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# C 20 cm 20 cm y b c x a what is the

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Unformatted text preview: ges A and B? A) C) FA FB FA B) FB D) Physics 1B - Summer Session 1 - C. Palmer FA FB FB August 5, 2013 FA 9   Electric force directions   The force is attractive from A (+,–)Toward A   The force is repulsive from B (–, –)Away from B A) C) FA FB FA B) FB D) Physics 1B - Summer Session 1 - C. Palmer FA FB FB August 5, 2013 FA 10   Superposition: FTotal = FA + FB   Field from A at P kE qA qC ˆ ˆ FA = 0 x + y 2 d AP   Field from B at P FB FTotal FA kE qB qC ˆ FB = x + 0y 2 kE qB qC kE qA qC dBP ˆ ˆ FTotal = x+ y 2 2 dBP d AP Physics 1B - Summer Session 1 - C. Palmer August 5, 2013 11   Vector form ˆ ˆ FTotal = +46 Nx − 120 Ny   Magnitude and Direction E Total = ( 46) + (120) 2 2 = 128N 120 tan (φ ) = tan(360 − θ ) = → θ = 291• 46 Physics 1B - Summer Session 1 - C. Palmer θwrt X φ FTotal, X FTotal August 5, 2013 FTotal,Y 12 Below are three collinear point particles. What is the total force on q2 (magnitude and direction)? x 30cm q1=2.0μC 50cm q3=- 3.0μC q2=- 2.0μC Physics 1B - Summer Session 1 - C. Palmer July 2, 2013 13 What is the force on q2 from q1 alone (magnitude and direction)? (Remember µ = 10-6 and ke= 8.99x109 N*m2/C2) A)  +4.0e7 N B)  +0.40 N x C)  -4.0e7 N D)  -0.40 N 30cm q1=2.0μC 50cm q3=- 3.0μC q2=- 2.0µC Physics 1B - Summer Session 1 - C. Palmer July 2, 2013 14 What is the force on q2 from q1 alone (magnitude and direction)? (Remember µ = 10-6) B) +0.4 N To the left (which is positive) is because of the attractive force between q2 and q1. 30cm q1=2.0μC 50cm q3=- 3.0μC q2=- 2.0µC Physics 1B - Summer Session 1 - C. Palmer x July 2, 2013 15 We know F1 on 2 and now need F3 on 2. The force will be to the left (+ in x) because both charges are negative. 2 −2 * 10 −6 C −3 * 10 −6 C ⎡ N*m ⎤ F3on 2 = 8.99 E9 ⎢ 2 ⎥ 2 ( 0.5m ) ⎣C ⎦ x = 0.22 N 30cm q1=2.0μC 50cm q3=- 3.0μC q2=- 2.0μC Physics 1B - Summer Session 1 - C. Palmer July 2, 2013 16 Now finally comes the superposition part. FTot = ∑ Fi on 2 = F1 on 2 + F3 on 2 = 0.40 N + 0.22 N = 0.62 N x 30cm q1=2.0μC 50cm q3=- 3.0μC q2=- 2.0μC Physics 1B - Summer Session 1 - C. Palmer July 2, 2013 17 Physics Concept Components Learning Outcome Electric Force and Fields Charge and its analogy to mass Coulomb’s Law and its analogy to gravity Electric Fields Electric Flux Gauss’s Law Electric Potential Capacitance Evaluate the forces of charged objects on one another. Evaluate value and magnitude of electric fields at arbitrary points in space. Evaluate the potential differences between charged objects (surfaces and points). Discern between electric potential and electric potential energy Physics 1B - Summer Session 1...
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## This note was uploaded on 09/10/2013 for the course PHYS 2B 2b taught by Professor Hirsch during the Summer '10 term at UCSD.

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