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Physics 2054C – Spring 2002

Physics 2054C – Spring 2002. Review for Quiz on Wed. Start Chapter 17. Housekeeping. Lab Sections. Most People Assigned Let me know if you are in a lab section and need to change. Provide a 1 st and 2 nd choice. CAPA. Need names of those without CAPA logins. Other Questions?.

leonard-townsend
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Physics 2054C – Spring 2002

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  1. Physics 2054C – Spring 2002 Review for Quiz on Wed. Start Chapter 17 Larry Dennis

  2. Housekeeping • Lab Sections. • Most People Assigned • Let me know if you are in a lab section and need to change. • Provide a 1st and 2nd choice. • CAPA. • Need names of those without CAPA logins. • Other Questions?

  3. F = kQ1Q2 R2 Electric Forces (Coulomb’s Law) • Like Charges Repel. • Opposite Charges Attract. - +

  4. - F = kQ1Q2 = ma R2 + Coulomb’s Law & Newton’s Law • Both apply • Electrical forces cause accelerations. • It is a vector.

  5. Electric Fields • Related to the Electric Force. • Removes the dependence on one of the charges. • E = F/Q1 = kQ2/R2

  6. Rules for Drawing Electric Field Lines • The lines must begin on a positive charges and terminate on a negative charges (or at infinity). • The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge. • No two field lines can cross.

  7. Superposition • Add the effects of multiple charges. • Fnet = F1 + F2 + F3 . • Calculate each of the forces as if the other charges were not present. • Enet = E1 + E2 + E3 . • Calculate each of the electric fields as if the other charges were not present.

  8. L R 0.6 m Sample Problem • Two charges are sitting on a table separated by a distance of 0.60 m. The sign and magnitude of these two charges (L and R) are unknown. We are going to use a positive test charge (Q = 5.0 x 10-6 C) to determine which charge is positive and which charge is negative. • If the charges L and R repel then they must both positive or both be negative. • If the charges attract (the case shown below), then they must be of opposite sign. • For now we will assume the attract one another.

  9. Q L R 0.6 m Sample Problem (continued) • We place our test charge Q midway between the two charges and observe that it moves to the right (towards R). From this we can determine that: • Q could be opposite R (attract), same as L (repel). Or • R could be much less than L and both are positive, but these charges would repel one another. Or • L could be much less than R and both are negative, but again, L and R would repel one another. Case 1 is our only choice.

  10. This is the one! L R Case 1: R negative & L positive L R Case 2: R positive & L negative Sample Problem (continued) • What is the direction of the electric field at the midpoint? • (recall) F = Q E and since Q>0, F and E are parallel. • Because the positive charge, Q, moved to the right, the electric field must be to the right.

  11. Q R 0.3 m Sample Problem (continued) • Suppose that we remove charge L and measure the force exerted on Q by R to be 8.0 N to the right. • What is the sign of R? • What is the sign of L? R must be negative since it attracts a positive charge! L must be positive since it must be opposite to R!

  12. Q R 0.3 m Sample Problem (continued) • What is the magnitude of the electric field at the point where Q is located due to charge R? F = Q*ER ER = FR/Q = 8.0 N / 5.0 x 10-6 C = 1.6 x 106 N/C

  13. Q R 0.3 m Sample Problem (continued) • What is the magnitude of the charge R? ER = kQR/D2 QR = ER * D2 /k = 1.6 x 106 N/C * (0.3 m)2 /9.0 x 109 N-m2 /C2 QR = 16 x 10-6 C (16 microcoulombs)

  14. Chapter 17 – Electrical Energy and Electrostatic Potential • Electric forces can be used to change the energy of matter. • Wab = -qVab = -q(Va – Vb) • Electric potential (often shortened to potential) is the change in potential energy per unit charge due to an electric field. • Measured in volts. • -Wab/q = (Va – Vb)

  15. Electric Potential Energy • PE = PEb = PEa = q(Vb – Va) Vab - + W = qV W = Fd W = (qE)d W = qV = qEd  E = V/d +q d

  16. Vab = 1500 V - + +q = .5 C d = 2 mm Electric Potential Energy • PE = PEb = PEa = q(Vb – Va) W = qV = .5 C * 1500 = 750 J W = qV = qEd  E = V/d = 1500 V / 0.002 m = 5.5 x 105 V/m = N/C

  17. The Electron Volt • Energy acquired by an electron as an electron moves through 1 volt. Vab - + W = qV W = 1.602 x 10-19 C * 1 Volt W = 1.602 x 10-19 Joules W = 1 eV (electron volt) +q d

  18. Next Time • Tutorials: • Mon. 12:15 – 3:15 • Tue. 9:00 – 12:00 • CAPA Due Wednesday Morning at 2:00 AM • Wed: First Quiz on Chapter 17 • Chapter 17 - Electrical Energy

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