1. Chapter 18.2 Capacitance and Potential

2. 1. A 5 μF capacitor is connected to a 12 volt battery
1. A 5 μF capacitor is connected to a 12 volt battery. What is the potential difference across the plates of the capacitor when it is fully charged?

3. 12 volts, a charged capacitor has the same potential difference across the plates as the source.

4. 2. What is the charge on one plate of the capacitor in problem 1 when it is fully charged? What is the net charge on the capacitor when it is fully charged?

5. q = VC q = 12 x 5 = 60 μC or q = 12(5 x 10-6)= 6 x 10-5 C But the net charge on a capacitor is always zero because the + and – plate have charges of equal magnitude.

6. 3. How much electrical potential energy is stored in the capacitor in problem 1?

7. PEE = ½ CV2 PEE = ½ 5(12)2 PEE = 360 μJ or, PEE = ½ (5x 10-6)(12)2 PEE = 3.6 x 10-4 J

8. 4. If the plates are separated by a distance of 0
4. If the plates are separated by a distance of m, what is the strength of the electric field across the plates?

9. Ed = V E(0.003) = 12 E = 4000 N/C

10. 5. A 9 μF capacitor is connected to a 9 volt battery
5. A 9 μF capacitor is connected to a 9 volt battery. What is the potential difference across the plates of the capacitor when it is fully charged?

11. 9 volts, a charged capacitor has the same potential difference across the plates as the source.

12. 6. What is the charge on one plate of the capacitor in problem 5 when it is fully charged? What is the net charge on the capacitor when it is fully charged?

13. q = VC q = 9 x 9 = 81 μC or q = 9(9 x 10-6)= 8.1 x 10-5 C But the net charge on a capacitor is always zero because the + and – plate have charges of equal magnitude.

14. 7. How much electrical potential energy is stored in the capacitor in problem 5?

15. PEE = ½ CV2 PEE = ½ 9(9)2 PEE = 364.5 μJ or, PEE = ½ (9 x 10-6)(9)2 PEE = 3.645 x 10-4 J

16. 8. If the plates are separated by a distance of 0
8. If the plates are separated by a distance of m, what is the strength of the electric field across the plates?

17. Ed = V E(0.005) = 9 E = 1800 N/C

18. 9. List four ways to increase the charge on a capacitor.

19. 1. Larger plates 2. Plates closer. together 3. Different dielectric 4
1. Larger plates 2. Plates closer together 3. Different dielectric 4. Increase voltage.

20. 10. What is the difference in electrical potential energy and electric potential?

21. Electric potential is electric potential energy divided by charge.

22. 11. A proton is moved 0.5 m in the direction of an electric field with a strength of 5000 N/C. What is the change in electric potential?

23. Ed = V 5000 x 0.5 = V 2500 V = V Moving in the direction of the field is a decrease of potential, so V = V.

24. 12. What is the change in electrical potential energy of the proton in problem 11?

25. PEE = qV PEE = (1.6 x 10-19)(2500) PEE = -4 x 10-16 J

26. 13. An electron is moved 0.8 m in the direction of an electric field with a strength of 3000 N/C. What is the change in electric potential?

27. Ed = V 3000 x 0.8 = V 2400 V = V Moving in the direction of the field is a decrease of potential, so V = V.

28. Remember! Potential (V) isn’t affected by whether the charge moving is positive or negative. Potential only refers to what would occur to a positive charge, so we don’t even consider the fact that the electron is negative!

29. 14. What is the change in electrical potential energy of the electron in problem 13?

30. V = EPE/q, so qV = EPE (1. 6 x10-19)2400 = 3
V = EPE/q, so qV = EPE (1.6 x10-19)2400 = 3.84 x J Since this is energy, it does matter that it is an electron (negative charge)! The electron does not want to move in the direction of the field, so the answer is x J

31. 15. A helium nucleus (composed of two protons and two neutrons) is moved 24 cm in the direction of an electric field with a strength of 9000 N/C. What is the change in electric potential?

32. Ed = V 9000 x 0.24 = V 2160 V = V Moving in the direction of the field is a decrease of potential, so V = V.

33. 16. What is the change in electric potential energy of the helium nucleus in problem 15?

34. V = EPE/q, so qV = EPE 2(1. 6 x10-19)2160 = 6
V = EPE/q, so qV = EPE 2(1.6 x10-19)2160 = 6.9 x J Since there are two protons, we must multiply the proton charge by 2. The proton moves in the direction of the field, so the answer is -6.9 x J