r. z. R. Linear charge density . Electric Field from a Ring of Charge. r. z. R. Linear charge density . Electric Field.
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Electric Field from a Ring of Charge Consider symmetry: all perpendicular fields cancel, only parallel components (cos q) r z R Linear charge density
Electric Field from a Ring of Charge r z R Linear charge density
Electric Field Suppose I put a charge close to the ring: Describe the motion of the particle: In what direction does the particle move? Does the particle move at a constant velocity ? Where does the particle stop ? r z q R Linear charge density
q2 = -1 C 10 cm 5 cm q1 = +1 C Dipoles • A dipole is a postive and negative charge separated by a distance d • Commonly found in molecules! Though the distances and charges are much smaller! Dipole moment is qd. It is a vector!
E-field from a Dipole: Limits If z is large, ie, far away from the dipole The field from a dipole weakens as one goes further away faster than a point charge.
q2 = -1 C q1 = +1 C E-field from a Dipole Dipole Axis d/2 What is the field along the dipole axis? d/2 Field from Point charge A dipole is just two point charges in a specific arrangment.
q2 = -1 C q1 = +1 C E-field from a Dipole II Field from 2 Point charges Dipole Axis z Field along the dipole axis d/2 d/2 Some algebra
Dipoles in an field - + In a uniform field, the force on each end is equal but opposite; no net force [as the charges are connected!] In a nonuniform field, the force on each end is not equal but opposite; So a net force
Dipoles in an field In a uniform field, the force on each end is equal but opposite; no net force [as the charges are connected!] - + There can be a torque!! If the dipole moment makes an angle with the field, t=dFsinq, where F is the force on 1 charge, and so t=dqEsinq=pEsinq What is the lever
Electric Flux • Electric Flux is the amount of electric field flowing through a surface • When electric field is at an angle, only the part perpendicular to the surface counts E • Multiply by cos En • For a non-constant electric field, or a curvy surface, you have to integrate over the surface E = EnA= EA cos • Usually you can pick your surface so that the integration doesn’t need to be done given a constant field
R Electric Flux • What is electric flux through surface surrounding a charge q? charge q Answer is always 4keq
Gauss’s Law charge q • Flux out of an enclosed region depends only on total charge inside A positive charge q is set down outside a sphere. Qualitatively, what is the total electric flux out of the sphere as a consequence? A) Positive B) Negative C) Zero D) It is impossible to tell from the given information
R Gauss’ Law and Coulumb’s • Suppose we had • measured the flux as: • From Gauss’ law: So Gauss’ law implies Coulomb’s law charge q • What if we lived in a Universe with a different number of physical dimensions?
Gauss’s Law charge q’’ charge q’ charge q
Gauss’s Law charge q charge 2q charge -2q charge -q How do we draw surfaces to contain the +2q charge and have flux?: Zero ? +3q/e0? -2q/e0 ? IMPOSSIBLE
Example • q/eo D) 2q/eo • -q/eo • 0 Figure 24-29. • What is the flux through the first surface? • What is the flux through the second surface? • What is the flux through the third surface? • What is the flux through the fourthsurface? • What is the flux through the fifth surface?
Practive Problem I A cube with 1.40 m edges is oriented as shown in the figure • Suppose there is a charge situated in the middle of • the cube. • What is the magnitude of the flux through the whole cube? • What is the magnitude of the flux through any one side? • q/eo D) q/6eo • q/4eo • 0
Practice Problem II A cube with 1.40 m edges is oriented as shown in the figure • Suppose the cube sits in a uniform electric field of 10i ? • What is the magnitude of the flux through the whole cube? • What is the magnitude of the flux through the top side? • How many sides have nonzero flux? • q/eo D) q/6eo • q/4eo • 0 • 2 D) 1 • 4 • 0