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iClicker Quiz (1) I have completed at least 50% of the reading and studyguide assignments associated with the lecture, as indicated on the course schedule. a) True b) False iClicker Quiz I have a copy of the text "Physics for Scientists and Engineers Vol 2" by Serway, and I have purchased the current course packet and the Formulas & Chapter Summaries booklet from the bookstore. a) True b) False iClicker Quiz I registered my quiz transmitter via the course website (not on the iclicker.com website). I realize that until I do so, my quiz scores will not be recorded. a) True b) False iClicker Quiz I have downloaded and studied my personal Physics 220 homework data sheet and the numeric response guide, and now understand how to correctly submit online homework responses. a) True b) False iClicker Quiz A cat slides down a rubber pole and falls into metal pail A, which rests on a wooden shelf. The impact breaks the shelf, causing metal pails B and C, which were in contact, to separate and fall to the floor. What is the final charge on pail C? a) Positive b) Negative c) Zero Coulomb’s torsion balance E31 E32 At what position x will the Coulomb forces acting on the charged gray bead be balanced? F=qE F E + F Depending on the sign of the charge, the force experienced by a charged particle in an electric field can either point with or against the field. Fields due to individual point charges Fields due to multiple charges iClicker Quiz The field magnitude is greatest smallest at at point: point:A, A, B, B, C? C? Electric monopoles We sometimes refer to the total charge Q of an extended charged object as its “monopole moment”. When you back far enough away, any charged object, regardless of its shape, looks like a simple point charge. In the expression for the electric field of a monopole, the vector r originates from the “center of charge”. kQrˆ E 2 r Electric dipoles r Equal and opposite charges of magnitude Q = |Q+| = |Q|. r p r r+ points from the negative charge to the positive charge. We now ask “What is the electric field produced by a dipole?” Define dipole moment: p Qr Like r+, the dipole moment vector p always points from the negative charge to the positive charge. E E E kQr k (Q)r 3 3 r r k 3(p rˆ )rˆ p r3 Limit : r 0 and Q such that p Qr is constant. Moments of a distribution (extra details for purists) (If you tell anyone that I mentioned this, I’ll deny it!) Monopole moment (scalar) q (r )dV dQ r Dipole moment (vector) p r (r ) dV Quadrupole moment (polar rank-2 tensor) ~ ~ Q (3 r r r 2 I ) (r ) dV Choose origin at the center of charge (like center of mass) The electric field of a charge distribution can be expressed as a series expansion involving successively higher moments of the distribution. ~ ~ qrˆ 3(p rˆ )rˆ p 5(rˆ Q rˆ ) Q rˆ Ek 2 k k ... 3 4 r r r Off-axis dipole field kQr k (Q)r kQ(r r ) E E E 3 r r3 r3 Q(r r ) a ˆi y ˆj a ˆi y ˆj Q(2a ˆi ) p For r a , r y kp kp E 3 3 r y r E p r p Serway 23.6 On-axis dipole field d = 2a +Q p E -Q p E x E E E kQr k (Q)r 3 3 r r 1 1 1 1 ˆ ˆ kQi 2 2 kQi 2 2 r ( x a) ( x a) r kQˆi 2 2 2 1 a / x 1 a / x x kQˆi 2a 2a 2 1 1 x x x 4a 2k (2aQˆi ) 2k p ˆ 3 kQi 3 3 x x x Binomial approximation (1 s) p 1 ps if s 1 Try using the general dipole formula (only for the purists among you) 3(p rˆ )rˆ p Ek r3 Perpendicular case On-axis case p p ˆi and rˆ ˆj 3( pˆi ˆj)ˆj ( pˆi ) Ek y3 0ˆj pˆi kp k 3 3 y y p p ˆi and rˆ ˆi 3( pˆi ˆi )ˆi pˆi Ek x3 2kpˆi 2kp 3 3 x x See interactive applet at http://lectureonline.cl.msu.edu/~mmp/applist/applets.htm Calculating the electric force/field due to a continuous charge distribution r qdQ r dF k q r3 dQ r dQ dQ r dE k 3 r qQ F dF ke L dx L / 2 (r x)2 L/2 L/2 qQ 1 ke L r x L / 2 Q dQ dx dx L ke qdQ qQ dx dF k ( r x) 2 L ( r x) 2 ke qQ 1 1 L r L / 2 r L / 2 L L (r ) (r ) qQ 2 2 ke L (r L )( r L ) 2 2 qQ ke 2 r L2 / 4 Find E at the center of a hemispherical shell of uniform charge density. The electricity vs. gravity analogy Coulomb’s Electric Force Law Q r q kQrˆ E 2 r kqQrˆ F qE 2 r qE F ma a m Newton’s Gravitational Force Law M r m GMrˆ g 2 r GmMrˆ F mg r2 F ma a g E-field perpendicular to conducting surfaces E-field perpendicular to conducting surfaces E = /0 = 4πk Electric field strength at the surface of a charged conductor is proportional to local surface charge density. F qE ma qE a m vx vx 0 axt vx 0 qE v y v y 0 a yt t { y0 0, v y 0 0} m qE 2 y y0 v y 0 t a y t t 2m 1 2 2 a constant v v0 at x x0 v0t at 1 2 2 v v 2a ( x x0 ) 2 2 0 F qE WE F ds q E ds Uniform field : E ds E ds E s E x K WE qEx