lecture13 - September 27, 2011 Physics for Scientists &...

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September 27, 2011 Physics for Scientists & Engineers 2, Chapter 25 1
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September 27, 2011 Physics for Scientists & Engineers 2, Chapter 25 2 F=q E F x =qE x E x = F x q = Fcos(62 o ) q = 54 10 6 cos(62 ) 3.1 10 6 = 8.178 N/C [ F x ] = F cos(62 o ) (Textbook, eq. 22.1, definition of E field) (x and y components of vector, many HW problems)
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September 27, 2011 Physics for Scientists & Engineers 2, Chapter 25 3 E x = V x = 2 x y 2 E y = V y = 2 xy z E z = V z = y ( x , y , z ) = ( 3,9,9) E x = 6 81 = 75; E y = 54 9 = 45; E z = 9 [ V / m ] E = E x 2 + E y 2 + E z 2 = 7731 = 87.93 V / m Textbook, eq. 23.15 and example in class Magnitude of E when components are given: Many HW problems
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September 27, 2011 Physics for Scientists & Engineers 2, Chapter 25 4 E = ρ r 3 ε 0 = 2.79 10 6 0.46 3 8.85 10 11 = 4833.9 V / m Textbook, eq. 22.20 and done in class Plug in numbers – done!
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September 27, 2011 Physics for Scientists & Engineers 2, Chapter 25 5 τ = pE sin( θ ); is max. at =90 Figure: max = ( = 90 ) = 114 10 28 Nm max = pE sin(90 ) p = max E = 141 10 28 14 = 1.007 10 27 Cm Textbook, eq. 22.11 torque was done in class
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September 27, 2011 Physics for Scientists & Engineers 2, Chapter 25 6 V B V A = kq 1 r B 1 r A = 8.99 10 9 2.43 10 6 1 2.03 1 5.15 = 6519.5 V = 6.520 kV Textbook, eq. 23.12 and done in class. For r>R, E field of uniformly charged sphere is the same as from point charge (eq. 22.22) -> potential must be same as for a point charge Equipotential lines are concentric circles (2d), Potential is the same for each point on a given circle
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September 27, 2011 Physics for Scientists & Engineers 2, Chapter 25 7 6. Equi.pot. lines at i closer together than at j (E=V/d, with d smaller at i) 7. Q1 is negative since the potential is negative 8. q=Vr/k – look at |V| =5V |Q 3 |>|Q 1 |>|Q 2 | because r 3 >r 1 >r 2 9. Same argument as above: |Q 2 |<|Q 1 | 10. k: E=|-1V-1V|/1.4cm not equal to 0 11. E field on + charge is in direction of force. E field points away from Q 3 force points up toward Q 2 r 1 r 3 r 2 Textbook, Example 23.6 Last problem of HW3 and done in class
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September 27, 2011 8 Yes, flux is 0 inside! Q i Q o Q = 1.6 Φ 0 ε 0 < 0 Q i + Q = 0.6 Φ 0 0 Q i = 2.2 Φ 0 0 > 0 Q + Q i + Q o = 0 Q o = 0.6 Φ 0 0 < 0 No, flux is NOT 0 inside! No, inner shell is positive No, see calculation above No, outer shell is negative No, if q=0 at r 3 , ϕ (r 3 )= ϕ (r 2 ) required – not the case, ϕ changes at r 3 No, outer shell is negatively charged If q was uniformly distributed, ϕ would increase between r 1 and r 2 Textbook, eq. 22.15 Law of Gauss (and second last problem of HW3)
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So far: Stationary electric charges and ±elds Electric circuits were introduced in the discussion of capacitors in Chapter 24, but it covered only situations involving fully charged capacitors, where the charge is at rest e world-changing impact of electricity is due to the properties of charges in motion , or electric current All electrical devices rely on some kind of current for their operation September 27, 2011 Physics for Scientists & Engineers 2, Chapter 25 9
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This note was uploaded on 02/03/2012 for the course MTH 235 taught by Professor Staff during the Fall '11 term at Michigan State University.

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lecture13 - September 27, 2011 Physics for Scientists &...

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