1 / 31

Chapter 18 Capacitance and Potential

Chapter 18 Capacitance and Potential. 1. What is the electric potential energy between a proton and an electron separated by a distance of 5.3 x 10 -11 m?.

brazeal
Download Presentation

Chapter 18 Capacitance and Potential

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 18 Capacitance and Potential

  2. 1. What is the electric potential energy between a proton and an electron separated by a distance of 5.3 x 10-11 m?

  3. PE = k q1 q2 / dEPE = (9 x 109)(1.6 x 10-19)(-1.6 x 10-19)/(5.3 x 10-11 )PE = -4.35 x 10-18 JThe charges are opposite so the energy is negative.

  4. 2. Two positive point charges are separated by a distance of 4 cm. If they are moved to a distance of 12 cm apart, by what factor does the electric potential energy between them change?

  5. PE = k q1 q2 / dThree times as far away, 1/3 the potential energy.This is NOT an inverse-square relationship.

  6. 3. A particle of charge 3 x 10-15 C moves 5 m in the direction of a uniform electric field of strength 4000 N/C. What is the change in electrical potential energy?

  7. PEE = qEdPEE = (3 x 10-15)4000(5)PEE = -6 x 10-11 J(The positive charge is moving in the direction it wants to go, so it is a decrease in potential energy.)

  8. 4. If you move the particle in problem 3 back to its starting position, what is the change in electrical potential energy for this second move?

  9. It is being moved in the direction it doesn’t want to go, so it is an increase in potential energy. +6 x 10-11 J

  10. 5. What is the strength of the electric field in problem 3?

  11. Duh, 4000 N/C.I wanted you to calculate it using Ed = V, but I messed up!

  12. 6. Two point charges, +2.7 μC and +3.8 μC, are 0.55 m apart. What is the electrical potential energy of this system of two charges?

  13. PE = k q1 q2 / dEPE = (9 x 109)(+2.7 x 10-6)(+3.8 x 10-6)/(0.55 m)PE = 0.168 JThe charges are the same so the energy is positive.

  14. 7. What would be the electrical potential energy of the system in problem 6 if both charges were negative instead of positive?

  15. The same, PE = 0.168 J, two negative charges would still produce a positive potential energy by PE = k q1 q2 / d.

  16. 8. A proton (q = 1.6 x 10-19) moves 3 meters along an electric field of 500 N/C. What is the potential difference between the proton’s beginning and final location?

  17. Ed = V500 x 3 = V1500 V = VMoving in the direction of the field is a decrease of potential, so -1500 V = V.

  18. 9. A uniform electric field of 1000 N/C is directed downward. If the potential at 10 m above the ground is 15,000 V, what is the potential at ground level? At 3 m above the ground?

  19. Ed = V, so the potential difference between 10 m and 0 m is 1000 x 10 = 10000 V. Since the field is directed downward, the potential must be 10000 V lower at the ground: 15000 -10000 = 5000 V. At 3 m above the ground the potential must be 1000 x 3 higher: 3000 + 5000 = 8000 V.

  20. 10. What is the electric potential 2 m from a point charge of 22 μC?

  21. V = k q/ dV = (9 x 109)(22 x 10-6) / 2V = 99000 VPotential is a scalar, so there is no direction associated with this answer.

  22. 11. Two point charges are separated by 3 meters. If the charges are 25 μC and 13 μC, what is the potential at the midway point between the two charges?

  23. V = k q/ dV1 = (9 x 109)(25 x 10-6) /1.5 V2 = (9 x 109)(13 x 10-6) /1.5V1 = 150000 V V2 = 78000 VV = 150000 + 78000 = 228000 V

  24. 12. Two point charges are separated by 4 meters. If the charges are +36 μC and -24 μC, what is the potential at the midway point between the two charges?

  25. V = k q/ dV+ = (9 x 109)(36 x 10-6) /2 V- = (9 x 109)(-24 x 10-6) /2V+ = 162000 V V- = -108000 VV = 162000 + -108000 = 54000 V

  26. 13. At what distance from a point charge of 16 μC is the potential 2 x 105 V?

  27. V = k q/ d2 x 105 = (9 x 109)(16 x 10-6) / dd = 0.72 m

  28. 14. At what distance from a point charge of 16 μC is the potential zero V?

  29. The potential is always zero at infinity.

More Related